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An Introduction to Mathematical Proofs

数学证明导论

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Textbooks in Mathematics

数学教科书

Series editors:

系列编辑:

Al Boggess and Ken Rosen

阿尔·博格斯和肯·罗森

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COMPLEX VARIABLES: A PHYSICAL APPROACH WITH APPLICATIONS, SECOND EDITION

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AN INTRODUCTION TO MATHEMATICAL PROOFS

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https://www.crcpress.com/Textbooks-in-Mathematics/book-series/CANDHTEXBOOMTH

https://www.crcpress.com/Textbooks-in-Mathematics/book-series/CANDHTEXBOOMTH

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An Introduction to Mathematical Proofs

数学证明导论

Nicholas A. Loehr

尼古拉斯·A·洛尔

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CRC Press

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This book is dedicated to my father Frank.

本书献给我的父亲弗兰克。

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Contents

内容

Preface

前言

1 Logic

1Logic

1.1 Propositions, Logical Connectives, and Truth Tables

1.1 命题、逻辑连接词和真值表

1.2 Logical Equivalences and IF-Statements

1.2 逻辑等价性和 IF 语句

1.3 IF, IFF, Tautologies, and Contradictions

1.3IF、IFF、重言式和矛盾式

1.4 Tautologies, Quantifiers, and Universes

1.4 重言式、量词和宇宙

1.5 Quantifier Properties and Useful Denials

1.5 量词的性质和有用的否定

1.6 Denial Practice and Uniqueness Statements

1.6 否认实践和独特性声明

2 Proofs

2个证明

2.1 Definitions, Axioms, Theorems, and Proofs

2.1 定义、公理、定理和证明

2.2 Proving Existence Statements and IF Statements

2.2 证明存在性语句和 IF 语句

2.3 Contrapositive Proofs and IFF Proofs

2.3 逆否命题证明和因果函数证明

2.4 Proofs by Contradiction and Proofs of OR-Statements

2.4 反证法和或语句的证明

2.5 Proofs by Cases and Disproofs

2.5 案例证明与反证

2.6 Proving Quantified Statements

2.6 证明量化陈述

2.7 More Quantifier Properties and Proofs (Optional)

2.7 量词的更多性质和证明(可选)

Review of Logic and Proofs

逻辑与证明回顾

3 Sets

3套

3.1 Set Operations and Subset Proofs

3.1集合运算和子集证明

3.2 Subset Proofs and Set Equality Proofs

3.2子集证明和集合相等性证明

3.3 Set Equality Proofs, Circle Proofs, and Chain Proofs

3.3集合等式证明、圆证明和链式证明

3.4 Small Sets and Power Sets

3.4 小型套装和电动套装

3.5 Ordered Pairs and Product Sets

3.5 有序对和产品集

3.6 General Unions and Intersections

3.6 一般工会和交叉口

3.7 Axiomatic Set Theory (Optional)

3.7 公理化集合论(选修)

4 Integers

4个整数

4.1 Recursive Definitions and Proofs by Induction

4.1 递归定义和归纳证明

4.2 Induction Starting Anywhere and Backwards Induction

4.2 从任意位置开始的归纳法和逆向归纳法

4.3 Strong Induction

4.3强感应

4.4 Prime Numbers and Integer Division

4.4 质数和整数除法

4.5 Greatest Common Divisors

4.5最大公约数

4.6 GCDs and Uniqueness of Prime Factorizations

4.6个最大公约数和质因数分解的唯一性

4.7 Consequences of Prime Factorization (Optional)

4.7 质因数分解的后果(可选)

Review of Set Theory and Integers

集合论与整数回顾

5 Relations and Functions

5. 关系与函数

5.1 Relations

5.1 人际关系

5.2 Inverses, Identity, and Composition of Relations

5.2 关系的逆元、恒等元和复合

5.3 Properties of Relations

5.3关系的性质

5.4 Definition of Functions

5.4 函数的定义

5.5 Examples of Functions and Function Equality

5.5 函数和函数等式的例子

5.6 Composition, Restriction, and Gluing

5.6 组成、限制和粘合

5.7 Direct Images and Preimages

5.7 直接像和原像

5.8 Injective, Surjective, and Bijective Functions

5.8 单射函数、满射函数和双射函数

5.9 Inverse Functions

5.9 反函数

6 Equivalence Relations and Partial Orders

6. 等价关系和偏序关系

6.1 Reflexive, Symmetric, and Transitive Relations

6.1 自反性、对称性和传递性关系

6.2 Equivalence Relations

6.2 等价关系

6.3 Equivalence Classes

6.3 等价类

6.4 Set Partitions

6.4 设置分区

6.5 Partially Ordered Sets

6.5 部分有序集

6.6 Equivalence Relations and Algebraic Structures (Optional)

6.6 等价关系和代数结构(可选)

7 Cardinality

7基数

7.1 Finite Sets

7.1 有限集

7.2 Countably Infinite Sets

7.2 可数无限集

7.3 Countable Sets

7.3 可数集

7.4 Uncountable Sets

7.4 不可数集

Review of Functions, Relations, and Cardinality

函数、关系和基数回顾

8 Real Numbers (Optional)

8. 实数(可选)

8.1 Axioms for R and Properties of Addition

8.1 R 的公理和加法的性质

8.2 Algebraic Properties of Real Numbers

8.2 实数的代数性质

8.3 Natural Numbers, Integers, and Rational Numbers

8.3 自然数、整数和有理数

8.4 Ordering, Absolute Value, and Distance

8.4 排序、绝对值和距离

8.5 Greatest Elements, Least Upper Bounds, and Completeness

8.5 最大元素、最小上界和完备性

Suggestions for Further Reading

延伸阅读建议

Index

指数

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Preface

前言

This book contains an introduction to mathematical proofs, including fundamental material on logic, proof methods, set theory, number theory, relations, functions, cardinality, and the real number system. The book can serve as the main text for a proofs course taken by undergraduate mathematics majors. No specific prerequisites are needed beyond familiarity with high school algebra. Most readers are likely to be college sophomores or juniors who have taken calculus and perhaps some linear algebra, but we do not assume any knowledge of these subjects. Anyone interested in learning advanced mathematics could use this text for self-study.

本书是对数学证明的入门介绍,内容涵盖逻辑、证明方法、集合论、数论、关系、函数、基数和实数系统等基础知识。本书可作为本科数学专业证明课程的主要教材。除高中代数基础外,无需其他特定先修知识。大多数读者可能是大学二年级或三年级的学生,他们可能学过微积分,也可能学过一些线性代数,但我们并不假定他们具备这些学科的知识。任何对高等数学感兴趣的人都可以使用本书进行自学。

Structure of the Book

本书结构

This book evolved from classes given by the author over many years to students at the College of William & Mary, Virginia Tech, and the United States Naval Academy. I have divided the book into 8 chapters and 54 sections, including three review sections. Each section corresponds very closely to the material I cover in a single 50-minute lecture. Sections are further divided into many short subsections, so that my suggested pacing can readily be adapted for classes that meet for 75 minutes, 80 minutes, or other time intervals. If the instructor omits all sections and topics designated as “optional,” it should be just possible to finish all of the core material in a semester class that meets for 2250 minutes (typically forty-five 50-minute meetings or thirty 75-minute meetings). More suggestions for possible course designs appear below.

本书源于作者多年来在威廉玛丽学院、弗吉尼亚理工大学和美国海军学院为学生讲授的课程。全书分为8章54节,其中包括3个复习节。每个节的内容都与我50分钟的单次授课内容高度吻合。每个节又细分为许多小节,因此我建议的教学进度可以根据75分钟、80分钟或其他课时的时长灵活调整。如果教师省略所有标有“可选”的章节和主题,那么在一个学期共计2250分钟(通常为45次50分钟的课或30次75分钟的课)的课程中,应该刚好可以完成所有核心内容的讲授。更多课程设计建议请参见下文。

I have tried to capture the best features of live mathematics lectures in the pages of this book. New material is presented to beginning students in small chunks that are easier to digest in a single reading or class meeting. The book maintains the friendly conversational style of a classroom presentation, without relinquishing the necessary level of precision and rigor. Throughout this text, you will find the personal pronouns “I” (the author), “you” (the reader), and “we” (the author and the reader, working together), reminding us that teaching and learning are fundamentally human activities. Teaching this material effectively can be as difficult as learning it, and new instructors are often unsure how much time to spend on the fundamentals of logic and proof techniques. The organization of this book shows at a glance how one experienced teacher of proofs allocates time among the various core topics. The text develops mathematical ideas through a continual cycle of examples, theorems, proofs, summaries, and reviews. A new concept may be introduced briefly via an example near the end of one section, then examined in detail in the next section, then recalled as needed in later sections. Every section ends with an immediate review of the key points just covered, and three review sections give detailed summaries of each major section of the book. The essential core material is supplemented by more advanced topics that appear in clearly labeled optional sections.

我力求在本书中捕捉到现场数学讲座的精华。本书将新知识以小块形式呈现给初学者,便于他们在一次阅读或课堂学习中消化吸收。本书保持了课堂讲解的友好对话风格,同时又不失必要的精确性和严谨性。贯穿全书,你会发现“我”(作者)、“你”(读者)和“我们”(作者和读者共同学习)这三个人称代词,提醒我们教与学本质上都是人类活动。有效地教授这些内容与学习它们一样困难,新教师常常不确定应该花多少时间讲解逻辑和证明技巧的基础知识。本书的结构清晰地展示了一位经验丰富的证明教师是如何在各个核心主题之间分配时间的。本书通过例题、定理、证明、总结和回顾的循环,逐步构建数学概念。新概念可能会在某一章节的结尾通过一个例子简要介绍,然后在下一章节中进行详细探讨,并在后续章节中根据需要再次提及。每节结尾都会立即回顾刚刚讨论的关键点,此外还有三个复习章节,分别对本书的每个主要章节进行详细总结。除了核心内容之外,本书还补充了一些更高级的主题,这些主题出现在清晰标注的选读章节中。

Contents of the Book

本书内容

Here is a detailed list of the topics covered in each chapter of the book:

以下是本书各章节涵盖的主题详细列表:

  1. Logic: propositions, logical connectives (NOT, AND, OR, XOR, IF, IFF), truth tables, logical equivalence, tautologies, contradictions, universal and existential quantifiers, translating and denying complex logical statements, uniqueness.
    逻辑:命题、逻辑连接词(非、与、或、异或、如果、如果)、真值表、逻辑等价、重言式、矛盾式、全称量词和存在量词、翻译和否定复杂的逻辑语句、唯一性。
  2. Proofs: ingredients in mathematical theories (definitions, axioms, inference rules, theorems, proofs), proof by example, direct proof, contrapositive proof, contradiction proof, proof by cases, generic-element proofs, proofs involving multiple quantifiers.
    证明:数学理论的组成部分(定义、公理、推理规则、定理、证明),举例证明、直接证明、逆否命题证明、矛盾证明、分情况证明、通用元素证明、涉及多个量词的证明。
  3. Set Theory: set operations (union, intersection, set difference), subset proofs, set equality proofs, circle proofs, chain proofs, power sets, ordered pairs, product sets, unions and intersections of indexed collections.
    集合论:集合运算(并集、交集、差集)、子集证明、集合相等证明、圆证明、链证明、幂集、有序对、乘积集、索引集合的并集和交集。
  4. Integers: recursive definitions, ordinary induction proofs, induction starting anywhere, backwards induction, strong induction, integer division with remainder, greatest common divisors, Euclid’s GCD algorithm, primes, existence and uniqueness of prime factorizations.
    整数:递归定义、普通归纳证明、任意起始归纳、逆归纳、强归纳、带余数的整数除法、最大公约数、欧几里得最大公约数算法、素数、素数分解的存在性和唯一性。
  5. Relations and Functions: relations, images, inverse of a relation, identity relation, composition of relations, formal definition of a function, function equality, operations on functions (pointwise operations, composition, restriction), direct images, preimages, injections, surjections, bijections, inverse functions.
    关系与函数:关系、像、关系的逆、恒等关系、关系的复合、函数的形式定义、函数相等性、函数运算(逐点运算、复合、限制)、直接像、原像、单射、满射、双射、反函数。
  6. Equivalence Relations and Partial Orders: reflexivity, symmetry, transitivity, equivalence relations, congruence modulo n , equivalence classes, set partitions, antisymmetry, partial orders, well-ordered sets.
    等价关系和偏序:自反性、对称性、传递性、等价关系、模 n 同余、等价类、集合划分、反对称性、偏序、良序集。
  7. Cardinality: finite sets, basic counting rules, countably infinite sets, countable sets, theorems on countability, uncountable sets, Cantor’s Theorem.
    基数:有限集、基本计数规则、可数无限集、可数集、可数性定理、不可数集、康托尔定理。
  8. Real Numbers (Optional): ordered field axioms for ℝ, algebraic properties, formal definition of ℕ and ℤ and ℚ, ordering properties, absolute value, distance, Least Upper Bound Axiom and its consequences (Archimedean ordering of ℝ, density of ℚ in ℝ, existence of real square roots, Nested Interval Theorem).
    实数(可选):ℝ 的有序域公理、代数性质、ℕ、ℤ 和 ℚ 的形式定义、排序性质、绝对值、距离、最小上界公理及其推论(ℝ 的阿基米德排序、ℚ 在 ℝ 中的密度、实数平方根的存在性、嵌套区间定理)。

Possible Course Designs

可能的课程设计

A standard three-credit (2250 minute) proofs class could cover most of the topics in Chapters 1 through 7 , which are essential for further study of advanced mathematics. When pressed for time, I have sometimes omitted or condensed the material on cardinality ( Chapter 7 ) or prime factorizations (last half of Chapter 4 ). Many variations of the standard course are also feasible. Instructors wishing to preview ideas from abstract algebra could supplement the standard core with the following optional topics:

一门标准的三学分(2250分钟)证明课程可以涵盖第1章至第7章的大部分内容,这些内容对于进一步学习高等数学至关重要。如果时间紧迫,我有时会省略或精简关于基数(第7章)或质因数分解(第4章后半部分)的内容。标准课程还有许多其他变体。希望预先介绍抽象代数概念的教师可以在标准核心课程之外,补充以下可选主题:

A course introducing ideas from advanced calculus could include these topics:

介绍高等微积分概念的课程可以包括以下主题:

A quarter-long (1500 minute) course focusing on basic proof methods might only cover Chapter 1 , Chapter 2 , and the early sections in Chapters 3 through 6 . A quarter course on set theory, aimed at students with some prior familiarity with logic and proofs, might cover all of Chapters 3 , 5 , 6 , and 7 .

一个为期一个季度(1500 分钟)的课程,重点讲解基本证明方法,可能只涵盖第 1 章、第 2 章以及第 3 章至第 6 章的前几节。一个为期一个季度的集合论课程,面向已经具备一些逻辑和证明知识的学生,可能涵盖第 3 章、第 5 章、第 6 章和第 7 章的全部内容。

Topics can also be studied in several different orders. Chapter 1 on logic must come first, and Chapter 2 on proof methods must come second. Thereafter, some flexibility is possible. Chapter 4 (on induction and basic number theory) can be covered before Chapter 3 (on sets) or omitted entirely. Chapter 6 (on equivalence relations) can be covered before the last five sections of Chapter 5 (on functions). Chapter 7 (on cardinality) requires material from Chapter 5 on bijections, but it does not rely heavily on Chapter 6 . Finally, the optional Chapter 8 (axiomatic development of the real numbers) could be covered anytime after Chapter 2 , with minor adjustments to avoid explicit mention of functions and relations. However, Chapter 8 is more challenging than it may appear at first glance. We are all so familiar with basic arithmetic and algebraic facts about real numbers that it requires considerable intellectual discipline to deduce these facts from the axioms without accidentally using a property not yet proved. Nevertheless, it is rewarding and instructive (albeit somewhat tedious) to work through this logical development of ℝ if time permits.

学习主题的顺序可以有多种选择。第一章(逻辑)必须先学,第二章(证明方法)必须后学。之后,学习顺序可以有一定的灵活性。第四章(归纳法和基础数论)可以在第三章(集合)之前学习,也可以完全省略。第六章(等价关系)可以在第五章(函数)的最后五节之前学习。第七章(基数)需要用到第五章关于双射的内容,但并不大量依赖第六章。最后,可选的第八章(实数的公理化推导)可以在第二章之后的任何时间学习,只需稍作调整,避免明确提及函数和关系即可。然而,第八章比乍看之下更具挑战性。我们都非常熟悉关于实数的基本算术和代数事实,因此,要从公理中推导出这些事实而不意外地使用尚未证明的性质,需要相当的思维能力。不过,如果时间允许的话,研究 ℝ 的这种逻辑发展过程是有益的,也是有启发性的(尽管有点枯燥)。

Book’s Approach to Key Topics

本书对关键主题的处理方法

This book adopts a methodical, detailed, and highly structured approach to teaching proof techniques and related mathematical topics. We start with basic logical building blocks and gradually assemble these ingredients to build more complex concepts. To give you a flavor of the teaching philosophy used here, the next few paragraphs describe my approach to explaining four key topics: proof-writing, functions, multiple quantifiers, and induction.

本书采用系统、详尽且结构严谨的方法教授证明技巧及相关数学主题。我们从基本的逻辑模块入手,逐步构建更复杂的概念。为了让您更好地了解本书的教学理念,接下来的几段将介绍我讲解四个关键主题的方法:证明写作、函数、多重量词和归纳法。

Skills for Writing Proofs

撰写证明的技巧

Like any other complex task, the process of writing a proof requires the synthesis of many small atomic skills. Every good proofs textbook develops the fundamental skill of breaking down a statement to be proved into its individual logical constituents, each of which contributes certain structure to the proof. For example, to begin a direct proof of a conditional statement “If P , then Q ,” we write: “Assume P is true; we must prove Q is true.” I explain this particular skill in great detail in this text, introducing explicit proof templates for dealing with each of the logical operators.

如同其他任何复杂任务一样,撰写证明的过程需要综合运用许多细小的技能。每一本优秀的证明教材都会培养一项基本技能:将待证明的命题分解成各个逻辑组成部分,每个组成部分都为证明贡献特定的结构。例如,要直接证明条件语句“如果P,则Q”,我们可以这样写:“假设P为真;我们必须证明Q为真。”本书将详细阐述这项技能,并介绍处理每个逻辑运算符的明确证明模板。

But there are other equally crucial skills in proof-writing: memorizing and expanding definitions; forming useful denials of complex statements; identifying the logical status of each statement and variable in a proof via appropriate status words; using known universal and existential statements in the correct way; memorizing and using previously proved theorems; and so on. I cover each of these skills on its own, in meticulous detail, before assembling the skills to build increasingly complex proofs. Remarkably, this reduces the task of writing many basic proofs into an almost completely automatic process. It is very rewarding to see students gain confidence and ability as they master the basic skills one at a time and thereby develop proficiency in proof-writing.

但证明写作中还有其他同样至关重要的技能:记忆和扩展定义;对复杂命题进行有效的否定;通过恰当的状态词识别证明中每个命题和变量的逻辑状态;正确使用已知的全称命题和存在命题;记忆和运用已证明的定理;等等。我会逐一详细讲解这些技能,然后再将它们组合起来,构建越来越复杂的证明。值得注意的是,这使得编写许多基础证明几乎变成了一个完全自动化的过程。看到学生们逐步掌握这些基本技能,从而提高证明写作的熟练度,并建立自信,这令人非常欣慰。

Here is an example to make the preceding ideas concrete. Consider a typical practice problem for beginning proof-writers: prove that for all integers x, if x is odd, then x + 5 is even. In the proof below, I have annotated each line with the basic skill needed to produce that line.

为了使前面的概念更加具体,这里举一个例子。考虑一个典型的证明题练习题,适合初学者:证明对于所有整数 x,如果 x 是奇数,则 x + 5 是偶数。在下面的证明中,我标注了每一行所需的基本技能。

Line in Proof

校样行

Skill Needed

所需技能

1. Let x 0 be a fixed, arbitrary integer.

1. 设 x 0 为一个固定的任意整数。

Prove ∀ statement using generic element.

使用通用元素证明 ∀ 命题。

2. Assume x 0 is odd; prove x 0 + 5 is even.

2. 假设 x 0 为奇数;证明 x 0 + 5 为偶数。

Prove an IF statement by direct proof.

用直接证明法证明一个 IF 语句。

3. We assumed there is k ∈ ℤ with x 0 = 2 k + 1.

3. 我们假设存在 k ∈ ℤ,使得 x 0 = 2k + 1。

Expand a memorized definition.

扩展记忆中的定义。

4. We will prove there is m ∈ ℤ with x 0 + 5 = 2 m .

4. 我们将证明存在 m ∈ ℤ,使得 x 0 + 5 = 2m。

Expand a memorized definition.

扩展记忆中的定义。

5. Doing algebra on the assumption gives:

5. 基于该假设进行代数运算可得:

Use logical status words.

使用符合逻辑的词语。

6. x 0 + 5 = (2 k + 1) + 5 = 2 k + 6 = 2( k + 3).

6. x 0 + 5 = (2k + 1) + 5 = 2k + 6 = 2(k + 3)。

Do basic algebraic manipulations.

进行基本的代数运算。

7. Choose m = k + 3, so x 0 + 5 = 2 m holds.

7. 选择 m = k + 3,使得 x 0 + 5 = 2m 成立。

Prove ∃ statement by giving an example.

通过举例证明命题。

8. Note m is in ℤ, being the sum of two integers.

8. 注意 m 在 ℤ 中,是两个整数之和。

Verify a variable is in the required set.

验证变量是否在所需的集合中。

Virtually every line in this proof is generated automatically using memorized skills; only the manipulation in line 6 requires a bit of creativity to produce the multiple of 2. Now, while many texts present a proof like this one, we seldom see a careful explanation of how the proof uses an assumed existential statement (line 3) to prove another existential statement (line 4) by constructing an example (lines 6 and 7) depending on the variable k in the assumption. This explanation may seem unnecessary in such a simple setting. But it is a crucial ingredient in understanding harder proofs in advanced calculus involving limits and continuity. There we frequently need to use an assumed multiply-quantified IF-statement to prove another multiply-quantified IF-statement. These proofs become much easier for students if they have already practiced the skill of using one quantified statement to prove another quantified statement in more elementary cases.

这个证明中的几乎每一行都是利用记忆技巧自动生成的;只有第 6 行的操作需要一些创造性才能得出 2 的倍数。虽然很多教材都提供了类似的证明,但我们很少看到对证明如何利用一个假定的存在命题(第 3 行)来证明另一个存在命题(第 4 行)的细致解释,而证明过程是通过构造一个取决于假设中变量 k 的例子(第 6 行和第 7 行)来实现的。在如此简单的情境下,这种解释似乎显得多余。但它对于理解高等微积分中涉及极限和连续性的更复杂证明至关重要。在这些证明中,我们经常需要使用一个假定的多重量化 IF 命题来证明另一个多重量化 IF 命题。如果学生已经在更基础的情况下练习过使用一个量化命题来证明另一个量化命题的技巧,那么这些证明对他们来说就会容易得多。

Similarly, there is not always enough prior coverage of the skill of memorizing and expanding definitions (needed to generate lines 3 and 4). This may seem to be a minor point, but it is in fact essential. Before writing this proof, students must have memorized the definition stating that “ x is even” means “there exists k ∈ ℤ with x = 2 k .” But to generate line 4 from this definition, k must be replaced by a new variable m (since k was already given a different meaning in line 3), and x must be replaced by the expression x 0 + 5 . I devote many pages to in-depth coverage of these separate issues, before integrating these skills into full proofs starting in Section 2.2 .

同样,对于记忆和扩展定义(生成第3行和第4行所必需的技能)的技能,以往的讲解往往不够充分。这看似微不足道,但实际上至关重要。在编写此证明之前,学生必须记住“x为偶数”的定义,即“存在k∈ℤ,使得x=2k”。但要从这个定义生成第4行,k必须替换为新变量m(因为k在第3行中已被赋予不同的含义),x必须替换为表达式x 0 + 5。我在2.2节开始将这些技能整合到完整的证明中,并用大量篇幅深入讲解了这些问题。

Functions

函数

A key topic in a proofs course is the rigorous definition of a function. A function is often defined to be a set of ordered pairs no two of which have the same first component. This definition is logically acceptable, but it causes difficulties later when studying concepts involving the codomain (set of possible outputs) for a function. Since the codomain cannot be deduced from the set of ordered pairs, great care is needed when talking about concepts that depend on the codomain (like surjectivity or the existence of a two-sided inverse). Furthermore, students accustomed to using the function notation y = f(x) find the ordered pair notation (x,y) f jarring and unpalatable. My approach includes the domain and codomain as part of the technical definition of a function; the set of ordered pairs by itself is called the graph of the function. This terminology better reflects the way most of us conceptualize functions and their graphs. The formal definition in Section 5.4 is preceded by carefully chosen examples (involving arrow diagrams, graphs in the Cartesian plane, and formulas) to motivate and explain the key elements of the technical definition. We introduce the standard function notations y = f(x) and f : X Y without delay, so students do not get bogged down with ordered triples and ordered pairs. Then we describe exactly what must be checked when a new function is introduced: single-valuedness and the fact that every x in the domain X has an associated output in the claimed codomain Y . We conclude with examples of formulas that do or do not give well-defined functions.

证明课程的一个关键主题是函数的严格定义。函数通常被定义为一组有序对,其中任意两个有序对的第一个分量都不相同。这个定义在逻辑上是可以接受的,但它会在后续学习涉及函数值域(所有可能输出的集合)的概念时造成困难。由于值域无法从有序对集合推导出来,因此在讨论依赖于值域的概念(例如满射性或是否存在双侧逆函数)时需要格外谨慎。此外,习惯于使用函数符号 y = f(x) 的学生会觉得有序对符号 (x,y) ∈ f 令人难以接受。我的方法是将定义域和值域作为函数技术定义的一部分;有序对集合本身被称为函数的图像。这种术语更能反映我们大多数人对函数及其图像的理解方式。第 5.4 节的正式定义之前,我们精心挑选了一些例子(包括箭头图、笛卡尔平面图和公式),以激发学生的兴趣并解释技术定义的关键要素。我们立即引入标准函数符号 y = f(x) 和 f : X → Y,以免学生陷入有序三元组和有序对的泥沼。然后,我们详细描述了引入新函数时必须检查的内容:单值性以及定义域 X 中的每个 x 在所声称的值域 Y 中都有一个对应的输出。最后,我们给出一些公式的例子,说明哪些公式能够给出定义良好的函数,哪些公式不能。

Multiple Quantifiers

多重量词

A hallmark of this book is its extremely careful and explicit treatment of logical quantifiers: ∀ (“for all”) and ∃ (“there exists”). The placement and relative ordering of these quantifiers has a big impact on the meaning of a logical statement. For example, the true statement ∀ x ∈ ℤ, ∃ y ∈ ℤ, y > x (“for every integer x there is a larger integer y ”) asserts something very different from the false statement ∃ y ∈ ℤ, ∀ x ∈ ℤ, y > x (“there exists an integer y larger than every integer x ”). However, these doubly quantified examples do not reveal the full complexity of statements with three or more nested quantifiers. Such statements are quite common in advanced calculus, as mentioned earlier.

本书的一大特色在于其对逻辑量词 ∀(“对于所有”)和 ∃(“存在”)的极其细致和明确的处理。这些量词的位置和相对顺序对逻辑语句的含义有着重大影响。例如,真命题 ∀x ∈ ℤ, ∃y ∈ ℤ, y > x(“对于每个整数 x,都存在一个更大的整数 y”)与假命题 ∃y ∈ ℤ, ∀x ∈ ℤ, y > x(“存在一个大于所有整数 x 的整数 y”)的含义截然不同。然而,这些双重量词的例子并不能完全揭示包含三个或更多嵌套量词的语句的复杂性。正如前文所述,这类语句在高等微积分中相当常见。

I give a very detailed explanation of multiple quantifiers in Sections 2.6 and 2.7 . After examining many statements containing two quantifiers, I introduce more complicated statements with as many as six quantifiers, focusing on the structural outline of proofs of such statements. Using these big examples is the best way to explain the main point: an existentially quantified variable may only depend on quantified variables preceding it in the given statement. Other examples examine disproofs of multiply quantified statements, where the proof-writer must first form a useful denial of the given statement (which interchanges existential and universal quantifiers). Many exercises develop these themes using important definitions from advanced calculus such as continuity, uniform continuity, convergence of sequences, and least upper bounds.

我在第 2.6 节和第 2.7 节中对多重量词进行了非常详细的解释。在考察了许多包含两个量词的语句之后,我引入了包含多达六个量词的更复杂的语句,重点阐述了此类语句证明的结构框架。使用这些大型示例是解释核心要点的最佳方式:一个存在量词变量只能依赖于给定语句中位于其前面的量词变量。其他示例考察了多重量词语句的反证,其中证明者必须首先对给定语句(其中存在量词和全称量词互换使用)形成一个有效的否定。许多练习运用高等微积分中的重要定义(例如连续性、一致连续性、序列收敛性和最小上界)来发展这些主题。

Induction

就职

Another vital topic in a proofs course is mathematical induction. Induction proofs are needed when working with recursively defined entities such as summations, factorials, powers, and sequences specified by a recursive formula. I discuss recursive definitions immediately before induction, and I carefully draw attention to the steps in an induction proof that rely on these definitions. Many expositions of induction do not make this connection explicit, causing some students to stumble at the point in the proof requiring the expansion of a recursive definition (for example, replacing a sum k = 1 n + 1 x k by [ k = 1 n x k ] + x n + 1 ).

数学归纳法是证明课程中另一个至关重要的主题。当处理递归定义的实体,例如求和、阶乘、幂以及由递归公式指定的序列时,就需要用到归纳证明。我会在讲解归纳法之前先讨论递归定义,并着重强调归纳证明中依赖于这些定义的步骤。许多关于归纳法的讲解并没有明确指出这种联系,导致一些学生在证明过程中遇到需要展开递归定义的地方时(例如,将求和式 ∑k=1n+1xk 替换为 [∑k=1nxk]+xn+1)会遇到困难。

Induction proofs are often formulated in terms of inductive sets : sets containing 1 that are closed under adding 1. Students are told to prove a statement ∀ n ∈ ℤ ≥0 , P(n) by forming the set S = { n ∈ ℤ ≥0 : P(n) is true} and checking that S is inductive. This extra layer of translation confuses many students and is not necessary. Inductive sets do serve an important technical purpose: they provide a rigorous construction of the set of natural numbers as the intersection of all inductive subsets of ℝ. I discuss this advanced topic in the optional final chapter on real numbers (see Section 8.3 ), but I avoid mentioning inductive sets in the initial treatment of induction. Instead, induction proofs are based on the Induction Axiom, which says that the statements P (1) and ∀ n ∈ ℤ ≥0 , P(n) P ( n + 1) suffice to prove ∀ n ∈ ℤ ≥0 , P(n) . This axiom is carefully motivated both with the visual metaphor of a chain of falling dominos and a more formal comparison to previously discussed logical inference rules.

归纳证明通常用归纳集来表述:归纳集是指包含元素 1 且对加 1 封闭的集合。例如,学生需要证明命题 ∀n ∈ ℤ ≥0 , P(n),方法是构造集合 S = {n ∈ ℤ ≥0 : P(n) 为真},并验证 S 是否为归纳集。这种额外的转换步骤会让很多学生感到困惑,而且并非必要。归纳集确实具有重要的技术意义:它们提供了一种严格的构造方法,将自然数集定义为 ℝ 的所有归纳子集的交集。我在可选的最后一章(实数部分,参见 8.3 节)中讨论了这个高级主题,但在归纳法的初始讲解中,我避免提及归纳集。相反,归纳证明基于归纳公理,该公理指出,命题 P(1) 和 ∀n ∈ ℤ ≥0 , P(n) ⇒ P(n + 1) 足以证明 ∀n ∈ ℤ ≥0 , P(n)。该公理的论证经过精心设计,既运用了多米诺骨牌倒下的视觉隐喻,也通过与先前讨论的逻辑推理规则的更正式的比较来加以说明。

Additional Pedagogical Features

其他教学特色

(a) Section Summaries and Global Reviews. Every section ends with a concise recap of the key points just covered. Each major part of the text (logic and proofs; sets and integers; relations, functions, and cardinality) ends with a global review summarizing the material covered in that part. These reviews assemble many definitions, theorem statements, and proof techniques in one place, facilitating memorization and mastery of this vast amount of information.

(a)章节总结和全局回顾。每节课结尾都附有对刚刚讲解的关键点的简明总结。教材的每个主要部分(逻辑与证明;集合与整数;关系、函数与基数)结尾都有全局回顾,总结该部分的内容。这些回顾将大量的定义、定理陈述和证明技巧集中在一起,便于记忆和掌握这些海量信息。

(b) Avoiding Logical Jargon. This text avoids ponderous terminology from classical logic (such as conjunction, disjunction, modus ponens, modus tollens, modus tollendo ponens, hypothetical syllogism, constructive dilemma, universal instantiation, and existential instantiation). I use only those terms from logic that are essential for mathematical work (such as tautology, converse, contrapositive, and quantifier). My exposition replaces antiquated Latin phrases like “modus ponens” by more memorable English names such as “the Inference Rule for IF.” Similarly, I refer to the hypothesis P and the conclusion Q of the IF-statement P Q , rather than calling P the antecedent and Q the consequent of this statement.

(b) 避免使用逻辑术语。本文避免使用古典逻辑中晦涩难懂的术语(例如合取、析取、肯定前件式、否定前件式、否定肯定前件式、假言三段论、构造性困境、全称实例化和存在实例化)。我仅使用对数学工作至关重要的逻辑术语(例如重言式、逆命题、逆否命题和量词)。我的阐述用更易于记忆的英文名称(例如“IF 推理规则”)取代了“肯定前件式”等古老的拉丁语短语。同样,对于 IF 语句 P ⇒ Q,我使用“假设”来指代假设 P 和结论 Q,而不是称 P 为该语句的前件,Q 为后件。

(c) Finding Useful Denials. This is one of the most crucial skills students learn in a proofs course. Every good textbook states the basic denial rules, but students do not always realize (and texts do not always emphasize) that the rules must be applied recursively to find a denial of a complex statement. I describe this recursive process explicitly in Section 1.5 (see especially the table on page 35). Section 1.6 reinforces this key skill with many solved sample problems and exercises.

(c) 寻找有用的否定。这是学生在证明课程中需要学习的最关键的技能之一。每本优秀的教科书都会阐述基本的否定规则,但学生并非总能意识到(教材也并非总能强调)这些规则必须递归地应用才能找到复杂命题的否定。我在第 1.5 节中详细描述了这个递归过程(尤其参见第 35 页的表格)。第 1.6 节通过大量的例题和练习巩固了这项关键技能。

(d) Annotated Proofs. Advanced mathematics texts often consist of a series of definitions, theorems, and proofs with little explanation given for how the author found the proofs. This text is filled with explicit annotations showing the reader how we are generating the lines of a proof, why we are proceeding in a certain way, and what the common pitfalls are. These annotations are clearly delineated from the official proof by enclosing them in square brackets. Many sample proofs are followed by commentary discussing important logical points revealed by the proof.

(d) 注释证明。高等数学教材通常由一系列定义、定理和证明组成,很少解释作者是如何得出这些证明的。本书则充满了清晰的注释,向读者展示我们如何生成证明的每一行,为什么我们采取某种方法,以及常见的陷阱是什么。这些注释用方括号括起来,与正式证明清晰区分开来。许多示例证明后都附有评注,讨论证明中揭示的重要逻辑要点。

(e) Disproofs Contrasted with Proofs by Contradiction. A very common student mistake is to confuse the disproof of a false statement P with a proof by contradiction of a true statement Q . This mistake occurs because of inattention to logical status words: the disproof of P begins with the goal of proving a denial of P , whereas a proof of Q by contradiction begins by assuming the denial of Q . We explicitly warn readers about this issue in Remark 2.60.

(e) 反证法与反证法的区别。学生常犯的一个错误是将对假命题 P 的反证法与对真命题 Q 的反证法混淆。这种错误源于对逻辑状态词的忽视:对 P 的反证法以证明 P 为否定命题为目标,而对 Q 的反证法则以假设 Q 为否定命题为目标。我们在注 2.60 中明确提醒读者注意这个问题。

(f) Set Definitions. New sets and set operations are often defined using set-builder notation. For example, the union of sets S and T is defined by writing S ∪ T = { x : x S or x T . This book presents these definitions in a format more closely matching how they arise in proofs, by explicitly stating what membership in the new set means. For instance, my definition of set union says that for all sets S and T and all objects x , the defined term x S T can be replaced by the definition text x S or x T at any point in a proof. This is exactly what the previous definition means, of course, but the extra layer of translation inherent in the set-builder notation causes trouble for many beginning students.

(f) 集合定义。新的集合和集合运算通常使用集合构造符号来定义。例如,集合 S 和 T 的并集定义为 S ∪ T = {x : x ∈ S 或 x ∈ T}。本书以一种更贴近证明中实际出现方式的格式来呈现这些定义,明确地说明了新集合的成员资格意味着什么。例如,我对集合并集的定义是:对于所有集合 S 和 T 以及所有对象 x,定义项 x∈S∪T 可以在证明中的任何位置替换为定义文本 x∈S 或 x∈T。这当然与之前的定义完全相同,但集合构造符号中固有的额外转换会给许多初学者带来困扰。

(g) Careful Organization of Optional Material. Advanced material and additional topics appear in clearly labeled optional sections. This organization provides maximum flexibility to instructors who want to supplement the material in the standard core, while signaling to readers what material may be safely skipped.

(g) 精心组织选修材料。进阶材料和补充主题列于清晰标注的选修章节中。这种组织方式既为希望补充标准核心教材内容的教师提供了最大的灵活性,又能引导读者了解哪些内容可以跳过。

Exercises, Errata, and Feedback

练习、勘误和反馈

The book contains more than 1000 exercises of varying scope and difficulty, which may be assigned as graded homework or used for self-study or review. I welcome your feedback about any aspect of this book, most particularly corrections of any errors that may be lurking in the following pages. Please send such communications to me by email at nloehr@vt.edu . I will post errata and other pertinent information on the book’s website.

本书包含1000多道练习题,难度和范围各不相同,既可作为作业布置并计分,也可用于自学或复习。我欢迎您就本书的任何方面提出反馈意见,尤其欢迎您指出书中可能存在的任何错误。请将此类反馈意见发送至我的邮箱:nloehr@vt.edu。我会在本书的网站上发布勘误表和其他相关信息。

Words of Thanks

致谢

Some pedagogical elements of this book were suggested by the exposition in A Transition to Advanced Mathematics by Smith, Eggen, and St. Andre. My debt to this excellent text will be evident to anyone familiar with it. My development as a mathematician and a writer has also been deeply influenced by the superb works of James Munkres, Joseph Rotman, J. Donald Monk, and the other authors listed in the Suggestions for Further Reading (see page 383). I thank all the editorial staff at CRC Press, especially Bob Ross and Jose Soto, and the anonymous reviewers whose comments greatly improved the quality of my original manuscript.

本书的部分教学理念借鉴了史密斯、埃根和圣安德烈合著的《高等数学的过渡》一书。任何熟悉此书的人都会明白我对这本优秀著作的感激之情。此外,詹姆斯·蒙克雷斯、约瑟夫·罗特曼、J·唐纳德·蒙克以及“延伸阅读建议”(见第383页)中列出的其他作者的杰出著作也对我的数学和写作发展产生了深远的影响。我衷心感谢CRC出版社的所有编辑人员,特别是鲍勃·罗斯和何塞·索托,以及匿名审稿人,他们的意见极大地提升了我原稿的质量。

I am grateful to many students, colleagues, friends, and family members who supported me during the preparation of this book. I especially thank my father Frank Loehr, my mother Linda Lopez, my stepfather Peter Lopez, Ken Zeger, Elizabeth Niese, Bill Floyd, Leslie Kay, and the students who took proofs classes from me over the years. Words cannot express how much I learned about teaching proofs from my students. Thank you all!

我衷心感谢在本书编写过程中给予我支持的众多学生、同事、朋友和家人。我尤其要感谢我的父亲弗兰克·洛尔、母亲琳达·洛佩兹、继父彼得·洛佩兹、肯·泽格、伊丽莎白·尼泽、比尔·弗洛伊德、莱斯利·凯,以及多年来跟随我学习证明课程的学生们。言语无法表达我从学生们身上学到了多少关于证明教学的知识。谢谢大家!

Very respectfully,

非常恭敬,

Nicholas A. Loehr

尼古拉斯·A·洛尔

#

1

Logic

逻辑

1.1  Propositions, Logical Connectives, and Truth Tables

1.1 命题、逻辑连接词和真值表

Many people despise mathematics, believing it to be nothing more than a confusing jumble of arcane formulas and mind-numbing computations. This depressing view of the subject is understandable, when we consider how math is presented in grade school and many calculus classes. But in truth, mathematics is a beautiful, intricately structured tower of knowledge built up from a small collection of basic statements (called axioms ) using the laws of logic . In this book, we shall study the foundation of this tower, as shown here:

许多人厌恶数学,认为它不过是一堆晦涩难懂的公式和令人头疼的计算。考虑到小学和许多微积分课程中数学的教学方式,这种对数学的悲观看法是可以理解的。但事实上,数学是一座精妙绝伦、结构复杂的知识塔,它由一小群基本陈述(称为公理)运用逻辑法则构建而成。本书将探讨这座知识塔的基石,如下图所示:


cardinality
基数 functions
函数 relations
关系 integers
整数 sets
集合 proofs
证明 logic
逻辑

Propositions

命题

We begin with propositional logic , which studies how the truth of a complex statement is determined by the truth or falsehood of its parts.

我们从命题逻辑开始,它研究的是如何通过复杂陈述的真假来决定该陈述的真假。

1.1. Definition: Propositions. A proposition is a statement that is either true or false, but not both.

1.1 定义:命题。命题是一个要么为真要么为假,但不能既为真又为假的陈述。

Many things we say are not propositions, as seen in the next example.

我们所说的很多话并不是命题,如下例所示。

1.2. Example. Which of these statements are propositions?

1.2. 示例。下列哪些陈述是命题?

(a) 7 is positive.

(a)7 为正数。

(b) 1 + 1 = 7.

(b)1 + 1 = 7。

(c) Memorize all definitions.

(c)记住所有定义。

(d) Okra tastes great.

(d)秋葵味道鲜美。

(e) Is it raining?

(e)下雨了吗?

(f) This sentence is false.

(f)这句话是错误的。

(g) Paris is a city and 2 + 2 is not 4, or Paris is not a city and 2 + 2 is 4.

(g)巴黎是一座城市,2 + 2 不等于 4,或者巴黎不是一座城市,2 + 2 等于 4。

Solution. Statement (a) is a true proposition. Statement (b) is a false proposition. Commands, opinions, and questions do not have a truth value, so statements (c) through (e) are not propositions. Statement (f) is an example of a paradox : if you assume this statement is true, then the statement itself asserts that it is false. If you instead assume the statement is false, then the statement is also true. Since propositions are not allowed to be both true and false, statement (f) is not a proposition. Finally, statement (g) is a false proposition, for reasons described below.

解答:陈述 (a) 为真命题。陈述 (b) 为假命题。命令、意见和疑问句没有真值,因此陈述 (c) 至 (e) 不是命题。陈述 (f) 是一个悖论的例子:如果你假设该陈述为真,那么该陈述本身就断言它是假的。如果你反过来假设该陈述为假,那么该陈述也为真。由于命题不允许既为真又为假,因此陈述 (f) 不是命题。最后,陈述 (g) 为假命题,原因如下所述。

Propositional Forms

命题形式

Statement (g) in the last example is a complex proposition built up from two shorter propositions using logical connective words such as AND, NOT, OR. The two shorter propositions are “Paris is a city” and “2 + 2 is 4.” Let us abbreviate the proposition “Paris is a city” by the letter P , and let us abbreviate the proposition “2 + 2 is 4” by the letter Q . Then statement (g) has the form

上例中的命题 (g) 是一个复杂的命题,它由两个较短的命题构成,并使用了诸如 AND、NOT、OR 之类的逻辑连接词。这两个较短的命题分别是“巴黎是一座城市”和“2 + 2 等于 4”。我们用字母 P 来缩写命题“巴黎是一座城市”,用字母 Q 来缩写命题“2 + 2 等于 4”。那么,命题 (g) 的形式为:

( P P AND ( NOT 不是 Q ) ) OR 或者 ( ( NOT 不是 P P ) AND Q ) .

We can create an even shorter expression to represent the logical form of statement (g) by introducing special symbols for the logical connectives. Propositional logic uses the six symbols shown in the following table, whose meaning is discussed in detail below.

我们可以通过引入逻辑连接词的特殊符号,创建一个更简洁的表达式来表示语句 (g) 的逻辑形式。命题逻辑使用下表所示的六个符号,其含义将在下文详细讨论。

Logical Symbol

逻辑符号

English Translation

英文翻译

~ P

~P

P is not true.

P 不成立。

P Q

P∧Q

P and Q .

P 和 Q。

P Q

P∨Q

P or Q (or both).

P 或 Q(或两者)。

P Q

P⊕Q

P or Q , but not both.

P 或 Q,但不能两者兼有。

P Q

P ⇒ Q

if P , then Q .

如果 P,则 Q。

P Q

P ⇔ Q

P if and only if Q .

当且仅当 Q 时,P 才成立。

Using these symbols, we can describe the logical form of statement (g) by the expression

利用这些符号,我们可以用表达式来描述语句 (g) 的逻辑形式。

( P P ( Q ) ) ( ( P P ) Q ) .

(1.1)

Any expression like this, which is built up by combining propositional variables (capital letters) using logical symbols, is called a propositional form .

任何像这样的表达式,都是通过逻辑符号组合命题变量(大写字母)而构成的,称为命题形式。

Definitions of NOT and AND

“非”和“与”的定义

In logic and mathematics, language is used in a very precise way that does not always coincide with how words are used in everyday conversation. In particular, before going further, we need to define the exact meaning of the logical connective words NOT, AND, OR, IF, etc. To do this, we give truth tables that show how to combine the truth values of propositions to obtain the truth value of a new proposition built from these using a logical connective. Here and below, italic capital letters such as P , Q , R are variables that represent propositions. The letters T and F stand for true and false , respectively.

在逻辑和数学中,语言的使用非常精确,这与日常对话中词语的使用方式并不总是一致。尤其是在深入探讨之前,我们需要明确逻辑连接词“非”、“与”、“或”、“如果”等的确切含义。为此,我们提供了真值表,它展示了如何组合命题的真值,从而得到由这些命题通过逻辑连接词构建的新命题的真值。在下文中,斜体大写字母(例如 P、Q、R)代表命题。字母 T 和 F 分别代表真和假。

1.3. Definition of NOT. For any proposition P , the truth value of ~ P (“not P ”) is determined by the following table.

1.3. 非的定义。对于任何命题 P,~P(“非 P”)的真值由下表确定。

P

P

~ P

~P

T

T

F

F

F

F

T

T

Remember this table by noting that ~ P always has the opposite truth value as P .

记住这张表格,注意 ~P 的真值总是与 P 的真值相反。

In our example above, where Q is the true proposition “2 + 2 is 4,” ~ Q is the false proposition “2 + 2 is not 4.” If R is the false proposition “9 is negative,” then ~ R is the true proposition “9 is not negative.”

在上面的例子中,如果 Q 是真命题“2 + 2 等于 4”,那么 ~Q 是假命题“2 + 2 不等于 4”。如果 R 是假命题“9 是负数”,那么 ~R 是真命题“9 不是负数”。

The remaining logical connectives combine two propositions to produce a new proposition. As seen in the next definition, we need a four-row truth table to list all possible truth values of the two propositions that we start with.

剩余的逻辑连接词将两个命题组合成一个新的命题。正如下一个定义所示,我们需要一个四行的真值表来列出这两个初始命题的所有可能真值。

1.4. Definition of AND. For any propositions P , Q , the truth value of P Q (“ P and Q ”) is determined by the following table.

1.4. AND 的定义。对于任意命题 P、Q,P∧Q(“P 且 Q”)的真值由下表确定。

P

P

Q

P Q

P∧Q

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

T

F

F

F

F

F

F

F

F

Remember this table by noting that P Q is true only when both inputs P , Q are true; in all other cases, P Q is false.

记住这张表格,注意 P∧Q 仅当输入 P 和 Q 都为真时才为真;在所有其他情况下,P∧Q 为假。

In our running example, where P is the true proposition “Paris is a city” and Q is the true proposition “2 + 2 is 4,” note that P ( Q ) is false, because P is true and ~ Q is false. Similarly, ( P ) Q is false, because ~ P is false and Q is true. In this example, for any proposition R , we can deduce that ( P ) R must be false no matter what the truth value of R is, since ~ P is already false. On the other hand, P R is true for true propositions R , and false for false propositions R .

在我们的示例中,P 是真命题“巴黎是一座城市”,Q 是真命题“2 + 2 等于 4”。注意,P∧(∼Q) 为假,因为 P 为真且 ~Q 为假。类似地,(∼P)∧Q 也为假,因为 ~P 为假且 Q 为真。在这个例子中,对于任何命题 R,我们可以推断 (∼P)∧R 必然为假,无论 R 的真值是什么,因为 ~P 本身就是假的。另一方面,对于真命题 R,P∧R 为真;对于假命题 R,P∧R 为假。

Definitions of OR and XOR

OR 和 XOR 的定义

We now come to the mathematical definition of the word OR. In everyday English, the word OR can be used in two different ways. In the inclusive usage of this word, “ P or Q ” means that at least one of the propositions P , Q is true (possibly both). In the exclusive usage, “ P or Q ” means that exactly one of the propositions P , Q is true and the other one is false. Logic uses two different symbols for these two usages of the word OR: ∨ stands for inclusive-OR , and stands for exclusive-OR (which is also abbreviated XOR). Remember this convention: in mathematics, the English word OR always means inclusive-OR , as defined by the truth table below. If you want to use XOR in mathematical English, you must use a longer phrase such as “ P or Q , but not both.”

现在我们来看“或”的数学定义。在日常英语中,“或”有两种不同的用法。在包含性用法中,“P 或 Q”表示命题 P 和 Q 中至少有一个为真(可能两者都为真)。在异或用法中,“P 或 Q”表示命题 P 和 Q 中恰好有一个为真,另一个为假。逻辑学使用两个不同的符号来表示“或”的这两种用法:∨ 表示包含性“或”,⊕ 表示异或(也称为 XOR)。请记住这个约定:在数学中,英语单词“或”始终表示包含性“或”,如下表所示。如果要在数学英语中使用“异或”,则必须使用更长的短语,例如“P 或 Q,但不能两者都为真”。

1.5. Definition of OR. For any propositions P , Q , the truth value of P Q (“ P or Q ”) is determined by the following table.

1.5. OR 的定义。对于任意命题 P、Q,P ∨ Q(“P 或 Q”)的真值由下表确定。

P

P

Q

P Q

P ∨ Q

T

T

T

T

T

T

T

T

F

F

T

T

F

F

T

T

T

T

F

F

F

F

F

F

Remember this table by noting that P Q is false only when both inputs P , Q are false; in all other cases, P Q is true. To avoid confusing the symbols ∧ and ∨, it may help to notice that ∧ (the symbol for AND) resembles the diagonal strokes in the capital letter A.

记住这张表格的要点是:当且仅当输入 P 和 Q 都为假时,P∨Q 才为假;在所有其他情况下,P∨Q 都为真。为了避免混淆符号 ∧ 和 ∨,可以注意到 ∧(表示“与”的符号)的形状类似于大写字母 A 的斜线。

To complete our analysis of statement (g) above, recall in this example that ( P ( Q ) ) is false and ( ( P ) Q ) is false. So the overall statement is false. On the other hand, P ∨(~ Q ), (~ P )∨ Q , and P Q all stand for true propositions in this example.

为了完成对上述陈述 (g) 的分析,请记住,在本例中,(P∧(∼Q)) 为假,((∼P)∧Q) 也为假。因此,整个陈述为假。另一方面,在本例中,P∨(~Q)、(~P)∨Q 和 P∨Q 都代表真命题。

1.6. Definition of XOR. For any propositions P , Q , the truth value of P Q (“ P XOR Q ”) is determined by the following table.

1.6. 异或的定义。对于任意命题 P、Q,P⊕Q(“P 异或 Q”)的真值由下表确定。

P

P

Q

P Q

P⊕Q

T

T

T

T

F

F

T

T

F

F

T

T

F

F

T

T

T

T

F

F

F

F

F

F

Remember this table by noting that P Q is true when the input propositions P , Q have different truth values; P Q is false when the input propositions P , Q have the same truth values.

记住这张表格,注意当输入命题 P 和 Q 的真值不同时,P⊕Q 为真;当输入命题 P 和 Q 的真值相同时,P⊕Q 为假。

1.7. Example. Let P be the false proposition “1 + 1 = 3,” and let Q be the true proposition “2 is positive.” Which propositions below are true?

1.7. 示例。设 P 为假命题“1 + 1 = 3”,Q 为真命题“2 是正数”。下列哪些命题为真?

(a) P Q ; (b) P ( Q ) ; (c) ( P ) Q ; (d) P Q ; (e) P ∨(~ Q ); (f) (~ P )∨ Q .

(a)P⊕Q;(b)P⊕(∼Q);(c)(∼P)⊕Q;(d)P∨Q;(e)P∨(~Q);(f)(~P)∨Q。

Solution . Using the truth tables for , ∨, and ∼, we find that (a) is true, (b) is false, (c) is false, (d) is true, (e) is false, and (f) is true. The different answers for (c) and (f) illustrate the distinction between exclusive-OR and inclusive-OR.

解答:利用⊕、∨和∼的真值表,我们发现(a)为真,(b)为假,(c)为假,(d)为真,(e)为假,(f)为真。(c)和(f)的不同答案说明了异或运算和同或运算的区别。

Truth Tables for Propositional Forms

命题形式的真值表

Given a complex propositional form built up from some variables using one or more logical connectives, we can determine the meaning of this form by making a truth table showing the truth value of the form for all possible combinations of truth values of the variables appearing in the form. We construct such a truth table step-by-step, making a column for each input variable and each smaller propositional form contained within the given form. We complete each column by using the defining truth tables (given above for ∼, ∧, ∨, , and given in later sections for ⇒ and ⇔) to combine the truth values in one or two previously completed columns. We illustrate this process in the following examples.

给定一个由若干变量通过一个或多个逻辑连接词构建而成的复杂命题形式,我们可以通过构建真值表来确定其含义。真值表列出了该命题形式中所有变量真值组合的真值。我们逐步构建这样的真值表,为每个输入变量以及给定命题形式中包含的每个子命题形式创建一列。我们使用定义真值表(上文给出了∼、∧、∨、⊕的定义真值表,后续章节将给出⇒和⇔的定义真值表)来合并先前已完成的一列或两列中的真值,从而完成每一列。以下示例说明了这一过程。

1.8 Example: Truth Table for ( P Q ) ( P ) . This form has two input variables, P and Q , so we need a four-row truth table. We make columns for P , Q , ~ P , P Q , and ( P Q ) ( P ) , and fill them in as shown here:

1.8 示例:(P∧Q)∨(∼P) 的真值表。此形式有两个输入变量 P 和 Q,因此我们需要一个四行的真值表。我们分别创建 P、Q、∼P、P∧Q 和 (P∧Q)∨(∼P) 列,并按如下方式填写:

P

P

Q

~ P

~P

P Q

P∧Q

( P Q ) ( P )

(P∧Q)∨(∼P)

T

T

T

T

F

F

T

T

T

T

T

T

F

F

F

F

F

F

F

F

F

F

T

T

T

T

F

F

T

T

F

F

F

F

T

T

F

F

T

T

Note, in particular, that the final column is completed by comparing the truth values in each row of columns 3 and 4. In row 2 (and only row 2), the two input truth values are both false, so the output in this row is false. All other rows must have output that is true, by definition of OR.

特别需要注意的是,最后一列是通过比较第 3 列和第 4 列每一行的真值来完成的。在第 2 行(且仅第 2 行),两个输入的真值均为假,因此该行的输出为假。根据“或”运算的定义,所有其他行的输出必须为真。

1.9 Example: Truth Table for P ( Q ( P ) ) . This form looks almost identical to the one in the last example, but the parentheses are in a different position. This changes the structure of the truth table, as shown here:

1.9 示例:P∧(Q∨(∼P)) 的真值表。此形式与上一个示例中的形式几乎相同,但括号的位置不同。这改变了真值表的结构,如下所示:

P

P

Q

~ P

~P

Q ∨(~ P )

Q∨(~P)

P ( Q ( P ) )

P∧(Q∨(∼P))

T

T

T

T

F

F

T

T

T

T

T

T

F

F

F

F

F

F

F

F

F

F

T

T

T

T

T

T

F

F

F

F

F

F

T

T

T

T

F

F

We fill in column 4 by looking for two Fs in columns 2 and 3 (which happens only in row 2), placing an F in that row, and filling the rest of column 4 with Ts. Then we fill in column 5 by looking for two Ts in columns 1 and 4 (which happens in row 1 only), placing a T in that row, and filling the rest of column 5 with Fs.

我们通过在第 2 列和第 3 列中查找两个 F(仅在第 2 行出现),将 F 填入第 4 列,然后用 T 填满第 4 列的剩余部分。然后,我们通过在第 1 列和第 4 列中查找两个 T(仅在第 1 行出现),将 T 填入第 5 列,然后用 F 填满第 5 列的剩余部分。

In these examples, observe that the truth tables for ( P Q ) ( P ) and P ( Q ( P ) ) are not identical, since the outputs in rows 3 and 4 differ in the two tables. This shows that these two propositional forms are not logically interchangeable with one another in all situations. More briefly, we say that the two forms are not logically equivalent. On the other hand, if two propositional forms A and B have truth tables whose outputs agree in every row, we say that A and B are logically equivalent and write A B . (We study logical equivalence in detail in the next few sections.) Notice, for instance, that P ( Q ( P ) ) is logically equivalent to P Q , since column 5 in the previous example agrees with the truth table for P Q in all four rows. So we could write P ( Q ( P ) ) P Q , enabling us to replace the complicated propositional form on the left by the shorter form P Q on the right.

在这些例子中,请注意 (P∧Q)∨(∼P) 和 P∧(Q∨(∼P)) 的真值表并不相同,因为两个真值表第 3 行和第 4 行的输出不同。这表明这两个命题形式并非在所有情况下都能逻辑互换。简而言之,我们称这两个形式逻辑不等价。另一方面,如果两个命题形式 A 和 B 的真值表每一行的输出都相同,则称 A 和 B 逻辑等价,记作 A≡B。(我们将在接下来的几节中详细讨论逻辑等价性。)例如,请注意 P∧(Q∨(∼P)) 与 P∧Q 逻辑等价,因为前一个例子中的第 5 列与 P∧Q 的真值表在所有四行都相同。因此,我们可以写成 P∧(Q∨(∼P))≡P∧Q,从而可以用右边的较短形式 P∧Q 替换左边复杂的命题形式。

1.10 Example: Translation of XOR. We have noted that P Q can be translated as “P or Q, but not both.” This English phrase is built up using the logical connectives OR, NOT, BUT, and BOTH (the last two words have the same meaning as AND). So a more literal encoding of the phrase “P or Q, but not both” as a propositional form is ( P Q ) ( ( P Q ) ) . The following truth table verifies that ( P Q ) ( ( P Q ) ) P Q . This logical equivalence justifies the use of this phrase as a translation of XOR.

1.10 示例:异或 (XOR) 的翻译。我们已经注意到 P⊕Q 可以翻译为“P 或 Q,但不能同时是两者”。这个英文短语由逻辑连接词 OR、NOT、BUT 和 BOTH 构成(后两个词与 AND 的含义相同)。因此,将短语“P 或 Q,但不能同时是两者”更字面地编码为命题形式为 (P∨Q)∧(∼(P∧Q))。下面的真值表验证了 (P∨Q)∧(∼(P∧Q))≡P⊕Q。这种逻辑等价性证明了使用该短语作为异或 (XOR) 翻译的合理性。

P

P

Q

P Q

P⊕Q

P Q

P ∨ Q

P Q

P∧Q

( P Q )

∼(P∧Q)

( P Q ) ( ( P Q ) )

(P∨Q)∧(∼(P∧Q))

T

T

T

T

F

F

T

T

T

T

F

F

F

F

T

T

F

F

T

T

T

T

F

F

T

T

T

T

F

F

T

T

T

T

T

T

F

F

T

T

T

T

F

F

F

F

F

F

F

F

F

F

T

T

F

F

Note that we combine columns 4 and 6 to complete column 7 using the definition of AND. The claimed logical equivalence follows from the complete agreement of columns 3 and 7 in every row. In contrast, since columns 3 and 4 disagree in row 1, we see that ( P Q ) ( P Q ) . In other words, inclusive-OR and exclusive-OR are not logically equivalent.

注意,我们利用“与”的定义,将第 4 列和第 6 列合并为第 7 列。所声称的逻辑等价性源于每一行中第 3 列和第 7 列的完全一致。相反,由于第 1 行中第 3 列和第 4 列不一致,因此 (P⊕Q)≢(P∨Q)。换句话说,“包含或”和“异或”在逻辑上并不等价。

Formal Definition of Propositional Forms (Optional)

命题形式的正式定义(可选)

In the main text, we informally defined a propositional form to be any expression built up by combining propositional variables using logical symbols. To give a precise, rigorous definition of propositional forms, we need a recursive definition (definitions of this type are studied in detail later). Specifically, propositional forms are defined recursively via the following rules:

在正文中,我们非正式地将命题形式定义为任何由逻辑符号组合命题变量构成的表达式。为了给出命题形式的精确、严谨的定义,我们需要一个递归定义(此类定义将在后文详细讨论)。具体而言,命题形式通过以下规则进行递归定义:

(a) A single capital italic letter is a propositional form.

(a)单个大写斜体字母是命题形式。

(b) If A is any propositional form, then ( A ) is a propositional form.

(b)如果 A 是任意命题形式,则(∼A)也是命题形式。

(c) If A and B are any propositional forms, then ( A B ) is a propositional form.

(c)如果 A 和 B 是任意命题形式,则(A∧B)也是命题形式。

(d) If A and B are any propositional forms, then ( A B ) is a propositional form.

(d)如果 A 和 B 是任意命题形式,则(A∨B)也是命题形式。

(e) If A and B are any propositional forms, then ( A B ) is a propositional form.

(e)如果 A 和 B 是任意命题形式,则(A⊕B)也是命题形式。

(f) If A and B are any propositional forms, then ( A B ) is a propositional form.

(f)如果 A 和 B 是任意命题形式,则(A⇒B)也是命题形式。

(g) If A and B are any propositional forms, then ( A B ) is a propositional form.

(g)如果 A 和 B 是任意命题形式,则(A⇔B)是命题形式。

(h) An expression is a propositional form only if it can be formed by applying rules (a) through (g) finitely many times.

(h)一个表达式是命题形式,当且仅当它可以由有限次应用规则(a)到(g)构成。

For example, rule (a) shows that P is a propositional form, as is Q . Then rule (b) shows that (~ P ) and (~ Q ) are propositional forms. By rule (c), ( P ( Q ) ) and ( ( P ) Q ) are propositional forms. By rule (d),

例如,规则 (a) 表明 P 是命题形式,Q 也是命题形式。那么规则 (b) 表明 (~P) 和 (~Q) 是命题形式。根据规则 (c),(P∧(~Q)) 和 ((~P)∧Q) 也是命题形式。根据规则 (d),

( ( P P ( Q ) ) ( ( P P ) Q ) )

(1.2)

is a propositional form, which is essentially the form studied in (1.1) .

是命题形式,本质上是(1.1)中研究的形式。

However, to be absolutely precise, the expression in (1.1) is not a propositional form. This is because (1.1) is missing the outermost pair of parentheses, which are required to be present when using rules (b) through (g). In practice, we often drop required parentheses to create abbreviated versions of propositional forms that are easier to read. For instance, we seldom write the outermost pair of parentheses. Logical equivalences called associativity rules (discussed in the next section) allow us to drop internal parentheses in an expression like (( P Q )∨ R ) or ( P ( Q R ) ) . We can erase even more parentheses using the precedence conventions that ∼ has highest precedence, followed by , followed by ∨. For example, with these conventions, we could drop all parentheses from (1.2) and abbreviate this form as P Q P Q . However, as shown in Examples 1.8 and 1.9 , the placement of parentheses often affects the logical meaning of a propositional form (which is one reason why the official definition requires them). In this text, we often write parentheses that could be deleted by the precedence conventions, to reduce the chance of confusion.

然而,严格来说,(1.1) 式并非命题形式。这是因为 (1.1) 式缺少最外层的一对括号,而使用规则 (b) 至 (g) 时,这对括号是必需的。在实践中,我们经常省略必要的括号,以创建更易于阅读的命题形式的简写形式。例如,我们很少写出最外层的一对括号。逻辑等价关系,即结合律(将在下一节讨论),允许我们省略表达式(例如 ((P ∨ Q)∨R) 或 (P∧(Q∧R)))中的内部括号。我们可以使用优先级约定来省略更多的括号,其中 ∼ 的优先级最高,其次是 ∧,最后是 ∨。例如,按照这些约定,我们可以省略(1.2)式中的所有括号,并将该形式简写为P∧∼Q∨∼P∧Q。然而,如例1.8和1.9所示,括号的位置通常会影响命题形式的逻辑含义(这也是官方定义要求使用括号的原因之一)。在本文中,我们经常会使用一些根据优先级约定可以省略的括号,以减少混淆的可能性。

1.11. Remark: Terminology for Propositional Forms. Let A and B be any propositional forms. In some logic texts, ( A ) is called the negation of A ; ( A B ) is called the conjunction of A and B ; ( A B ) is called the disjunction of A and B ; ( A B ) is called an implication or conditional with hypothesis (or antecedent ) A and conclusion (or consequent ) B ; ( A B ) is called the biconditional of A and B . We mostly avoid using these technical terms in this text, except for the term negation . Later, we study methods of finding propositional forms logically equivalent to ( A ) , which are called denials of A .

1.11. 说明:命题形式的术语。设 A 和 B 为任意命题形式。在一些逻辑教材中,(∼A) 称为 A 的否定;(A∧B) 称为 A 与 B 的合取;(A∨B) 称为 A 与 B 的析取;(A⇒B) 称为蕴涵或条件语句,其前提(或前件)为 A,结论(或后件)为 B;(A⇔B) 称为 A 与 B 的双条件语句。本文中,除“否定”一词外,我们尽量避免使用这些技术术语。稍后,我们将研究寻找与 (∼A) 逻辑等价的命题形式的方法,这些命题形式称为 A 的否定。

Section Summary

章节概要

1. Memorize the definitions of the logical connectives, as summarized in the following truth table.

1.记住逻辑连接词的定义,如下面的真值表所示。

NOT P

不是 P

P AND Q

P 和 Q

P OR Q

P 或 Q

P XOR Q

P XOR Q

P

P

Q

~ P

~P

P Q

P∧Q

P Q

P ∨ Q

P Q

P⊕Q

T

T

T

T

F

F

T

T

T

T

F

F

T

T

F

F

F

F

F

F

T

T

T

T

F

F

T

T

T

T

F

F

T

T

T

T

F

F

F

F

T

T

F

F

F

F

F

F

Note that NOT flips the truth value of the input; AND outputs true exactly when both inputs are true; OR outputs false exactly when both inputs are false; and XOR outputs true exactly when the two inputs differ.

注意 NOT 会反转输入的真值;AND 会在两个输入都为真时输出真;OR 会在两个输入都为假时输出假;XOR 会在两个输入不同时输出真。

2. Remember that in mathematics, “ P or Q ” always means P Q (inclusive-OR). To translate P Q (exclusive-OR) into English, you must use an explicit phrase such as “ P or Q , but not both.”

2. 请记住,在数学中,“P 或 Q”总是表示 P ∨ Q(包含或)。要将 P⊕Q(排除或)翻译成英语,您必须使用明确的短语,例如“P 或 Q,但不能同时是两者”。

3. Two propositional forms A and B are logically equivalent, denoted A B , when these forms have truth tables whose outputs agree in all rows of the truth table. If there is even a single row of disagreement, the forms are not logically equivalent. For instance, P Q P Q .

3.当两个命题形式 A 和 B 的真值表在所有行中都一致时,这两个命题形式逻辑等价,记作 A≡B。如果哪怕只有一行不一致,这两个命题形式就不逻辑等价。例如,P∨Q≢P⊕Q。

Exercises

练习

1. Is each statement true, false, or not a proposition? Explain.

1. 下列陈述是真、假还是不是命题?请解释。

(a) 1 + 1 = 2 or London is a city. (b) Math is fun. (c) France is a country, and Berlin is an ocean. (d) 0 < 1 or 1 < 0. (e) 0 < 1 or 1 < 0, but not both. (f) Chopin was the greatest classical composer. (g) 2 is not even 1 , or 2 is positive. (h) 15 is odd or positive, but not both.

(a) 1 + 1 = 2 或 伦敦是一座城市。(b) 数学很有趣。(c) 法国是一个国家,柏林是一片海洋。(d) 0 < 1 或 1 < 0。(e) 0 < 1 或 1 < 0,但两者不能同时成立。(f) 肖邦是最伟大的古典音乐作曲家。(g) 2 不是偶数,或者 2 是正数。(h) 15 是奇数或正数,但两者不能同时成立。

2. Is each statement true, false, or not a proposition? Explain.

2. 下列每个陈述是真、假还是不是命题?请解释。

(a) It is not the case that 2 is even or 2 is positive. (b) The following statement is a true proposition. (c) 0 1 x 2 d x = 1 / 3 and 2 5 = 32. (d) When do fish sleep? (e) The following statement is a true proposition. (f) The previous statement is a false proposition. (g) I am tired.

(a) 2 不是偶数,也不是正数。(b) 以下命题为真。(c) ∫01x2dx=1/3 且 2 5 = 32。(d) 鱼什么时候睡觉?(e) 以下命题为真。(f) 上一个命题为假。(g) 我累了。

3. Construct truth tables for the following propositional forms.

3. 为下列命题形式构造真值表。

(a) ( P ) Q (b) ( P Q ) ( ( P ) ( Q ) ) (c) ( P ( Q ) ) ( ( Q ) P ) (d) ( P Q ) R (e) P ( Q R ) (f) P ( Q R )

(a) (∼P)∧Q (b) (P∧Q)∨((∼P)∧(∼Q)) (c) (P∨(∼Q))⊕((∼Q)∨P) (d) (P∧Q)∨R (e) P∧(Q∨R) (f) P∨(Q⊕R)

4. (a) Is ~( P Q ) logically equivalent to (~ P )∨(~ Q )? Explain with a truth table. (b) Give English translations of the two propositional forms in (a).

4.(a) ~(P ∨ Q) 与 (~P)∨(~Q) 逻辑等价吗?用真值表解释。(b) 给出 (a) 中两个命题形式的英文翻译。

5. Assume P is a true proposition, Q is a false proposition, and R is an arbitrary proposition. Say as much as you can about the truth or falsehood of each proposition below.

5. 假设 P 为真命题,Q 为假命题,R 为任意命题。请尽可能详细地说明以下每个命题的真假。

(a) P ( Q ) (b) (~ P )∨ Q (c) P Q (d) P ( Q ) (e) R P (f) Q R (g) Q R (h) R ( R ) (i) R ∨(~ R ) (j) R R (k) R ( R ) .

(a) P∧(∼Q) (b) (~P)∨Q (c) P⊕Q (d) P⊕(∼Q) (e) R∨P (f) Q∨R (g) Q⊕R (h) R∧(∼R) (i) R∨(~R) (j) R⊕R (k) R⊕(∼R).

6. (a) Make a truth table for the propositional form ( P ( Q ) ) ( ( P ) Q ) . Be sure to show the columns for all intermediate forms. (b) Find a short propositional form that is logically equivalent to the form in (a). (c) Use (b) to write an English sentence with the same logical meaning as statement (g) in Example 1.2 .

6.(a) 为命题形式 (P∧(∼Q))∨((∼P)∧Q) 构建真值表。务必显示所有中间形式的列。(b) 找到一个与 (a) 中的形式逻辑等价的简短命题形式。(c) 利用 (b) 的结果,写出一个与例 1.2 中的语句 (g) 逻辑含义相同的英文句子。

7. (a) Make a truth table for P ( ( Q ) P ) . (b) Make a truth table for ( P ( Q ) ) P . (c) Are the propositional forms in (a) and (b) logically equivalent? Why? (d) Find a short propositional form that is logically equivalent to the form in (a).

7.(a) 为 P∧((∼Q)∨P) 写出真值表。(b) 为 (P∧(∼Q))∨P 写出真值表。(c) (a) 和 (b) 中的命题形式逻辑等价吗?为什么?(d) 找到一个与 (a) 中的形式逻辑等价的简短命题形式。

8. Which propositional forms (if any) are logically equivalent? Justify your answer with truth tables.

8.哪些命题形式(如果有的话)在逻辑上等价?用真值表证明你的答案。

(a) P (b) ~ P (c) P P (d) P P (e) P P (f) ( P Q ) P .

(a)P (b)~P (c)P ∨ P (d)P∧P (e)P⊕P (f)(P∧Q)∨P。

9. Which propositional forms (if any) are logically equivalent? Explain.

9.哪些命题形式(如果有的话)在逻辑上等价?请解释。

(a) P Q (b) Q P (c) ( P Q ) P (d) ( P Q ) .

(a)P∧Q (b)Q∧P (c)(P∧Q)∧P (d)∼(P⊕Q)。

10. Which propositional forms (if any) are logically equivalent? Explain.

10.哪些命题形式(如果有的话)在逻辑上是等价的?请解释。

(a) ( P Q ) (b) ( P ) ( Q ) (c) ( ( P ) Q ) ( P ( Q ) ) (d) P ( Q ) (e) P ∨(~ Q ) (f) ( P ( Q ) ) ( ( P ) Q ) .

(a) ∼(P∧Q) (b) (∼P)∧(∼Q) (c) ((∼P)∧Q)∨(P∧(∼Q)) (d) P⊕(∼Q) (e) P∨(~Q) (f) (P∨(∼Q))∧((∼P)∨Q).

11. Which propositional forms (if any) are logically equivalent? Explain.

11.哪些命题形式(如果有的话)在逻辑上是等价的?请解释。

(a) P ∨( Q R ) (b) P ( Q R ) (c) P ( Q R ) (d) P ( Q R ) (e) ( P Q )∨ R (f) ( P Q ) R (g) ( P Q ) R (h) ( P Q ) R .

(a)P∨(Q∨R) (b)P⊕(Q∨R) (c)P∨(Q⊕R) (d)P⊕(Q⊕R) (e)(P∨Q)∨R (f)(P⊕Q)∨R (g)(P∨Q)⊕R (h)(P⊕Q)⊕R。

12. Let x 0 be a fixed integer. Let P be the proposition “ x 0 < 8,” let Q be “0 < x 0 ,” and let R be “ x 0 = 0.” Encode each statement as a propositional form involving P , Q , and R .

12.设 x 0 为一个固定的整数。设 P 为命题“x 0 < 8”,Q 为“0 < x 0 ”,R 为“x 0 = 0”。将每个语句编码为包含 P、Q 和 R 的命题形式。

(a) x 0 is strictly positive. (b) x 0 ≥ 8. (c) x 0 is less than zero. (d) x 0 is nonnegative. (e) 0 < x 0 < 8. (f) 0 ≤ x 0 < 8. (g) | x 0 − 4| ≥ 4.

(a) x 0 严格为正。(b) x 0 ≥ 8。(c) x 0 小于零。(d) x 0 非负。(e) 0 < x 0 < 8。(f) 0 ≤ x 0 < 8。(g) |x 0 − 4| ≥ 4。

13. Let P be the statement “1 + 1 = 2,” let Q be “0 = 1,” and let R be “71 is prime.” Convert each propositional form into an English sentence without parentheses . Be sure that the sentence has the precise logical structure encoded by the form.

13.设 P 为命题“1 + 1 = 2”,Q 为命题“0 = 1”,R 为命题“71 是素数”。将每个命题形式转换为不含括号的英文句子。确保句子具有与命题形式精确一致的逻辑结构。

(a) ( P Q ) R (b) P ( Q R ) (c) ( P Q ) (d) ( P ) Q (e) ( P Q ) R (f) P ( Q R ) (g) ( ( P ) ( Q ) ) ( R )

(a) (P∧Q)∨R (b) P∧(Q∨R) (c) ∼(P∧Q) (d) (∼P)∧Q (e) (P∧Q)∧R (f) P∧(Q∧R) (g) ((∼P)∧(∼Q))⊕(∼R)

14. Explain why each statement is not a proposition.

14.解释为什么每个陈述都不是命题。

(a) The last train leaves in five minutes. (b) We should raise the minimum wage. (c) Will you marry me? (d) My father’s name is Frank. (e) She is smarter than he is. (f) Do your homework. (g) It is cold here. (h) This statement is true.

(a) 末班车还有五分钟就开了。(b) 我们应该提高最低工资。(c) 你愿意嫁给我吗?(d) 我父亲名叫弗兰克。(e) 她比他聪明。(f) 做你的作业。(g) 这里很冷。(h) 这句话是正确的。

15. Find propositional forms using only the connectives ∨, , and ∼ that have the truth tables shown below. Try to use as few connectives as possible.

15.仅使用连接词∨、∧和∼,找出满足以下真值表的命题形式。尽量使用最少的连接词。

P

P

Q

R

(a)

(一个)

(b)

(b)

(c)

(c)

(d)

(d)

T

T

T

T

T

T

F

F

F

F

F

F

F

F

T

T

T

T

F

F

T

T

T

T

F

F

T

T

T

T

F

F

T

T

T

T

F

F

F

F

T

T

T

T

F

F

F

F

T

T

T

T

T

T

T

T

F

F

T

T

T

T

F

F

F

F

T

T

F

F

F

F

T

T

F

F

T

T

F

F

F

F

T

T

F

F

F

F

T

T

T

T

F

F

F

F

F

F

F

F

F

F

F

F

T

T

F

F

F

F

T

T

16. (a) Find a simple verbal description characterizing when ( ( A B ) ( C D ) ) E is true. (b) Does your answer to (a) change if we rearrange parentheses in the expression? What if we reorder A , B , C , D , and E ?

16.(a) 用简单的文字描述 ((A⊕B)⊕(C⊕D))⊕E 成立的条件。(b) 如果重新排列表达式中的括号,(a) 的答案会改变吗?如果重新排列 A、B、C、D 和 E 的顺序呢?

17. Find a propositional form involving variables P , Q , R that is true precisely when exactly one of P , Q , R is true. Try to use as few logical connectives as possible.

17. 找到一个包含变量 P、Q、R 的命题形式,使得该命题为真当且仅当 P、Q、R 中恰好有一个为真。尽量少用逻辑连接词。

18. Find a propositional form involving variables P , Q , R that is true when at least two of the inputs are true, and false otherwise. Try to use as few logical connectives as possible.

18. 找出包含变量 P、Q、R 的命题形式,当至少两个输入为真时该命题为真,否则为假。尽量少用逻辑连接词。

19. (a) How many truth tables are possible for propositional forms containing two variables P and Q ? (b) For each possible truth table in (a), find a short propositional form having that truth table.

19.(a) 包含两个变量 P 和 Q 的命题形式有多少种可能的真值表?(b) 对于 (a) 中的每一种可能的真值表,找到一个具有该真值表的简短命题形式。

1.2 Logical Equivalences and IF-Statements

1.2 逻辑等价性和 IF 语句

You have probably been using the word IF for most of your life, but do you really know what this word means? The official answer, given in this section, may surprise you. We also look more closely at logical equivalence, developing transformation rules for propositional forms analogous to some basic laws of algebra. Everything is proved using our favorite proof technique, truth tables.

你可能一生中大部分时间都在使用“如果”这个词,但你真的了解它的含义吗?本节给出的官方答案或许会让你感到惊讶。我们还会更深入地探讨逻辑等价性,并推导出类似于代数基本定律的命题形式转换规则。所有内容都将使用我们最喜欢的证明技巧——真值表——进行证明。

Logical Equivalence

逻辑等价性

We first recall the definition of logically equivalent propositional forms from the last section.

我们首先回顾上一节中逻辑等价命题形式的定义。

1.12. Definition: Logically Equivalent Propositional Forms. Two propositional forms A and B are logically equivalent when the truth tables for A and B have outputs that agree in every row. We write A B when A and B are logically equivalent; we write A B when A and B are not logically equivalent.

1.12. 定义:逻辑等价命题形式。当两个命题形式 A 和 B 的真值表每一行的输出都相同时,称 A 和 B 逻辑等价。当 A 和 B 逻辑等价时,记作 A≡B;当 A 和 B 逻辑不等价时,记作 A≢B。

Our first theorem lists some logical equivalences that can often be used to replace complicated propositional forms by shorter ones.

我们的第一个定理列出了一些逻辑等价关系,这些关系通常可以用来用更短的命题形式来代替复杂的命题形式。

1.13. Theorem on Logical Equivalence. For all propositional forms P , Q , and R , the following logical equivalences hold:

1.13. 逻辑等价定理。对于所有命题形式 P、Q 和 R,以下逻辑等价关系成立:

(a) Commutative Laws: P Q Q P , P Q Q P , and P Q Q P .

(a)交换律:P∧Q≡Q∧P,P∨Q≡Q∨P,P⊕Q≡Q⊕P。

(b) Associative Laws: P ( Q R ) ( P Q ) R , P ∨( Q R ) ≡ ( P Q )∨ R , and P ( Q R ) ( P Q ) R .

(b)结合律:P∧(Q∧R)≡(P∧Q)∧R,P∨(Q∨R)≡(P∨Q)∨R,以及P⊕(Q⊕R)≡(P⊕Q)⊕R。

(c) Distributive Laws: P ( Q R ) ( P Q ) ( P R ) , P ( Q R ) ( P Q ) ( P R ) , and P ( Q R ) ( P Q ) ( P R ) .

(c)分配律:P∧(Q∨R)≡(P∧Q)∨(P∧R),P∨(Q∧R)≡(P∨Q)∧(P∨R),以及 P∧(Q⊕R)≡(P∧Q)⊕(P∧R)。

(d) Idempotent Laws: P P P and P P P .

(d)幂等定律:P∧P≡P 和 P∨P≡P。

(e) Absorption Laws: P ( P Q ) P and P ( P Q ) P .

(e)吸收定律:P∧(P∨Q)≡P 和 P∨(P∧Q)≡P。

Negation Laws: ~(~ P ) ≡ P , ( P Q ) ( P ) ( Q ) , and ( P Q ) ( P ) ( Q ) .

否定定律:~(~P) ≡ P,∼(P∧Q)≡(∼P)∨(∼Q),以及∼(P∨Q)≡(∼P)∧(∼Q)。

The names of the first three laws indicate the analogy between these logical equivalences and certain algebraic properties of real numbers. For example, the commutative laws for propositional forms resemble the commutative laws x + y = y + x and x · y = y · x for real numbers x , y ; similarly, observe the resemblance between the distributive laws listed above and the algebraic distributive law x · ( y + z ) = ( x · y ) + ( x · z ).

前三条定律的名称表明了这些逻辑等价关系与实数的某些代数性质之间的类比。例如,命题形式的交换律类似于实数 x 和 y 的交换律 x + y = y + x 和 x · y = y · x;类似地,可以观察到上面列出的分配律与代数分配律 x · (y + z) = (x · y) + (x · z) 之间的相似性。

All parts of this theorem are proved by constructing truth tables for the propositional forms on each side of each equivalence, and verifying that the truth tables agree in every row. We illustrate the technique for a few parts of the theorem, asking you to prove the other parts in the exercises.

本定理的所有部分均可通过为每个等价式两侧的命题形式构造真值表,并验证真值表每一行都一致来证明。我们将用此方法证明定理的几个部分,其余部分则需在练习中证明。

1.14 Proof of the Idempotent Laws. The truth table is shown here:

1.14 幂等定律的证明。真值表如下所示:

P

P

P

P

P P

P∧P

P P

P ∨ P

T

#

T

T

T

T

T

T

T

F

F

F

F

F

F

F

F

We list the column for P twice to make it easier to compute P P and P P . We compute the last two columns from the first two columns using the definition of AND and OR, respectively. Since column 3 and column 2 agree in all rows, P P P . Since column 4 and column 2 agree in all rows, P P P .

我们将 P 列列出两次,以便于计算 P∧P 和 P ∨ P。我们分别利用 AND 和 OR 的定义,根据前两列计算后两列。由于第 3 列和第 2 列在所有行中都相同,因此 P∧P≡P。由于第 4 列和第 2 列在所有行中都相同,因此 P ∨ P ≡ P。

1.15 Proof of ( P Q ) ( P ) ( Q ) . For this equivalence, we need a four-row truth table:

1.15 证明 ∼(P∨Q)≡(∼P)∧(∼Q)。为了证明此等价性,我们需要一个四行真值表:

P

P

Q

P Q

P ∨ Q

~( P Q )

~(P ∨ Q)

~ P

~P

~ Q

~Q

( P ) ( Q )

(∼P)∧(∼Q)

T

T

T

T

T

T

F

F

F

F

F

F

F

F

T

T

F

F

T

T

F

F

F

F

T

T

F

F

F

F

T

T

T

T

F

F

T

T

F

F

F

F

F

F

F

F

F

F

T

T

T

T

T

T

T

T

Columns 4 and 7 agree in every row, so the logical equivalence is proved. This equivalence and its companion ( P Q ) ( P ) ( Q ) are called de Morgan’s laws . In words, the law we just proved says that the negation of an OR-statement “ P or Q ” is equivalent to the statement “(not P ) and (not Q ),” whereas the dual law tells us that “not ( P and Q )” is logically equivalent to “(not P ) or (not Q ).” For a specific example, the statement “it is false that 10 is prime or 10 is odd” may be simplified to the statement “10 is not prime and 10 is not odd” without changing the truth value (both statements are true). Similarly, the statement that “John has brown hair AND blue eyes” is false precisely when John’s hair is not brown OR John’s eyes are not blue. In §1.5 , we will say a lot more about simplifying statements that begin with NOT.

第 4 列和第 7 列每一行都一致,因此逻辑等价性得证。这个等价性及其对应的等价性 ∼(P∧Q)≡(∼P)∨(∼Q) 被称为德摩根定律。用文字表述,我们刚刚证明的定律指出,OR 语句“P 或 Q”的否定等价于语句“(非 P) 且 (非 Q)”,而对应的定律则指出,“非 (P 且 Q)”逻辑等价于“(非 P) 或 (非 Q)”。举个具体的例子,“10 是质数或 10 是奇数为假”这个语句可以简化为“10 不是质数且 10 不是奇数”,而真值不变(两个语句都为真)。类似地,“约翰有棕色头发且蓝色眼睛”这个语句为假,当且仅当约翰的头发不是棕色或约翰的眼睛不是蓝色。在 §1.5 中,我们将更详细地讨论如何简化以 NOT 开头的语句。

1.16. Proof of Associativity of XOR. This law has three input variables, so we need a truth table with eight rows, as shown below. One quick way to create the eight possible input combinations is to write 4 Ts followed by 4 Fs in the column for P ; then write 2 Ts, 2 Fs, 2 Ts, 2 Fs in the column for Q ; then write alternating Ts and Fs in the column for R . This pattern extends to truth tables with even more variables. In general, if there are n distinct input variables that stand for propositions, the truth table has 2 n rows.

1.16. 异或运算结合律的证明。该定律有三个输入变量,因此我们需要一个八行的真值表,如下所示。一种快速创建八种可能输入组合的方法是:在 P 列中写 4 个真 (T) 后跟 4 个假 (F);然后在 Q 列中写 2 个真 (T)、2 个假 (F)、2 个真 (T)、2 个假 (F);最后在 R 列中交替写真 (T) 和假 (F)。这种模式可以扩展到包含更多变量的真值表。一般来说,如果有 n 个不同的输入变量代表命题,则真值表有 2^n 行。

P

P

Q

R

Q R

Q⊕R

P ( Q R )

P⊕(Q⊕R)

P Q

P⊕Q

( P Q ) R

(P⊕Q)⊕R

T

T

T

T

T

T

F

F

T

T

F

F

T

T

T

T

T

T

F

F

T

T

F

F

F

F

F

F

T

T

F

F

T

T

T

T

F

F

T

T

F

F

T

T

F

F

F

F

F

F

T

T

T

T

T

T

F

F

T

T

T

T

F

F

F

F

T

T

F

F

F

F

T

T

F

F

T

T

T

T

T

T

T

T

F

F

F

F

T

T

T

T

T

T

F

F

T

T

F

F

F

F

F

F

F

F

F

F

F

F

F

F

The quickest way to fill in this truth table is to look for disagreements between the inputs of each XOR (being sure to refer to the correct columns), writing Ts in those rows, and writing Fs in all other rows. Columns 5 and 7 agree in all eight rows, so the associative law for is proved. The other associative laws and distributive laws are proved by similar eight-row truth tables.

填写此真值表的最快方法是找出每个异或运算的输入之间的不一致之处(务必参考正确的列),在这些行中填写真(T),在所有其他行中填写假(F)。第 5 列和第 7 列在所有八行中都一致,因此 ⊕ 的结合律得证。其他结合律和分配律可以通过类似的八行真值表来证明。

What IF Means

“如果”的含义

We are now ready to define the precise logical meaning of the word IF. Be warned that the usage of IF in everyday conversational English often does not match the definition we are about to give. Nevertheless, this definition is universally used in logic, mathematics, and all technical communication. When discussing the statement P Q , which is read “if P then Q ,” we call P the hypothesis and Q the conclusion of the IF-statement.

现在我们准备定义“如果”一词的精确逻辑含义。需要注意的是,日常英语口语中“如果”的用法通常与我们即将给出的定义不符。然而,这个定义在逻辑学、数学和所有技术交流中都被普遍使用。在讨论命题 P ⇒ Q(读作“如果 P 则 Q”)时,我们称 P 为假设,Q 为结论。

1.17. Definition of IF. For any propositions P and Q , the truth value of P Q (“if P then Q ”) is determined by the following table.

1.17. IF 的定义。对于任意命题 P 和 Q,P ⇒ Q(“如果 P 则 Q”)的真值由下表确定。

P

P

Q

P Q

P ⇒ Q

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

T

T

T

F

F

F

F

T

T

Remember this table by noting that P Q is false only when the hypothesis P is true and the conclusion Q is false; in all other situations , the IF-statement is true. In particular, when P is false, the IF-statement P Q is automatically true, whether Q is true or false; this fact often confuses beginners.

记住这张表格,注意只有当假设 P 为真且结论 Q 为假时,P ⇒ Q 才为假;在所有其他情况下,该 IF 语句都为真。尤其需要注意的是,当 P 为假时,无论 Q 为真还是假,IF 语句 P ⇒ Q 都自动为真;这一点常常会让初学者感到困惑。

1.18. Example. True or false?

1.18. 示例。对还是错?

(a) If 1 + 1 = 3, then 1 + 1 = 5.

(a)如果 1 + 1 = 3,那么 1 + 1 = 5。

(b) If 0 ≠ 0, then 0 = 0.

(b)如果 0 ≠ 0,则 0 = 0。

(c) If 3 > 2, then London is the capital of England.

(c)如果 3 > 2,那么伦敦是英格兰的首都。

(d) If 9 is odd, then 9 is even.

(d)如果 9 是奇数,那么 9 就是偶数。

Solution. Statement (a) has the form P Q , where P is the false statement “1 + 1 = 3,” and Q is the false statement “1 + 1 = 5.” By definition of IF, statement (a) is true , no matter how unintuitive this statement may sound. Statement (b) has the form (~ Q ) ⇒ Q , where Q is the true proposition “0 = 0.” Since the hypothesis ~ Q is false, we can see that (b) is true without even reading the statement following THEN. Statement (c) is true (according to row 1 of the truth table), even though the two parts of the statement have no causal relation to each other. Finally, statement (d) is false because “9 is odd” is true, but “9 is even” is false.

解答。命题 (a) 的形式为 P ⇒ Q,其中 P 是假命题“1 + 1 = 3”,Q 是假命题“1 + 1 = 5”。根据“如果”的定义,命题 (a) 为真,无论它听起来多么反直觉。命题 (b) 的形式为 (~Q) ⇒ Q,其中 Q 是真命题“0 = 0”。由于假设 ~Q 为假,我们甚至无需阅读“那么”后面的语句即可看出 (b) 为真。命题 (c) 为真(根据真值表第一行),即使该命题的两个部分之间没有因果关系。最后,命题 (d) 为假,因为“9 是奇数”为真,而“9 是偶数”为假。

These examples show that we must not rely upon an intuitive impression of the content of an IF-statement to determine its truth value; we must instead combine the truth values of the inputs according to the defining truth table. As shown in (c), there need not be any cause-and-effect relationship between the hypothesis and conclusion of an IF-statement.

这些例子表明,我们不能仅凭对 IF 语句内容的直觉来判断其真值;而必须根据定义的真值表,将各个输入的真值组合起来。如图 (c) 所示,IF 语句的假设和结论之间不一定存在因果关系。

It is fair to ask why IF is defined in such an apparently unnatural way. Most people agree that “if true then true” should be considered true, whereas “if true then false” should be considered false. Why, though, is “if false then anything” defined to be true? One possible answer draws an analogy between IF-statements, as they are used in logic, and IF-statements appearing in legal contracts or judicial proceedings. Suppose a painting company issues a contract to a homeowner containing the statement: “if you pay us $1000 by May 1, then we will paint your house by June 1.” Under what conditions would we say that the company has violated this contract? For example, suppose the company does not paint the house because it was not paid on time. Both clauses of the IF-statement are false, yet no one would say that the company violated the contract in this situation. Similarly, supposing that the money is paid on May 3 and the house is painted on time, the company has not violated the contract even though the first part of the IF-statement is false.

我们有理由质疑,为什么“如果”的定义如此不合常理。大多数人认同“如果为真,则为真”应被视为真,“如果为真,则为假”应被视为假。然而,为什么“如果为假,则任何事”被定义为真呢?一种可能的解释是,将逻辑中使用的“如果”语句与法律合同或司法程序中出现的“如果”语句进行类比。假设一家油漆公司与房主签订了一份合同,其中包含这样一条条款:“如果您在5月1日前支付我们1000美元,我们将在6月1日前粉刷您的房屋。”在什么情况下我们会认为该公司违反了合同?例如,假设该公司因为没有按时收到款项而没有粉刷房屋。在这种情况下,“如果”语句的两个条件都是假的,但没有人会认为该公司违反了合同。同样地,假设款项在 5 月 3 日支付,房屋也按时粉刷完毕,即使 IF 语句的第一部分是错误的,该公司也没有违反合同。

We can give another, more mathematical motivation for the definition of IF by considering the following statement: “for all real numbers x , if x > 2 then x 2 > 4.” (This statement uses variables and quantifiers, which are discussed later.) Most people would judge this statement to be true, on intuitive grounds, and this judgment is correct. But once you agree this statement should be true, you are forced to agree that it should be true after replacing each x in the statement by any specific real number. In particular, we see that each of these statements must be true:

我们可以通过考虑以下语句,为“如果”的定义提供另一种更具数学性的解释:“对于所有实数 x,如果 x > 2,则 x > 4。”(该语句使用了变量和量词,我们将在后面讨论。)大多数人会凭直觉判断该语句为真,而这种判断是正确的。但是,一旦你认同该语句为真,你就必然会认同,将语句中的每个 x 替换为任何特定的实数后,该语句仍然为真。具体来说,我们可以看到以下每个语句都必然为真:

“if 3 > 2 then 3 2 > 4” (illustrating row 1 of the truth table);

•“如果 3 > 2,则 3 2 > 4”(说明真值表的第 1 行);

“if −4 > 2 then (− 4) 2 > 4” (illustrating row 3 of the truth table);

•“如果 −4 > 2 则 (− 4) 2 > 4”(说明真值表的第 3 行);

“if 0 > 2 then 0 2 > 4” (illustrating row 4 of the truth table).

•“如果 0 > 2,则 0 2 > 4”(说明真值表的第 4 行)。

Thus the whole truth table for IF is forced upon us, granting that row 2 of the truth table ought to be false.

因此,IF 的整个真值表被强加给我们,前提是真值表的第 2 行应该是假的。

In mathematics, a statement “if P then Q ” is not making any claims about the truth of the conclusion Q in the event that the hypothesis P does not hold; so in this event, we should not declare the IF-statement to be false. But since propositions must have some truth value, we are forced to call the IF-statement true in such situations.

在数学中,“如果 P 则 Q”这样的陈述并没有对假设 P 不成立时结论 Q 的真假做出任何断言;因此,在这种情况下,我们不应该断言这个“如果”陈述为假。但由于命题必须具有真值,我们不得不在这种情况下称这个“如果”陈述为真。

Converse and Contrapositive

逆命题和逆否命题

Given an IF-statement P Q , we can form two related IF-statements called the converse and contrapositive of the original statement.

给定一个 IF 语句 P ⇒ Q,我们可以形成两个相关的 IF 语句,分别称为原语句的逆命题和逆否命题。

1.19. Definition: Converse and Contrapositive. For any propositional forms P and Q : the converse of P Q is Q P ; the contrapositive of P Q is ( Q ) ( P ) .

1.19. 定义:逆命题和逆否命题。对于任意命题形式 P 和 Q:P⇒Q 的逆命题是 Q⇒P;P⇒Q 的逆否命题是 (∼Q)⇒(∼P)。

1.20. Example. The converse of “if 9 is negative, then 9 is odd” is “if 9 is odd, then 9 is negative.” Observe that the first statement is true, whereas the converse is false. The contrapositive of “if 9 is negative, then 9 is odd” is “if 9 is not odd, then 9 is not negative,” which is true.

1.20. 例。“如果9是负数,那么9是奇数”的逆命题是“如果9是奇数,那么9是负数”。请注意,第一个命题是正确的,而它的逆命题是错误的。“如果9是负数,那么9是奇数”的逆否命题是“如果9不是奇数,那么9不是负数”,这个命题是正确的。

The converse of “if 2 = 3, then 4 = 9” is “if 4 = 9, then 2 = 3.” Both the original statement and its converse are true. The contrapositive of “if 2 = 3, then 4 = 9” is “if 4 ≠ 9, then 2 ≠ 3,” which is also true.

“如果 2 = 3,那么 4 = 9”的逆命题是“如果 4 = 9,那么 2 = 3”。原命题及其逆命题都成立。“如果 2 = 3,那么 4 = 9”的逆否命题是“如果 4 ≠ 9,那么 2 ≠ 3”,这也成立。

In both examples, the contrapositive had the same truth value as the original statement, but this was not always the case for the converse. The next theorem explains what happens in general.

在这两个例子中,逆否命题的真值与原命题相同,但逆命题并非总是如此。下一个定理解释了一般情况下会发生什么。

1.21. Theorem on IF. Let P and Q be distinct propositional variables.

1.21. 关于 IF 的定理。设 P 和 Q 是不同的命题变量。

(a) Non-equivalence of Converse: P Q Q P .

(a)逆命题不等价:P⇒Q≢Q⇒P。

(b) Equivalence of Contrapositive: P Q ( Q ) ( P ) .

(b)逆否命题的等价性:P⇒Q≡(∼Q)⇒(∼P)。

(c) Elimination of IF: P Q ( P ) Q .

(c)消除 IF:P⇒Q≡(∼P)∨Q。

(d) Denial of IF: ( P Q ) P ( Q ) .

(d)否定 IF:∼(P⇒Q)≡P∧(∼Q)。

Part (a) says that a general IF-statement is not logically equivalent to its converse. Part (b) says that a general IF-statement is logically equivalent to its contrapositive. Part (c) says that IF-statements are essentially special kinds of OR-statements in which the first input to the OR has been negated. We can use (c) to convert IF-statements to OR-statements and vice versa. Part (d) shows that the negation of an IF-statement is logically equivalent to an AND-statement. Parts (b), (c), and (d) hold for any propositional forms P and Q , not just individual letters. For instance, ( A B ) ( C D ) ( ( A B ) ) ( C D ) follows by replacing P by A B and replacing Q by C D in part (c). Memorize all parts of this theorem , as they will be used constantly throughout our study of logic and proofs.

(a) 部分指出,一般的 IF 语句与其逆命题逻辑不等价。(b) 部分指出,一般的 IF 语句与其逆否命题逻辑等价。(c) 部分指出,IF 语句本质上是特殊的 OR 语句,其中 OR 语句的第一个输入被取反。我们可以利用 (c) 部分将 IF 语句转换为 OR 语句,反之亦然。(d) 部分表明,IF 语句的否定逻辑等价于 AND 语句。(b)、(c) 和 (d) 部分适用于任何命题形式 P 和 Q,而不仅仅是单个字母。例如,(A∨B)⇒(C⊕D)≡(∼(A∨B))∨(C⊕D) 可以通过将 (c) 部分中的 P 替换为 A ∨ B,将 Q 替换为 C⊕D 得到。记住这个定理的所有部分,因为它们将在我们学习逻辑和证明的过程中不断用到。

To prove the theorem, consider the following truth table:

为了证明该定理,请考虑以下真值表:

P

P

Q

P Q

P ⇒ Q

Q P

问 ⇒ 答

~ Q

~Q

~ P

~P

(~ Q ) ⇒ (~ P )

(~Q)⇒ (~P)

(~ P )∨ Q

(~P)∨Q

T

T

T

T

T

T

T

T

F

F

F

F

T

T

T

T

T

T

F

F

F

F

T

T

T

T

F

F

F

F

F

F

F

F

T

T

T

T

F

F

F

F

T

T

T

T

T

T

F

F

F

F

T

T

T

T

T

T

T

T

T

T

T

T

Column 3 does not match column 4 in rows 2 and 3, proving (a). Column 3 matches column 7 in all rows, proving (b). Column 3 also matches column 8 in all rows, proving (c). We can prove (d) by adding a few more columns to this truth table. Alternatively, (d) follows from (c) and the Negation Laws, since

第 2 行和第 3 行的第 3 列与第 4 列不匹配,证明 (a)。第 3 列与所有行的第 7 列匹配,证明 (b)。第 3 列也与所有行的第 8 列匹配,证明 (c)。我们可以通过向真值表中添加几列来证明 (d)。或者,(d) 可由 (c) 和否定定律得出,因为

( P P Q ) ( ( P P ) Q ) ( ( P P ) ) ( Q ) P P ( Q ) .

(1.3)

Five Remarks on Logical Equivalence (Optional)

关于逻辑等价性的五点说明(可选)

(1) The Theorem on Logical Equivalence shows that many of the laws of logic have an algebraic character analogous to algebraic laws for arithmetical operations on real numbers. To make this analogy more explicit, we introduce the convention of representing true (T) by 1 and false (F) by 0. The symbols 0 and 1 are called bits . In this notation, the defining truth tables for the six logical operators (including IFF, discussed in the next section) look like this:

(1)逻辑等价定理表明,许多逻辑定律具有类似于实数运算代数定律的代数特性。为了更明确地表达这种类比,我们引入用 1 表示真 (T)、用 0 表示假 (F) 的约定。符号 0 和 1 被称为比特。在这种表示法下,六个逻辑运算符(包括下一节讨论的 IFF)的定义真值表如下所示:

P

P

Q

~ P

~P

P Q

P∧Q

P Q

P ∨ Q

P Q

P⊕Q

P Q

P ⇒ Q

P Q

P ⇔ Q

0

0

1

0

0

0

1

1

0

1

1

0

1

1

1

0

1

0

0

0

1

1

0

0

1

1

0

1

1

0

1

1

If we think of P and Q as bits (the actual integers 0 and 1, rather than propositions), we see that ~ P = 1 − P ; P Q = P Q = min ( P , Q ) , the product or minimum of P and Q ; P Q = max ( P , Q ), the maximum of P and Q ; and P Q is the mod-2 sum of P and Q . (We discuss modular addition in detail later.) We can also think of , ⇒, and ⇔ as relations comparing their two inputs. In this interpretation, P Q is true precisely when P Q , so models non-equality of bits; P Q is true precisely when P = Q , so ⇔ models equality of bits; and P Q is true precisely when P Q , so ⇒ models ordering of bits. Now the various identities in the Theorem on Logical Equivalences translate into identities for these algebraic operations on bits. For instance, supposing we already know that x ( yz ) = ( xy ) z for all integers x , y , z , we can deduce the associativity law P ( Q R ) ( P Q ) R for propositional forms.

如果我们把 P 和 Q 看作比特(实际的整数 0 和 1,而不是命题),那么 ~P = 1 − P;P∧Q=P⋅Q=min(P,Q),即 P 和 Q 的乘积或最小值;P ∨ Q = max(P, Q),即 P 和 Q 的最大值;P⊕Q 是 P 和 Q 的模 2 和。(我们稍后会详细讨论模加法。)我们还可以将 ⊕、⇒ 和 ⇔ 看作是比较两个输入的关系。在这种解释下,P⊕Q 为真当且仅当 P ≠ Q,因此 ⊕ 表示比特不相等;P ⇔ Q 为真当且仅当 P = Q,因此 ⇔ 表示比特相等;P ⇒ Q 为真当且仅当 P ≤ Q,因此 ⇒ 表示比特的顺序。现在,逻辑等价定理中的各种恒等式可以转化为这些位代数运算的恒等式。例如,假设我们已经知道对于所有整数 x、y、z,都有 x(yz) = (xy)z,那么我们可以推导出命题形式的结合律 P∧(Q∧R)≡(P∧Q)∧R。

(2) Logical equivalence of propositional forms (denoted ≡) is similar to, but not the same as, equality of forms (denoted =). Thinking of propositional forms A and B as strings of symbols, A = B means that A has exactly the same symbols as B , in the same order. For example, P Q is not equal to Q P since the first symbols do not match. Yet, as we know, P Q Q P since the two truth tables agree. For most purposes of logic, two logically equivalent forms are interchangeable with each other (since their truth values must agree), even though such forms are often not equal as strings.

(2) 命题形式的逻辑等价性(记为≡)类似于形式的相等性(记为=),但并不相同。将命题形式A和B视为符号串,A=B表示A和B包含完全相同的符号,且顺序也相同。例如,P∨Q不等于Q∨P,因为前几个符号不匹配。然而,我们知道P∨Q≡Q∨P,因为这两个命题的真值表一致。在大多数逻辑应用中,两个逻辑等价的形式可以互换使用(因为它们的真值必须一致),即使它们作为字符串时通常并不相等。

(3) Although logical equivalence of propositional forms (denoted ≡) is not the same concept as logical equality (denoted =), these two concepts share many common properties. For example, given any objects x , y , z whatsoever, the following facts are true about equality:

(3)虽然命题形式的逻辑等价(记为≡)与逻辑相等(记为=)并非同一概念,但二者有很多共同的性质。例如,对于任意对象x、y、z,关于相等性,以下事实成立:

x x = x x ; if 如果 x x = y then 然后 y = x x ; if 如果 x x = y and y = z z then 然后 x x = z z .

These facts are called reflexivity , symmetry , and transitivity of equality. Three analogous facts also hold for logical equivalence: given any propositional forms A , B , C ,

这些事实被称为等式的自反性、对称性和传递性。逻辑等价性也具有三个类似的性质:给定任意命题形式 A、B、C,

A 一个 A 一个 ; if 如果 A 一个 B B then 然后 B B A 一个 ; if 如果 A 一个 B B and B B C C then 然后 A 一个 C C .

(1.4)

These facts follow from the corresponding facts for equality, applied to the rows of the truth tables. For instance, if the output in every row of A ’s truth table equals the corresponding output of B ’s truth table, then the output in every row of B ’s truth table equals the corresponding output of A ’s truth table, which is why the second fact is true. The text implicitly uses the reflexivity, symmetry, and transitivity of logical equivalence in many places. For instance, the derivation (1.3) has the form A B C D , which is really an abbreviation for “ A B and B C and C D .” Using transitivity once, we see that A C . Using transitivity again, we see that A D , which is the required conclusion. Similarly, symmetry of ≡ is needed whenever we want to use a known logical equivalence in reverse, e.g., to replace ( P Q ) ( P R ) by P ( Q R ) by invoking the Distributive Law. We will have much more to say about the properties of reflexivity, symmetry, and transitivity in Chapter 6 , when we study equivalence relations.

这些事实源于等式的相应事实,并应用于真值表的每一行。例如,如果A真值表每一行的输出都等于B真值表的对应输出,那么B真值表每一行的输出也都等于A真值表的对应输出,这就是第二个事实为真的原因。文中多处隐含地运用了逻辑等价的自反性、对称性和传递性。例如,推导式(1.3)的形式为A≡B≡C≡D,实际上是“A≡B且B≡C且C≡D”的缩写。运用一次传递性,我们得出A≡C。再次运用传递性,我们得出A≡D,这正是所需的结论。类似地,当我们想要反向使用已知的逻辑等价关系时,例如,通过运用分配律将 (P∧Q)∨(P∧R) 替换为 P∧(Q∨R),就需要用到 ≡ 的对称性。关于自反性、对称性和传递性的性质,我们将在第六章学习等价关系时进行更详细的讨论。

(4) The logical equivalences in this section’s theorems apply to all propositional forms P , Q , R , which means that we can replace each letter P , Q , R (wherever it appears in an identity) by a longer propositional form. For instance, replacing P by A B and Q by B C in the second negation law, we see that ( ( A B ) ( B C ) ) ( ( A B ) ) ( ( B C ) ) . In contrast, consider the non-logical equivalence asserted in part (a) of the Theorem on IF. In that theorem, P and Q are distinct propositional variables (as opposed to general propositional forms), and we have ( P Q ) ( Q P ) as shown in the proof. But for some choices of the form P and the form Q , the logical equivalence will hold; for instance, suppose P and Q are both R . It can be checked that when P and Q are logically equivalent forms, ( P Q ) ≡ ( Q P ), because in this case we must always be in row 1 or row 4 of the truth table shown in the proof. (In fact, using terminology from the next section, P Q and Q P are both tautologies in this situation.) For example, (( A B ) ⇒ ( B A )) is logically equivalent to (( B A ) ⇒ ( A B )). On the other hand, when P and Q are not logically equivalent, there must exist some assignment of truth values to the propositional variables appearing in P and Q that cause P and Q themselves to have opposite truth values. In this row of the truth table, P Q and Q P disagree, so that these forms are not logically equivalent. For instance, ( ( A B ) ( A B ) ) and ( ( A B ) ( A B ) ) are not logically equivalent, as we see by considering the row of the truth table where A and B are both true.

(4) 本节定理中的逻辑等价性适用于所有命题形式 P、Q、R,这意味着我们可以将每个字母 P、Q、R(无论它出现在等价关系中的哪个位置)替换为更长的命题形式。例如,在第二否定法则中,将 P 替换为 A ⇒ B,将 Q 替换为 B⊕C,我们可以看到 ∼((A⇒B)∧(B⊕C))≡(∼(A⇒B))∨(∼(B⊕C))。相比之下,考虑关于 IF 的定理 (a) 部分中断言的非逻辑等价性。在该定理中,P 和 Q 是不同的命题变量(而不是一般的命题形式),并且我们有 (P⇒Q)≢(Q⇒P),如证明所示。但是对于某些形式的 P 和形式 Q 的选择,逻辑等价性仍然成立;例如,假设 P 和 Q 都是 R。可以验证,当 P 和 Q 是逻辑等价形式时,(P ⇒ Q) ≡ (Q ⇒ P),因为在这种情况下,我们必然总是位于证明中所示真值表的第 1 行或第 4 行。(事实上,使用下一节的术语,在这种情况下,P ⇒ Q 和 Q ⇒ P 都是重言式。)例如,((A ∨ B) ⇒ (B ∨ A)) 逻辑等价于 ((B ∨ A) ⇒ (A ∨ B))。另一方面,当 P 和 Q 不逻辑等价时,必然存在某种赋予 P 和 Q 中出现的命题变量的真值,使得 P 和 Q 本身具有相反的真值。在真值表的这一行中,P ⇒ Q 和 Q ⇒ P 的真值不一致,因此这些形式不逻辑等价。例如,((A∨B)⇒(A⊕B)) 和 ((A⊕B)⇒(A∨B)) 在逻辑上并不等价,我们可以通过考虑真值表中 A 和 B 都为真的那一行来看出这一点。

(5) In algebra, if we know two variables represent the same number, we can replace one variable by the other in longer expressions. We can express this fact symbolically by substitution rules such as the following: for all numbers x , y , z , if y = z then x + y = x + z and xy = xz . There are analogous substitution rules for logically equivalent propositional forms. For instance, given any propositional forms A , B , and C , if B C , then ( A B ) ( A C ) , ( A B ) ( A C ) , ( A B ) ( A C ) , ( B A ) ( C A ) , and so on. We freely use these substitution rules without justification in the main text, but it is possible to prove each rule by comparing truth tables. As an example, assume B C , and consider a fixed row in the truth tables for A B and A C . Suppose that, for the values of the input propositional variables appearing in this row, A is true and B is false. On one hand, A B must be false in this row. On the other hand, since B and C were assumed to be logically equivalent, C is false in this row, so A C is also false in this row. Thus, A B and A C have the same truth value (namely, false) in this row. Considering the other possible cases, we see that A B and A C have the same truth value in every row of the truth table, so they are logically equivalent. We can summarize the cases in a meta-truth table that looks like this:

(5) 在代数中,如果我们知道两个变量代表同一个数,就可以在较长的表达式中用一个变量替换另一个变量。我们可以用替换规则来符号化地表达这个事实,例如:对于任意数 x、y、z,如果 y = z,则 x + y = x + z 且 xy = xz。对于逻辑等价的命题形式,也有类似的替换规则。例如,给定任意命题形式 A、B 和 C,如果 B≡C,则 (A∨B)≡(A∨C),(A∧B)≡(A∧C),(A⇒B)≡(A⇒C),(B⇒A)≡(C⇒A),等等。我们在正文中随意使用这些替换规则而没有给出理由,但可以通过比较真值表来证明每条规则。例如,假设 B≡C,并考虑 A⇒B 和 A⇒C 真值表中的固定行。假设对于本行中出现的输入命题变量的值,A 为真,B 为假。一方面,A⇒B 在本行中必然为假。另一方面,由于假设 B 和 C 逻辑等价,C 在本行中为假,因此 A⇒C 在本行中也为假。所以,A⇒B 和 A⇒C 在本行中具有相同的真值(即假)。考虑其他可能的情况,我们发现 A⇒B 和 A⇒C 在真值表的每一行中都具有相同的真值,因此它们逻辑等价。我们可以将这些情况总结在一个元真值表中,如下所示:

A

一个

B

B

C

C

A B

A⇒B

A C

A⇒C

T

T

T

T

T

T

T

T

T

T

T

T

F

F

F

F

F

F

F

F

F

F

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

T

T

T

We call this a meta-truth table because the underlying propositional variables appearing in A , B , and C are not explicitly listed. (For example, these forms might involve five propositional variables P , Q , R , S , and T , so that the actual truth table would have 32 rows.) However, to prove the logical equivalence of A B and A C , we only need to know the truth values of the propositional forms A , B , and C . Row 2 of the table is an abbreviation for the discussion preceding the table, and the other rows cover the remaining cases. The key point is that the table has four rows, not eight, because we have assumed that B and C are logically equivalent forms, hence always have the same truth value. You can construct similar meta-truth tables to prove the other substitution rules.

我们称之为元真值表,因为A、B和C中出现的底层命题变量并未明确列出。(例如,这些形式可能涉及五个命题变量P、Q、R、S和T,因此实际的真值表将有32行。)然而,为了证明A⇒B和A⇒C的逻辑等价性,我们只需要知道命题形式A、B和C的真值即可。表的第2行是对前面讨论的简要概括,其余行涵盖了剩余的情况。关键在于该表只有四行而不是八行,因为我们假设B和C是逻辑等价的形式,因此它们的真值始终相同。您可以构建类似的元真值表来证明其他替换规则。

Section Summary

章节概要

1. The logical operators satisfy the commutative, associative, distributive, idempotent, and absorption laws listed in the Theorem on Logical Equivalence. They also obey these negation rules:

1. 逻辑运算符满足逻辑等价定理中列出的交换律、结合律、分配律、幂等律和吸收律。它们也遵循以下否定规则:

( P P Q ) ( P P ) ( Q ) , ( P P Q ) ( P P ) ( Q ) ,
( P P ) P P , ( P P Q ) P P ( Q ) .

We check all these logical equivalences by verifying that the truth tables for each side agree in every row.

我们通过验证每一方的真值表在每一行都一致来检验所有这些逻辑等价性。

2. P Q is false when P is true and Q is false; P Q is true in all other situations.

2.当 P 为真且 Q 为假时,P ⇒ Q 为假;在所有其他情况下,P ⇒ Q 为真。

3. The converse of P Q is Q P , which is not logically equivalent to P Q . The contrapositive of P Q is (~ Q ) ⇒ (~ P ), which is logically equivalent to P Q . We have P Q ≡ (~ P )∨ Q , and hence ( P Q ) P ( Q ) .

3. P ⇒ Q 的逆命题是 Q ⇒ P,它与 P ⇒ Q 逻辑不等价。P ⇒ Q 的逆否命题是 (~Q) ⇒ (~P),它与 P ⇒ Q 逻辑等价。我们有 P ⇒ Q ≡ (~P)∨Q,因此 ∼(P⇒Q)≡P∧(∼Q)。

Exercises

练习

1. Make a truth table for each propositional form.

1. 为每个命题形式制作真值表。

(a) ( P ∨(~ Q )) ⇒ P (b) P ( Q ( P ) ) (c) ( P Q ) ( P Q ) .

(a) (P∨(~Q)) ⇒ P (b) P⇒(Q∧(∼P)) (c) (P⊕Q)⇒(P∨Q).

2. Make a truth table for each propositional form.

2.为每个命题形式制作真值表。

(a) ( P Q ) ( Q R ) (b) ( P Q ) ⇒ ( Q R ) (c) ( ( P Q ) ) ( ( R P ) ) .

(a)(P⇒Q)∧(Q⇒R) (b)(P ⇒ Q) ⇒ (Q∨R) (c)(∼(P⊕Q))⇒(∼(R∧P))。

3. Prove the negation laws ~(~ P ) ≡ P and ( P Q ) ( P ) ( Q ) using truth tables.

3.利用真值表证明否定定律~(~P)≡P和∼(P∧Q)≡(∼P)∨(∼Q)。

4. Use truth tables to prove the commutative laws for , ∨, and in the Theorem on Logical Equivalences.

4.利用真值表证明逻辑等价定理中∧、∨和⊕的交换律。

5. Prove the associative laws for and ∨.

5.证明∧和∨的结合律。

6. Prove the absorption laws in part (e) of the Theorem on Logical Equivalences.

6.证明逻辑等价定理(e)部分中的吸收定律。

7. Prove the distributive laws in part (c) of the Theorem on Logical Equivalences.

7.证明逻辑等价定理(c)部分中的分配律。

8. Prove or disprove: ( P ) ( Q P ) ( Q ) .

8.证明或反驳:(∼P)⊕(Q⊕P)≡(∼Q)。

9. True or false?

9.对还是错?

(a) If Mars is a planet, then 2 + 2 = 4.

(a)如果火星是行星,那么 2 + 2 = 4。

(b) If 2 · 2 ≠ 4, then π is negative.

(b)如果 2 · 2 ≠ 4,则 π 为负数。

(d) If Paris is a city, then Europe is an ocean.

(d)如果巴黎是一座城市,那么欧洲就是一片海洋。

(d) If 0 = 0, then 5 is even.

(d)如果 0 = 0,那么 5 是偶数。

(e) If some trees have roots, then all crows are white.

(e)如果有些树有根,那么所有的乌鸦都是白色的。

(f) If 3 + 3 ≠ 6, then 0 1 x 4 + 1 d x = π 2 / 4 .

(f)如果 3 + 3 ≠ 6,则 ∫0∞1x4+1dx=π2/4。

(g) If 1 + 1 = 3, then 1 + 1 = 2, but the converse is false.

(g)如果 1 + 1 = 3,则 1 + 1 = 2,但反过来不成立。

10. Assume P is a true proposition, Q is a false proposition, and R is an arbitrary proposition. Say as much as you can about the truth or falsehood of each proposition below.

10. 假设 P 为真命题,Q 为假命题,R 为任意命题。请尽可能详细地说明以下每个命题的真假。

(a) P R (b) Q R (c) R P (d) R Q (e) R R (f) R ⇒ (~ R ).

(a)P ⇒ R (b)Q ⇒ R (c)R ⇒ P (d)R ⇒ Q (e)R ⇒ R (f)R ⇒ (~R)。

11. Let x 0 be a fixed integer, and let P be the statement:

11.设 x 0 为一个固定的整数,P 为以下语句:

“if x 0 is even and x 0 is prime, then x 0 = 2.”

“如果 x 0 是偶数且 x 0 是质数,则 x 0 = 2。”

(a)Write the converse of P .

(a)写出 P 的逆问题。

(b)Write the contrapositive of P .

(b)写出P的逆否命题。

(c)Write the converse of the contrapositive of P .

(c)写出命题 P 的逆否命题的逆命题。

(d)Suppose x 0 is 8. Is P true or false? What about the converse of P ?

(d)假设x 0 为8。P为真还是假?P的逆命题呢?

12. The inverse of P Q is defined to be (~ P ) ⇒ (~ Q ). Decide whether the inverse of P Q is logically equivalent to: (a) P Q ; (b) the converse of P Q ; (c) the contrapositive of P Q .

12. P ⇒ Q 的逆命题定义为 (~P) ⇒ (~Q)。判断 P ⇒ Q 的逆命题是否逻辑等价于:(a) P ⇒ Q;(b) P ⇒ Q 的逆命题;(c) P ⇒ Q 的逆否命题。

13. Write the converse and the contrapositive of each statement.

13.写出下列各命题的逆命题和逆否命题。

(a)If f is continuous or monotonic, then f is integrable.

(a)如果 f 是连续的或者单调的,那么 f 是可积的。

(b)If g is not continuous, then g is not differentiable.

(b)如果 g 不连续,则 g 不可微。

(c)If p is prime, then if p is even, then p = 2.

(c)如果 p 是质数,那么如果 p 是偶数,那么 p = 2。

14. Use known logical equivalences to show that ( P R ) ( Q ) is logically equivalent to ( P ( Q R ) ) without making a truth table.

14.利用已知的逻辑等价关系证明 (P∧R)⇒(∼Q) 与 ∼(P∧(Q∧R)) 逻辑等价,而不列出真值表。

15. (a) Use known logical equivalences to show that P ⇒ ( Q R ) is logically equivalent to ( P Q ) R without making a truth table. (b) Is ( P Q ) ⇒ R logically equivalent to P ⇒ ( Q R )? Explain.

15.(a) 利用已知的逻辑等价关系,证明 P ⇒ (Q ⇒ R) 与 (P∧Q) ⇒ R 逻辑等价,无需列出真值表。(b) (P ⇒ Q) ⇒ R 与 P ⇒ (Q ⇒ R) 逻辑等价吗?请解释。

16. Write the converse of ( P Q ) ⇒ R and the contrapositive of ( P Q ) ⇒ R .

16.写出 (P ⇒ Q) ⇒ R 的逆命题和 (P ⇒ Q) ⇒ R 的逆否命题。

17. Use the Theorem on IF to replace each propositional form with an equivalent form that does not use ⇒.

17.利用 IF 定理,将每个命题形式替换为不使用 ⇒ 的等价形式。

( P Q ) ( R S ) (b) P ⇒ ((~ Q ) ⇒ P ) (c) [ P ⇒ ( Q R )] ⇒ [( P Q ) ⇒ ( P R )]

(P∧Q)⇒(R∨S) (b) P ⇒ ((~Q) ⇒ P) (c) [P ⇒ (Q ⇒ R)] ⇒ [(P ⇒ Q) ⇒ (P ⇒ R)]

18. Which proposed distributive laws are correct? Prove your answers.

18.哪些分配律是正确的?请证明你的答案。

(a) P ( Q R ) ( P Q ) ( P R ) .

(a)P∨(Q⊕R)≡(P∨Q)⊕(P∨R)。

(b) P ( Q R ) ( P Q ) ( P R ) .

(b)P⇒(Q∧R)≡(P⇒Q)∧(P⇒R)。

(c) P ⇒ ( Q R ) ≡ ( P Q )∨( P R ).

(c)P ⇒ (Q∨R) ≡ (P ⇒ Q)∨(P ⇒ R)。

(d) P ( Q R ) ( P Q ) ( P R ) .

(d)P⇒(Q⊕R)≡(P⇒Q)⊕(P⇒R)。

(e) P ( Q R ) ( P Q ) ( P R ) .

(e)P∧(Q⇒R)=(P∧Q)⇒(P∧R)。

(d) P ∨( Q R ) ≡ ( P Q ) ⇒ ( P R ).

(d)P∨(Q ⇒ R) ≡ (P ∨ Q) ⇒ (P∨R)。

(g) P ( Q R ) ( P Q ) ( P R ) .

(g)P⊕(Q⇒R)≡(P⊕Q)⇒(P⊕R)。

19. Prove A ( B C D E ) ( A B ) ( A C ) ( A D ) ( A E ) without using truth tables.

19. 不使用真值表证明 A∧(B∨C∨D∨E)≡(A∧B)∨(A∧C)∨(A∧D)∨(A∧E)。

20. Use logical equivalences from this section to find four different IF-statements that are logically equivalent to (~ P )∨(~ Q )∨ R .

20.利用本节中的逻辑等价性,找出四个与 (~P)∨(~Q)∨R 逻辑等价的不同 IF 语句。

21. The FOIL Rule in algebra says that for all real numbers a , b , c , d , ( a + b ) · ( c + d ) = ( ac + ad ) + ( bc + bd ). (a) Prove this rule using laws of algebra. (b) Formulate three analogous FOIL rules for logical connectives, and prove these rules without truth tables using the Theorem on Logical Equivalences.

21. 代数中的 FOIL 法则指出,对于任意实数 a、b、c、d,(a + b) · (c + d) = (ac + ad) + (bc + bd)。(a) 用代数定律证明该法则。(b) 为逻辑连接词构造三个类似的 FOIL 法则,并利用逻辑等价定理在不使用真值表的情况下证明这些法则。

22. Use known logical equivalences (not truth tables) to prove: ( A B ) ( C D ) ( ( A C ) ) ( A D ) ( C B ) ( B D ) .

22.使用已知的逻辑等价关系(而不是真值表)证明:(A⇒B)∧(C⇒D)≡(∼(A∨C))∨(A⇒D)∨(C⇒B)∨(B∧D)。

23. True or false? Explain.

23.判断题:对还是错?请解释。

(a) The contrapositive of P Q always has the same truth value as P Q .

(a)P ⇒ Q 的逆否命题总是与 P ⇒ Q 具有相同的真值。

(b) The converse of P Q always has the same truth value as the negation of P Q .

(b)P ⇒ Q 的逆命题与 P ⇒ Q 的否定命题的真值总是相同的。

(c) If P Q is false, then the converse must be true.

(c)如果 P ⇒ Q 为假,那么其逆命题必定为真。

(d) If P Q is true, then the converse must be true.

(d)如果 P ⇒ Q 为真,那么其逆命题也必定为真。

(e) If the hypothesis and conclusion of a conditional statement have the same truth value, then the conditional statement must be true.

(e)如果一个条件语句的假设和结论具有相同的真值,那么这个条件语句必然为真。

24. Give specific examples of propositional forms A and B with each property below. Explain why your examples work. (a) A B is logically equivalent to A B . (b) A B B A , but A B A B . (c) A B is logically equivalent to ( A B ) .

24. 请给出命题形式 A 和 B 的具体例子,并解释以下每个性质为何成立。(a) A∨B 逻辑等价于 A⊕B。(b) A⇒B≡B⇒A,但 A⇒B≢A∧B。(c) A∨B 逻辑等价于 ∼(A∧B)。

25. Prove that logical equivalence of propositional forms is reflexive and transitive, as stated in (1.4) .

25.证明命题形式的逻辑等价性是自反的和传递的,如(1.4)所述。

26. Prove the substitution rules for propositional forms given in Remark 5 on page 14.

26.证明第 14 页备注 5 中给出的命题形式的替换规则。

1.3 IF, IFF, Tautologies, and Contradictions

1.3 IF、IFF、重言式和矛盾式

Now that we know what IF means, it is time to talk about other English words and phrases that have the same logical meaning as an IF-statement. We also introduce the new logical connective IFF (pronounced “if and only if”). In everyday life, people often say IF when they mean IFF, so it is vital to understand the correct technical usage of these words. We close the section by introducing tautologies (propositional forms that always evaluate to true) and contradictions (propositional forms that always evaluate to false).

既然我们已经了解了“如果”的含义,接下来我们将讨论其他与“如果”语句具有相同逻辑含义的英语单词和短语。我们还会介绍新的逻辑连接词“IFF”(读作“if and only if”)。在日常生活中,人们常常用“如果”来指代“IFF”,因此理解这些词语的正确技术用法至关重要。本节最后,我们将介绍重言式(总是为真的命题形式)和矛盾式(总是为假的命题形式)。

Translations of P Q

P ⇒ Q 的翻译

The most straightforward English translation of the propositional form P Q is “if P , then Q .” However, there are many other English phrases with the same logical meaning as P Q . We begin by considering some translations of IF that often cause trouble. By definition:

命题形式 P ⇒ Q 最直接的英文翻译是“如果 P,那么 Q”。然而,还有许多其他英文短语与 P ⇒ Q 具有相同的逻辑含义。我们首先来探讨一些经常引起问题的“如果”的翻译。根据定义:

P if Q means Q P .

•PifQ 表示 Q⇒P。

P only if Q means P Q .

•P仅当Q意味着P⇒Q时。

The phrase “ P if Q ” is obtained by changing the order of the clauses in “if Q then P .” In both phrases, the proposition Q immediately follows the word IF, so Q is the hypothesis and P is the conclusion of the conditional statement. However, replacing IF by ONLY IF changes the meaning: in the sentence “ P only if Q ,” P is the hypothesis and Q is the conclusion! We explain this counterintuitive phenomenon through an example. Suppose a professor says to a student, “You will pass my class ONLY IF you take the final exam.” This statement means the same thing as, “If you do not take the final exam, then you will not pass the class.” We recognize this as the contrapositive of the statement, “If you pass the class, then you took the final exam.” The same reasoning holds in general: “ P only if Q ” really means “if Q is not true, then P is not true,” which can be encoded as (~ Q ) ⇒ (~ P ). This is the contrapositive of P Q , so this statement is logically equivalent to P Q .

“P 如果 Q”这句话是通过改变“如果 Q 那么 P”中子句的顺序得到的。在这两个短语中,命题 Q 都紧跟在“如果”之后,因此 Q 是条件语句的假设,P 是结论。然而,如果将“如果”替换为“仅当”,意思就发生了变化:在“P 仅当 Q”这句话中,P 变成了假设,Q 变成了结论!我们用一个例子来解释这种反直觉的现象。假设一位教授对学生说:“你只有参加期末考试才能通过我的课。”这句话的意思与“如果你不参加期末考试,那么你就不会通过这门课”相同。我们将其视为“如果你通过了这门课,那么你参加了期末考试”的逆否命题。同样的道理也适用于一般情况:“P 仅当 Q”实际上意味着“如果 Q 不为真,那么 P 也不为真”,这可以编码为 (~Q) ⇒ (~P)。这是 P ⇒ Q 的逆否命题,因此该命题在逻辑上等价于 P ⇒ Q。

Now consider the meaning of the words NECESSARY and SUFFICIENT. By definition:

现在来思考一下“必要”和“充分”这两个词的含义。根据定义:

P is sufficient for Q means P Q .

•P 足以满足 Q 意味着 P⇒Q。

P is necessary for Q means Q P .

•P 是 Q 所必需的,意味着 Q⇒P。

For example, “Differentiability of f is sufficient for continuity of f ” means “If f is differentiable, then f is continuous.” On the other hand, the word NECESSARY (like the word ONLY) has a hidden negating effect. The statement “Taking the final exam is necessary for passing the course” means, “If you do not take the final exam, then you will not pass the course,” which in turn has the same meaning as, “If you pass the course, then you took the final exam.” In general, saying “ P is necessary for Q ” means that if P does NOT hold, then Q does NOT hold, which can be encoded as (~ P ) ⇒ (~ Q ). This is the contrapositive of Q P , so this statement has the same logical meaning as Q P .

例如,“f 的可微性是 f 连续性的充分条件”意味着“如果 f 可微,则 f 连续”。另一方面,“必要”(如同“唯一”)一词具有隐含的否定作用。“参加期末考试是课程及格的必要条件”意味着“如果你不参加期末考试,那么你就不会通过这门课程”,这与“如果你通过了这门课程,那么你参加了期末考试”的含义相同。一般来说,“P 是 Q 的必要条件”意味着如果 P 不成立,则 Q 也不成立,这可以编码为 (~P) ⇒ (~Q)。这是 Q ⇒ P 的逆否命题,因此该语句与 Q ⇒ P 具有相同的逻辑含义。

One way to remember the correct meaning of NECESSARY is to note that “ P is necessary for Q ” is the converse of “ P is sufficient for Q .” Similarly, “ P only if Q ” is the converse of “ P if Q .” We must take care, however, because the passive voice (and similar constructions that change the order of phrases in a sentence) can reverse the positions of P and Q . For example, all of the following phrases can be used to translate P Q :

记住“NECESSARY”正确含义的一个方法是注意到“P是Q的必要条件”是“P是Q的充分条件”的逆句。类似地,“P仅当Q”是“P当Q”的逆句。但是,我们必须注意,因为被动语态(以及其他改变句子中短语顺序的类似结构)可能会颠倒P和Q的位置。例如,以下所有短语都可以用来翻译为P ⇒ Q:

P is sufficient for Q .

•P 足以满足 Q。

For Q to be true, it is sufficient that P be true.

•要使 Q 为真,P 为真即可。

A sufficient condition for Q is P .

•Q 的一个充分条件是 P。

Q is necessary for P .

•Q 是 P 的必要条件。

A necessary condition for P is Q .

•P 的必要条件是 Q。

For P to be true, it is necessary that Q be true.

•要使 P 为真,Q 必须为真。

Notice the key preposition FOR, which can help you locate the correct hypothesis and conclusion in each phrase. Yet other possible translations of P Q include:

注意关键介词 FOR,它可以帮助你在每句话中找到正确的假设和结论。P ⇒ Q 的其他可能翻译包括:

P implies Q .

•P蕴含Q。

Q is implied by P .

•Q 由 P 隐含。

When P is true, Q is true.

•当P为真时,Q也为真。

Q whenever P .

•Q 当 P 时。

Q follows from P .

•Q 由 P 推导而来。

Q is a consequence of P .

•Q 是 P 的结果。

In particular, when reading a form such as P Q aloud, you may read the symbol ⇒ as the word IMPLIES. But it is wrong to read ⇒ as the word IF, since “ P if Q ” means Q P , not P Q . It is correct, but potentially confusing, to pronounce the symbol ⇒ as ONLY IF. My personal recommendation is to read P Q as “if P , then Q .” But you must still be familiar with all of the variants above, since they occur frequently in mathematical writing.

特别是,在朗读 P ⇒ Q 这类表达式时,你可能会把符号 ⇒ 读作“暗示”(IMPLIES)。但把 ⇒ 读作“如果”(IF)是错误的,因为“P 如果 Q”的意思是 Q ⇒ P,而不是 P ⇒ Q。把符号 ⇒ 读作“仅当”(ONLY IF)是正确的,但可能会造成混淆。我个人建议把 P ⇒ Q 读作“如果 P,那么 Q”。但你仍然需要熟悉以上所有变体,因为它们在数学写作中经常出现。

While we are discussing translation issues, we should mention the logical meaning of a few other common English phrases:

在讨论翻译问题时,我们应该提及一些其他常用英语短语的逻辑含义:

P but Q ” means P Q .

•“P 但 Q”表示 P∧Q。

“Although P , Q ” means P Q .

•“虽然 P,Q”表示 P∧Q。

P , Q are both true” means P Q .

•“P、Q 都为真”表示 P∧Q。

“Neither P nor Q ” means ( P ) ( Q ) , or equivalently ~( P Q ).

•“既非 P 也非 Q”表示 (∼P)∧(∼Q),或等价地 ~(P ∨ Q)。

The words BUT and ALTHOUGH suggest some kind of contrast between the two clauses joined by these words, but this contrast is ignored in logic. For logical purposes, BUT and ALTHOUGH (and BOTH) have exactly the same meaning as AND.

“但是”和“虽然”这两个词暗示着它们所连接的两个分句之间存在某种对比,但在逻辑中,这种对比会被忽略。从逻辑角度来看,“但是”和“虽然”(以及“两者”)与“并且”的含义完全相同。

There are some other English phrases whose mathematical meaning does not seem to be completely standardized. I assert that “either P or Q ” means P Q (inclusive-OR), “ P provided that Q ” means Q P , and “ P unless Q ” means P Q , but not all mathematical writers agree with these conventions.

还有一些英语短语的数学含义似乎尚未完全标准化。我认为“either P or Q”表示 P ∨ Q(包含或),“P provided that Q”表示 Q ⇒ P,“P unless Q”表示 P ∨ Q,但并非所有数学家都认同这些约定。

What IFF Means

IFF是什么意思

Now that we understand IF and its many synonyms, it is time to define the related logical connective IFF (pronounced “if and only if” and symbolized by ⇔). The formal definition of this word appears in the truth table below.

现在我们已经了解了 IF 及其众多同义词,接下来需要定义相关的逻辑连接词 IFF(读作“如果且仅当”,符号为 ⇔)。该词的正式定义如下表所示。

1.22. Definition of IFF. For any propositions P and Q , the truth value of P Q (“ P iff Q ”) is determined by the following table.

1.22. IFF 的定义。对于任意命题 P 和 Q,P ⇔ Q(“P 当且仅当 Q”)的真值由下表确定。

P

P

Q

P Q

P ⇔ Q

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

T

F

F

F

F

F

#

F

T

T

Remember this table by noting that P Q is true exactly when the two inputs P , Q have the same truth values; P Q is false when the inputs P , Q have different truth values. Comparing the defining truth tables for IF and IFF, we see that P Q P Q .

记住这张表,注意 P ⇔ Q 为真当且仅当两个输入 P 和 Q 的真值相同;P ⇔ Q 为假当且仅当两个输入 P 和 Q 的真值不同。比较 IF 和 IFF 的定义真值表,我们发现 P⇒Q≢P⇔Q。

1.23 Example True or false?

1.23 示例:对还是错?

(a) 1 + 1 = 3 only if 1 + 1 = 2.

(a)1 + 1 = 3 当且仅当 1 + 1 = 2。

(b) 0 < 1 if 1 < 0.

(b)如果 1 < 0,则 0 < 1。

(c) 0 < 1 iff 1 < 0.

(c)0 < 1 当且仅当 1 < 0。

(d) A necessary condition for 7 to be even is that 5 be odd.

(d)7 为偶数的必要条件是 5 为奇数。

(e) For 7 to be even, it is sufficient that 5 be odd.

(e)要使 7 为偶数,只需 5 为奇数即可。

(f) 5 is negative iff −5 is positive.

(f)5 为负数当且仅当 −5 为正数。

Solution. The suggested approach is to encode each English sentence symbolically, then refer to the definitions of IF and IFF. Thus, (a) becomes “1 + 1 = 3 ⇒ 1 + 1 = 2,” which is true since F ⇒ T is true. Similarly, (b) becomes “1 < 0 ⇒ 0 < 1,” which is true. But (c) is “0 < 1 ⇔ 1 < 0,” which is false since T ⇔ F is false. (d) says that “7 is even ⇒ 5 is odd,” which is true. (e) says that “5 is odd ⇒ 7 is even,” which is false. Finally, (f) is true since both inputs to the ⇔ operator are false.

解答:建议的方法是将每个英文句子进行符号编码,然后参考 IF 和 IFF 的定义。因此,(a) 变为“1 + 1 = 3 ⇒ 1 + 1 = 2”,这是正确的,因为 F ⇒ T 为真。类似地,(b) 变为“1 < 0 ⇒ 0 < 1”,这是正确的。但是 (c) 变为“0 < 1 ⇔ 1 < 0”,这是错误的,因为 T ⇔ F 为假。(d) 表示“7 是偶数 ⇒ 5 是奇数”,这是正确的。(e) 表示“5 是奇数 ⇒ 7 是偶数”,这是错误的。最后,(f) 为真,因为 ⇔ 运算符的两个输入均为假。

1.24. Theorem on IFF. For all propositional forms P and Q :

1.24. 关于因果函数的定理。对于所有命题形式 P 和 Q:

(a) Reduction to IF: P Q ( P Q ) ( Q P ) .

(a)简化为 IF:P⇔Q≡(P⇒Q)∧(Q⇒P)。

(b) Reduction to AND and OR: P Q ( P Q ) ( ( P ) ( Q ) ) .

(b)简化为 AND 和 OR:P⇔Q≡(P∧Q)∨((∼P)∧(∼Q))。

(c) Symmetry of IFF: P Q Q P .

(c)IFF 的对称性:P ⇔ Q ≡ Q ⇔ P。

(d) Denial of IFF: ( P Q ) P Q .

(d)否定 IFF:∼(P⇔Q)≡P⊕Q。

(e) Negating Both Sides: P Q ≡ (~ P ) ⇔ (~ Q ).

(e)两边取反:P ⇔ Q ≡ (~P) ⇔ (~Q)。

Part (a) relates IFF to IF and explains why the symbol ⇔ is used for IFF. Part (b) lets us replace IFF by a form built from other logical connectives.

(a) 部分将 IFF 与 IF 联系起来,并解释了为什么 IFF 使用符号 ⇔。(b) 部分允许我们用由其他逻辑连接词构建的形式来替换 IFF。

As with previous theorems, we prove these logical equivalences with truth tables. As a sample, we prove parts (a) and (e) as shown:

与之前的定理一样,我们用真值表证明这些逻辑等价性。例如,我们证明 (a) 和 (e) 部分,如下所示:

P

P

Q

P Q

P ⇔ Q

P Q

P ⇒ Q

Q P

问 ⇒ 答

( P Q ) ( Q P )

(P⇒Q)∧(Q⇒P)

~ P

~P

~ Q

~Q

(~ P ) ⇔ (~ Q )

(~P)⇔(~Q)

T

T

T

T

T

T

T

T

T

T

T

T

F

F

F

F

T

T

T

T

F

F

F

F

F

F

T

T

F

F

F

F

T

T

F

F

F

F

T

T

F

F

T

T

F

F

F

F

T

T

F

F

F

F

F

F

F

F

T

T

T

T

T

T

T

T

T

T

T

T

T

T

Columns 3 and 6 agree, proving (a); columns 3 and 9 agree, proving (e).

第 3 列和第 6 列一致,证明了 (a);第 3 列和第 9 列一致,证明了 (e)。

1.25. Translations of IFF. We can translate P Q by several English phrases, including:

1.25. IFF 的翻译。我们可以用几个英语短语来翻译 P ⇔ Q,包括:

(a) P iff Q .

(a)P 当且仅当 Q。

(b) P if and only if Q .

(b)P 当且仅当 Q。

(c) P is necessary and sufficient for Q .

(c)P 是 Q 的必要且充分条件。

(d) P when and only when Q .

(d)P 当且仅当 Q。

(e) P precisely when Q .

(e)P 恰好当 Q 时。

(f) P is equivalent to Q .

(f)P 等价于 Q。

(g) P has the same truth value as Q .

(g)P 与 Q 具有相同的真值。

Regarding (b), note that “ P if and only if Q ” is short for “ P if Q , and P only if Q ,” which translates to ( Q P ) ( P Q ) . We know this is equivalent to P Q by part (a) of the theorem and commutativity of . A similar analysis applies to the phrases “necessary and sufficient” and “when and only when.”

关于 (b),请注意,“P 当且仅当 Q”是“P 如果 Q,且 P 仅当 Q”的简写,即 (Q⇒P)∧(P⇒Q)。根据定理的 (a) 部分以及 ∧ 的交换律,我们知道这等价于 P ⇔ Q。类似的分析也适用于“必要且充分”和“当且仅当”。

From part (e) of the Theorem on IFF, negating both sides of an IFF-statement does not deny the entire statement, but actually produces a logically equivalent statement. To deny an IFF-statement, we must replace the IFF by XOR, as stated in part (d) of the theorem. By negating each side of the equivalences in the Theorem on IFF, or by constructing more truth tables, we obtain the following equivalences for XOR.

根据因果相合定理的 (e) 部分,否定一个因果相合命题的两侧并不能否定整个命题,而是会产生一个逻辑等价的命题。要否定一个因果相合命题,我们必须用异或运算替换它,正如定理的 (d) 部分所述。通过否定因果相合定理中等价关系的每一侧,或者构造更多的真值表,我们可以得到以下异或运算的等价关系。

1.26 Theorem on XOR (Optional). For all propositional forms P , Q , R :

1.26 异或定理(可选)。对于所有命题形式 P、Q、R:

(a) Reduction to AND and OR: P Q ( P ( Q ) ) ( Q ( P ) ) .

(a)简化为 AND 和 OR:P⊕Q≡(P∧(∼Q))∨(Q∧(∼P))。

(b) Negating Both Sides: P Q ( P ) ( Q ) .

(b)两边取反:P⊕Q≡(∼P)⊕(∼Q)。

(c) Second Reduction to AND and OR: P Q ( P Q ) ( ( P Q ) ) .

(c)第二个简化为与和或:P⊕Q≡(P∨Q)∧(∼(P∧Q))。

(d) Commutativity of XOR: P Q Q P .

(d)异或运算的交换律:P⊕Q≡Q⊕P。

(e) Denial of XOR: ( P Q ) P Q .

(e)异或的否定:∼(P⊕Q)≡P⇔Q。

(f) Reduction to IFF: P Q P ( Q ) ( P ) Q .

(f)简化为 IFF:P⊕Q≡P⇔(∼Q)≡(∼P)⇔Q。

(g) Associativity of XOR: P ( Q R ) ( P Q ) R .

(g)异或运算的结合律:P⊕(Q⊕R)≡(P⊕Q)⊕R。

Part (a) shows that we can translate P Q as “exactly one of P and Q is true.” Part (c), which we proved in an earlier section, justifies the translation of P Q as “ P or Q , but not both.” Part (f) indicates how to transform XOR-statements into IFF-statements by negating one of the inputs.

(a) 部分表明我们可以将 P⊕Q 翻译为“P 和 Q 中恰好有一个为真”。(c) 部分(我们在前面的章节中已证明)证明了将 P⊕Q 翻译为“P 或 Q,但不能同时为真”的合理性。(f) 部分指出了如何通过对其中一个输入取反来将异或语句转换为倒数当量语句。

Tautologies and Contradictions

同义反复和矛盾

Certain propositional forms, called tautologies and contradictions, play a special role in logic and proofs. These are defined as follows.

某些命题形式,称为重言式和矛盾式,在逻辑和证明中扮演着特殊角色。它们的定义如下。

1.27. Definition: Tautologies and Contradictions. A propositional form A is called a tautology iff every row of the truth table for A has output true. A propositional form B is called a contradiction iff every row of the truth table for B has output false.

1.27. 定义:重言式和矛盾式。命题形式 A 被称为重言式,当且仅当 A 的真值表中每一行的输出都为真。命题形式 B 被称为矛盾式,当且仅当 B 的真值表中每一行的输出都为假。

Note that most propositional forms have a mixture of true and false outputs, so most propositional forms are neither tautologies nor contradictions .

请注意,大多数命题形式都有真假混合的输出,因此大多数命题形式既不是重言式也不是矛盾式。

1.28 Example. The propositional form R ∨(~ R ) is a tautology, whereas the form R ( R ) is a contradiction, as shown in the following truth table:

1.28 例。命题形式 R∨(~R) 是重言式,而形式 R∧(~R) 是矛盾式,如下真值表所示:

R

~ R

~R

R ∨(~ R )

R∨(~R)

R ( R )

R∧(∼R)

T

T

F

F

T

T

F

F

F

F

T

T

T

T

F

F

Intuitively, “ R or not R ” must be true, regardless of the truth value of R , whereas “ R and not R ” must be false. More generally, for any propositional form A (not just the individual variable R ), A ( A ) is a tautology, whereas A ( A ) is a contradiction.

直观地说,“R 或非 R”必然为真,无论 R 的真值如何;而“R 且非 R”必然为假。更一般地,对于任何命题形式 A(而不仅仅是单个变量 R),A∨(∼A) 是重言式,而 A∧(∼A) 是矛盾式。

1.29. Example. Classify each propositional form as a tautology, a contradiction, or neither. (a) P ⇒ ( Q P ) (b) ( P Q ) ⇒ P (c) ( P Q ) ( P Q )

1.29. 示例。将下列命题形式分类为重言式、矛盾式或两者都不是。(a) P ⇒ (Q ⇒ P) (b) (P ⇒ Q) ⇒ P (c) (P⊕Q)∧(P⇔Q)

Solution. In each case, we make a truth table for the given form and see if every output is true (for a tautology) or false (for a contradiction). The truth table for (a) and (b) is shown here:

解法。对于每种情况,我们为给定的形式构建真值表,并检查每个输出是真(对于重言式)还是假(对于矛盾式)。(a) 和 (b) 的真值表如下所示:

P

P

Q

Q P

问 ⇒ 答

P ⇒ ( Q P )

P ⇒ (Q ⇒ P)

P Q

P ⇒ Q

( P Q ) ⇒ P

(P ⇒ Q) ⇒ P

T

T

T

T

T

T

T

T

T

T

T

T

T

T

F

F

T

T

T

T

F

F

T

T

F

F

T

T

F

F

T

T

T

T

F

F

F

F

F

F

T

T

T

T

T

T

F

F

Column 4 is true in every row, so (a) is a tautology. Column 6 has a mixture of Ts and Fs, so (b) is neither a tautology nor a contradiction. For (c), we make the following truth table:

第 4 列在每一行都为真,因此 (a) 是重言式。第 6 列真假参半,因此 (b) 既不是重言式也不是矛盾式。对于 (c),我们构造如下真值表:

P

P

Q

P Q

P⊕Q

P Q

P ⇔ Q

( P Q ) ( P Q )

(P⊕Q)∧(P⇔Q)

T

T

T

T

F

F

T

T

F

F

T

T

F

F

T

T

F

F

F

F

F

F

T

T

T

T

F

F

F

F

F

F

F

F

F

F

T

T

F

F

The final output is always false, so (c) is a contradiction.

最终输出始终为假,因此(c)是矛盾的。

Section Summary

章节概要

1. All of the following phrases mean P Q : if P then Q ; P implies Q ; Q if P ; P only if Q ; when P , Q ; Q whenever P ; P is sufficient for Q ; Q is necessary for P ; and other variations involving passive voice or changes in word order.

1.下列所有短语都表示 P ⇒ Q:如果 P 则 Q;P 蕴含 Q;Q 如果 P;P 仅当 Q;当 P 时,Q;Q 只要 P;P 是 Q 的充分条件;Q 是 P 的必要条件;以及其他涉及被动语态或词序变化的变体。

2. P Q (“ P iff Q ”) is true when P and Q are both true or both false; P Q is false when exactly one of P and Q is true. Some useful equivalences involving IFF include:

2. P ⇔ Q(“P 当且仅当 Q”)当且仅当 P 和 Q 都为真或都为假时为真;当且仅当 P 和 Q 中只有一个为真时,P ⇔ Q 为假。一些涉及 IFF 的有用等价关系包括:

P P Q ( P P Q ) ( Q P P ) , P P Q ( P P Q ) ( ( P P ) ( Q ) ) ,
P P Q ( P P ) ( Q ) Q P P , ( P P Q ) P P Q .

3. P only if Q ” is the converse of “ P if Q .” “ P is necessary for Q ” is the converse of “ P is sufficient for Q .” So, P Q can be translated “ P if and only if Q ” or “ P is necessary and sufficient for Q .” Other possible translations include: P iff Q ; P precisely when Q ; P is equivalent to Q .

3.“P 仅当 Q”是“P 如果 Q”的逆命题。“P 是 Q 的必要条件”是“P 是 Q 的充分条件”的逆命题。因此,P ⇔ Q 可以翻译为“P 当且仅当 Q”或“P 是 Q 的必要且充分条件”。其他可能的翻译包括:P 当且仅当 Q;P 当且仅当 Q;P 等价于 Q。

4. In logic, the words BUT, ALTHOUGH, and BOTH have the same meaning as AND. “Neither P nor Q ” means ( P ) ( Q ) .

4.在逻辑学中,BUT、ALTHOUGH 和 BOTH 与 AND 具有相同的含义。“既非 P 也非 Q”表示 (∼P)∧(∼Q)。

5. A tautology is a propositional form that has output true in all rows of its truth table. A contradiction is a propositional form that has output false in all rows of its truth table. Most propositional forms are neither tautologies nor contradictions.

5. 重言式是指真值表中所有行都为真的命题形式。矛盾式是指真值表中所有行都为假的命题形式。大多数命题形式既不是重言式也不是矛盾式。

Exercises

练习

1. Use truth tables to prove parts (b), (c), and (d) of the Theorem on IFF.

1.利用真值表证明关于因果函数定理的(b)、(c)和(d)部分。

2. Which expressions below are logically equivalent to the statement “if P then Q ”?

2.下列哪些表达式在逻辑上等价于语句“如果 P 则 Q”?

P Q

P ⇒ Q

Q P

问 ⇒ 答

P Q

P ⇔ Q

P implies Q .

P蕴含Q。

(~ P ) ⇒ (~ Q )

(~P)⇒ (~Q)

(~ Q ) ⇒ (~ P )

(~Q)⇒ (~P)

(~ P ) ⇔ (~ Q )

(~P)⇔(~Q)

( P Q ) ( ( P ) ( Q ) )

(P∧Q)∨((∼P)∧(∼Q))

(~ P )∨ Q

(~P)∨Q

( P Q ) .

∼(P⊕Q)。

P if Q .

P 如果 Q。

Q if P .

Q 如果 P。

P iff Q .

P iff Q。

P is sufficient for Q .

P 足以满足 Q。

P is necessary for Q .

P 是 Q 的必要条件。

P is nec. and suff. for Q .

P 是 Q 的必要条件和充分条件。

( P Q ) ( Q P )

(P⇒Q)∧(Q⇒P)

Q P

Q ⇔ P

3. Which expressions listed in the previous exercise are logically equivalent to the statement “ P if and only if Q ”?

3.上一题中列出的哪些表达式在逻辑上等价于语句“P 当且仅当 Q”?

4. Use truth tables to classify each propositional form below as a tautology, a contradiction, or neither.

4.使用真值表将下列每个命题形式分类为重言式、矛盾式或两者都不是。

(a) P ⇔ (~ P ) (b) P ⇒ (~ P ) (c) ( P Q ) ( ( P ) Q ) (d) Q ( ( P Q ) ) .

(a) P ⇔ (~P) (b) P ⇒ (~P) (c) (P⊕Q)⇔((∼P)⇔Q) (d) Q∧(∼(P⇒Q)).

5. Is each propositional form a tautology, a contradiction, or neither? Explain.

5. 下列各命题形式是重言式、矛盾式,还是两者都不是?请解释。

(a) ( P Q ) Q (b) ( P Q ) ⇒ Q (c) ~( Q ⇒ ( P Q )) (d) ( P Q ) ⇔ ( Q P ) (e) ( P Q ) ( P ( Q ) ) (f) ( P Q ) ( P ( Q ) ) .

(a) (P∧Q)⇒Q (b) (P ∨ Q) ⇒ Q (c) ~(Q ⇒ (P ∨ Q)) (d) (P ⇔ Q) ⇔ (Q ⇔ P) (e) (P⇔Q)∧(P⇔(∼Q)) (f) (P⇒Q)∧(P⇒(∼Q)).

6. Rewrite each statement, replacing all logical connective words (such as IF, BUT, SUFFICIENT, and so on) with appropriate logical symbols (such as ⇒).

6.重写每个语句,将所有逻辑连接词(如 IF、BUT、SUFFICIENT 等)替换为适当的逻辑符号(如 ⇒)。

(a). If x 0 is even but prime, then x 0 = 2.

(a)如果 x 0 是偶数但质数,则 x 0 = 2。

[ Answer. “( x 0 is even x 0 is prime) ⇒ x 0 = 2.”]

[答案:“(x 0 为偶数 ∧ x 0 为质数) ⇒ x 0 = 2。”]

(b) f is continuous only if f is bounded.

(b)f 连续当且仅当 f 有界。

(c) x is real whenever x is rational.

(c)当 x 是有理数时,x 也是实数。

(d) A necessary condition for m to be perfect is that m be even.

(d)m 为完全体的必要条件是 m 为偶数。

(e) X is compact iff X is both closed and bounded.

(e)X 是紧致的当且仅当 X 既是闭集又是有界集。

(f) Although x > 2 implies x 2 > 4, the converse is not true. [Don’t use the word “converse” in your answer.]

(f)虽然 x > 2 蕴含 x 2 > 4,但其逆命题不成立。[答案中不要使用“逆命题”一词。]

(g) When f has compact domain, continuity of f is equivalent to uniform continuity of f .

(g)当 f 具有紧致区域时,f 的连续性等价于 f 的一致连续性。

(h) Continuity of f is sufficient but not necessary for integrability of f .

(h)函数 f 的连续性是函数 f 可积性的充分条件,但不是必要条件。

7. Write seven different English sentences with the same meaning as “If A C ¯ and B C ¯ are equal in length, then A = B .”

7. 用英语写出七个与“如果 AC¯ 和 BC¯ 长度相等,那么 ∠A=∠B”意思相同的句子。

8. Write five different English sentences equivalent to “ x 0 is odd iff x 0 + 3 is even.”

8.写出五个不同的英文句子,等价于“x 0 为奇数当且仅当 x 0 + 3 为偶数”。

9. Rewrite each statement in the standard form “If P , then Q .”

9. 将每个语句改写成标准形式“如果 P,则 Q”。

(a) G cyclic implies G is commutative.

(a)G 是循环的,意味着 G 是交换的。

(b) Normality of X is sufficient for regularity of X .

(b)X 的正态性是 X 的正则性的充分条件。

(c) Measurability of f is necessary for continuity of f .

(c)f 的可测量性是 f 连续性的必要条件。

(d) f is integrable only if f is bounded.

(d)f 可积当且仅当 f 有界。

(e) For K to be path-connected, it is necessary that K be connected.

(e)为了使 K 是路径连通的,K 必须是连通的。

(f) A sufficient condition for T to be linear is that T be orthogonal.

(f)T 为线性的充分条件是 T 为正交的。

10. True or false? (It may help to first rewrite complex statements in symbolic form.)

10.判断题:对还是错?(先将复杂的语句改写成符号形式可能会有所帮助。)

(a) Columbus discovered America in 1942 iff World War I ended in 1981.

(a)哥伦布于 1942 年发现美洲,当且仅当第一次世界大战于 1981 年结束。

(b) Mongolia is a vegetable if Paris is a city.

(b)如果巴黎是一座城市,那么蒙古就是一种蔬菜。

(c) Mongolia is a vegetable iff Paris is a city.

(c)蒙古是一种蔬菜,当且仅当巴黎是一座城市。

(d) Snakes are not reptiles whenever 2 + 2 = 4.

(d)当 2 + 2 = 4 时,蛇不是爬行动物。

(e) 3 is neither even nor positive.

(e)3 既不是偶数也不是正数。

(f) Grand pianos are heavier than oboes only if cannonballs are lighter than feathers.

(f)只有当炮弹比羽毛轻时,三角钢琴才比双簧管重。

(g) For Venus to be a planet, it is necessary that 1 + 1 = 3.

(g)要使金星成为行星,必须满足 1 + 1 = 3。

(h) 1 + 1 = 2 is a sufficient condition for anvils to float.

(h)1 + 1 = 2 是铁砧漂浮的充分条件。

(i) Although whales are mammals, they live in the ocean.

(i)虽然鲸鱼是哺乳动物,但它们生活在海洋中。

(j) January has 32 days only if March has 31 days.

(j)只有当三月有 31 天时,一月才有 32 天。

(k) For Hillary Clinton to be 45th President of the U.S., it is sufficient that George Washington was the 1st President of the U.S.

(k)希拉里·克林顿成为美国第45任总统,只要乔治·华盛顿是美国第1任总统即可。

11. For each sentence below, write a logically equivalent sentence using the word NECESSARY.

11.对于下面的每个句子,用“NECESSARY”一词写一个逻辑上等价的句子。

(a) If x is rational, then x is real.

(a)如果 x 是有理数,那么 x 就是实数。

(b) For f to be continuous, it is sufficient that f be differentiable.

(b)为了使函数 f 连续,只需使函数 f 可微即可。

(c) det ( B ) 0 implies B is invertible.

(c)det(B)≠0意味着B是可逆的。

(d) G is cyclic only if G is commutative.

(d)G 是循环的当且仅当 G 是交换的。

(e) A sufficient condition for T to be isosceles is that T be equilateral.

(e)T 为等腰三角形的充分条件是 T 为等边三角形。

12. For each sentence below, write a logically equivalent sentence using the phrase ONLY IF.

12.对于下面的每个句子,用短语“ONLY IF”写一个逻辑上等价的句子。

(a) Whenever n ends in the digit 8, n is not prime.

(a)当n以数字8结尾时,n不是质数。

(b) If F is a field, then F is an integral domain.

(b)如果 F 是一个域,那么 F 就是一个整环。

(c) A necessary condition for a sequence ( a n ) to converge is that ( a n ) be bounded.

(c)序列(a n )收敛的必要条件是(a n )有界。

(d) Having a zero column is sufficient for a matrix B to be singular.

(d)矩阵 B 具有零列足以保证它是奇异的。

(e) K is compact or K is connected.

(e)K 是紧凑的或 K 是连通的。

13. Prove that IFF is transitive by showing that [ ( P Q ) ( Q R ) ] ( P R ) is a tautology. (b) Is the converse of (a) a tautology? Explain.

13.证明因果律正则化简是传递的,方法是证明 [(P⇔Q)∧(Q⇔R)]⇒(P⇔R) 是重言式。(b) (a) 的逆命题是重言式吗?解释原因。

14. Is IFF associative? In other words, is ( P Q ) ⇔ R logically equivalent to P ⇔ ( Q R )? Explain.1

14. 因果关系是否满足结合律?换句话说,(P ⇔ Q) ⇔ R 与 P ⇔ (Q ⇔ R) 逻辑等价吗?请解释。

15. True or false?

15.对还是错?

(a) For 57 to be negative, it is necessary and sufficient that George Bush was the 1st President of the United States.

(a)要使 57 为否定,必要且充分条件是乔治·布什是美国第一任总统。

(b) Peking is the capital of Venezuela only if Boise is the capital of Idaho.

(b)只有当博伊西是爱达荷州的首府时,北京才是委内瑞拉的首都。

(c) 3 is odd if and only if the positivity of 7 implies that 9 is even.

(c)3 为奇数当且仅当 7 为正数意味着 9 为偶数。

(d) Whenever pigs can fly or fish have gills, it is false that money grows on trees iff Los Angeles is located in California.

(d)只要猪会飞或者鱼有鳃,如果洛杉矶位于加利福尼亚州,那么钱从树上长出来的说法就是错误的。

(e) If R is paracompact, then red is a color.

(e)如果 R 是拟紧集,那么红色是一种颜色。

16. True or false? Explain.

16.判断题:对还是错?请解释。

(a) Every propositional form is either a tautology or a contradiction.

(a)每个命题形式要么是重言式,要么是矛盾式。

(b) The negation of P Q is logically equivalent to “ P or Q .”

(b)P ⇔ Q 的否定在逻辑上等价于“P 或 Q”。

(c) ~( P Q ) is logically equivalent to ( P ) ( Q ) .

(c)~(P ⇒ Q)在逻辑上等价于(∼P)⊕(∼Q)。

(d) P is necessary for Q ” is logically equivalent to the converse of “ P is sufficient for Q .”

(d)“P 是 Q 的必要条件”在逻辑上等价于“P 是 Q 的充分条件”的逆命题。

(e) P only if Q ” is logically equivalent to the contrapositive of “ Q if P .”

(e)“P 仅当 Q”在逻辑上等价于“Q 如果 P”的逆否命题。

17. (a) Prove parts (a) and (b) of the Theorem on XOR using the Theorem on IFF and known logical equivalences (not truth tables). (b) Prove the rest of the Theorem on XOR by any convenient technique.

17.(a) 利用正则表达式定理和已知的逻辑等价关系(而非真值表)证明异或定理的(a)和(b)部分。(b) 用任何方便的方法证明异或定理的其余部分。

18. Suppose A , B , and C are propositional forms such that B C . (a) Prove the substitution rule ( A B ) ( A C ) using a meta-truth table (see Remark 5 on page 1.2). (b) Deduce from (a) that ( B A ) ( C A ) without using a truth table.

18. 假设 A、B 和 C 是命题形式,且 B≡C。(a)使用元真值表证明替换规则 (A⇔B)≡(A⇔C)(参见 1.2 页的备注 5)。(b)不使用真值表,从 (a) 推导出 (B⇔A)≡(C⇔A)。

19. Let A and B be any propositional forms. (a) Suppose A B . Show that A B is a tautology. (b) Now suppose A B is a tautology. Show that A B .

19.设 A 和 B 为任意命题形式。(a) 假设 A≡B。证明 A⇔B 是重言式。(b) 现在假设 A⇔B 是重言式。证明 A≡B。

20. For each propositional form, find a logically equivalent form that only uses the logical symbols ~ and ∨. (a) P Q (b) ( P Q ) (c) P Q (d) P Q .

20.对于每个命题形式,找到一个逻辑等价的形式,该形式只使用逻辑符号 ~ 和 ∨。(a)P ⇒ Q(b)∼(P∧Q)(c)P∧Q(d)P⊕Q。

21. For each propositional form, find a logically equivalent form that only uses the connectives ~ and . (a) ~( P Q ) (b) P Q (c) P Q (d) P Q (e) P Q .

21.对于每个命题形式,找到一个逻辑等价的形式,该形式只使用连接词 ~ 和 ∧。(a) ~(P ⇒ Q) (b) P ⇒ Q (c) P ∨ Q (d) P ⇔ Q (e) P⊕Q。

22. For each propositional form, find a logically equivalent form that only uses the connectives ~ and ⇒. (a) P Q (b) P Q (c) P Q (d) P Q .

22.对于每个命题形式,找到一个逻辑等价的形式,该形式只使用连接词 ~ 和 ⇒。(a)P ∨ Q(b)P∧Q(c)P ⇔ Q(d)P⊕Q。

23. Write the converse and the contrapositive of each statement. Give the truth value of the statement, its converse, and its contrapositive.

23.写出下列各命题的逆命题和逆否命题。并给出该命题、其逆命题和其逆否命题的真值。

(a) If milk is a beverage, then chowder is a soup.

(a)如果牛奶是饮料,那么浓汤就是汤。

(b) If 9 is negative, then 10 is odd.

(b)如果 9 是负数,那么 10 就是奇数。

(c) If 1 + 1 = 5, then the cosine function is continuous.

(c)如果 1 + 1 = 5,则余弦函数是连续的。

(d) Sugar is sweet if lemons are sour.

(d)如果柠檬是酸的,那么糖就是甜的。

(e) A necessary condition for pigs to fly is that eagles oink.

(e)猪会飞的必要条件是老鹰会哼哼。

(f) For π to be negative, it is sufficient that −1000 be positive.

(f)要使 π 为负数,只需 −1000 为正数即可。

(g) An argon atom is lighter than a helium atom whenever a sodium atom is heavier than a hydrogen atom.

(g)当钠原子比氢原子重时,氩原子比氦原子轻。

(h) If 7 is positive iff 8 is odd, then 9 is a square iff 10 is even.

(h)如果 7 是正数当且仅当 8 是奇数,那么 9 是平方数当且仅当 10 是偶数。

(i) The oddness of 13 implies that 5 > 11 only if the liquidity of water at room temperature is a sufficient condition for fish to have lungs.

(i)13 的奇特性意味着 5 > 11 仅当室温下水的流动性是鱼拥有肺的充分条件时才成立。

24. Show that any propositional form using only the logical connectives ∼ and cannot be logically equivalent to P Q .

24.证明任何仅使用逻辑连接词∼和⊕的命题形式都不能逻辑等价于P∧Q。

1.4 Tautologies, Quantifiers, and Universes

1.4 重言式、量词和宇宙

Elementary logic has two main parts: the logic of propositions and the logic of quantifiers. After a few more examples to round out our treatment of propositional logic, we begin our discussion of quantifiers in the second part of this section. Intuitively, quantifiers allow us to extend the propositional connectives AND and OR to apply to variables that can range over potentially infinite sets of values.

初等逻辑主要包含两部分:命题逻辑和量词逻辑。在通过几个例子完善我们对命题逻辑的阐述之后,我们将在本节的第二部分开始讨论量词。直观地说,量词允许我们将命题连接词“与”和“或”扩展到可以取值于潜在无限集合的变量。

More on Tautologies and Contradictions

关于同义反复和矛盾的更多内容

Recall that a tautology is a propositional form that is always true, regardless of the truth values of the variables appearing in the form; similarly, a contradiction is a propositional form that is always false. Recall also that two propositional forms are logically equivalent iff they have the same truth value for all possible values of the input variables.

回想一下,重言式是一种命题形式,无论其中出现的变量取何真值,它始终为真;类似地,矛盾式是一种命题形式,它始终为假。还要记住,两个命题形式逻辑等价当且仅当它们对于所有可能的输入变量值都具有相同的真值。

1.30. Example. Show that the following propositional form is a tautology:

1.30. 示例。证明下列命题形式是重言式:

( P P Q ) ( ( P P R ) ( Q R ) ) .

Solution. Call the given propositional form A . We make an eight-row truth table, shown here:

解:将给定的命题形式记为 A。我们列出如下的八行真值表:

P

P

Q

R

P Q

P ⇔ Q

P R

P∧R

Q R

Q∧R

( P R ) ( Q R )

(P∧R)⇔(Q∧R)

A

一个

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

F

F

T

T

F

F

F

F

T

T

T

T

T

T

F

F

T

T

F

F

T

T

F

F

F

F

T

T

T

T

F

F

F

F

F

F

F

F

F

F

T

T

T

T

F

F

T

T

T

T

F

F

F

F

T

T

F

F

T

T

F

F

T

T

F

F

F

F

F

F

F

F

T

T

T

T

F

F

F

F

T

T

T

T

F

F

F

F

T

T

T

T

F

F

F

F

F

F

T

T

F

F

F

F

T

T

T

T

We fill in the Ts in columns 5 and 6 by looking for two Ts in the appropriate input columns; the remaining entries in these columns are Fs. We fill in columns 4 and 7 by looking for agreement between the two input columns, putting Ts in these rows, and putting Fs elsewhere. Finally, we fill column 8 by looking for places where column 4 is true and column 7 is false. No such places exist, so column 8 is all Ts. This shows that A is a tautology.

我们通过在相应的输入列中查找两个“T”来填充第5列和第6列;这两列中的其余条目均为“F”。我们通过查找两个输入列是否一致来填充第4列和第7列,将“T”填入这两行,将“F”填入其他行。最后,我们通过查找第4列为真而第7列为假的情况来填充第8列。不存在这样的情况,因此第8列全部为“T”。这表明A是一个重言式。

1.31. Example. Let A be any propositional form, and let C be any contradiction.

1.31. 例。设 A 为任意命题形式,C 为任意矛盾。

(a) Show that A C is a contradiction.

(a)证明 A∧C 是矛盾的。

(b) Show that A C A .

(b)证明 A∨C≡A。

(c) Show that ( ( A ) C ) A is a tautology.

(c)证明((∼A)⇒C)⇒A 是一个重言式。

Solution. We cannot make the actual, complete truth tables for the propositional forms appearing here, since we do not know which specific propositional variables appear within the forms A and C . However, we can still make a table showing the possible truth values of various expressions built from A and C . The key point is to recall that C must be false, whereas (without further information about A ), A might be true or false. So we prepare the following two-row table:

解答:我们无法为这里出现的命题形式构建完整的真值表,因为我们不知道 A 和 C 中具体包含哪些命题变量。但是,我们仍然可以构建一个表格,列出由 A 和 C 构成的各种表达式的可能真值。关键在于要记住 C 必须为假,而(在没有关于 A 的更多信息的情况下)A 可能为真也可能为假。因此,我们准备以下两行表格:

A

一个

C

C

A C

A∧C

A C

A∨C

A

∼A

( ( A ) C )

((∼A)⇒C)

( ( A ) C ) A

((∼A)⇒C)⇒A

T

T

F

F

F

F

T

T

F

F

T

T

T

T

F

F

F

F

F

F

F

F

T

T

F

#

F

T

T

For any assignment of truth values to the propositional variables appearing within A and C , A must be true or false, whereas C must be false. Column 3 therefore proves that A C must also be false in all situations, so this form is a contradiction. Comparison of columns 1 and 4 shows that A C always has the same truth value as A , hence these two forms are logically equivalent. Finally, the last three columns show that the form in (c) is always true, so this form is a tautology. We return to this tautology later, using it to justify a method of proof called proof by contradiction.

对于 A 和 C 中出现的命题变量的任何真值赋值,A 必须为真或为假,而 C 必须为假。因此,第 3 列证明了 A∧C 在所有情况下也必须为假,所以这种形式是矛盾式。比较第 1 列和第 4 列可知,A∨C 的真值始终与 A 相同,因此这两种形式逻辑等价。最后,最后三列表明 (c) 中的形式始终为真,因此这种形式是重言式。我们稍后会回到这个重言式,并用它来证明一种称为反证法的证明方法。

1.32. Remark. By arguments similar to those in the last example, you can establish the following facts. Let A and B be any propositional forms, and let T be any tautology.

1.32. 备注。通过与上一个例子类似的论证,你可以证明以下事实。设 A 和 B 为任意命题形式,T 为任意重言式。

(a) A T A .

(a)A∧T≡A。

(b) A T is a tautology.

(b)A∨T 是同义反复。

(c) A is a tautology iff A is a contradiction.

(c)A 是重言式当且仅当 ∼A 是矛盾式。

(d) A B iff A B is a tautology.

(d)A≡B 当且仅当 A⇔B 是重言式。

Part (d) lets us harvest many tautologies from the logical equivalences proved earlier. For instance, the logical equivalence P Q Q P leads to the tautology ( P Q ) ( Q P ) .

(d)部分允许我们从先前证明的逻辑等价关系中得出许多重言式。例如,逻辑等价关系 P∧Q≡Q∧P 导出重言式 (P∧Q)⇔(Q∧P)。

Variables and Quantifiers

变量和量词

Quantified logic studies the truth or falsehood of statements containing variables such as x and y . To see how this differs from propositional logic, consider the following example.

量化逻辑研究包含变量(例如 x 和 y)的陈述的真假。为了了解它与命题逻辑有何不同,请考虑以下示例。

1.33. Example. True or false?

1.33. 示例。对还是错?

(a) For all real numbers x , x 2 = x .

(a)对于所有实数 x,x 2 = x。

(b) There is a real number x satisfying x 2 = x .

(b)存在实数 x 满足 x 2 = x。

(c) x 2 = x (where x is a variable varying over all real numbers).

(c)x 2 = x(其中 x 是一个在所有实数上变化的变量)。

(d) x 0 2 = x 0 (where x 0 is a constant , designating one particular real number).

(d)x02=x0(其中 x 0 是一个常数,表示一个特定的实数)。

Intuitively, (a) is false, since the real number x = 2 does not satisfy x 2 = x (as 2 2 = 4 ≠ 2). Similarly, (b) is true, since x = 1 satisfies 1 2 = 1 (note x = 0 works also). Statement (c) is a bit problematic: if the variable x has not been assigned a particular value, but is varying through all real numbers, then (c) has no truth value (it is not a proposition). Put another way, (c) is true for some values of the variable x and false for other values of x , so statement (c) has no definite truth value of its own. In contrast, the constant x 0 in (d) represents one, particular, fixed real number. So x 0 2 = x 0 is a proposition, although we cannot tell without more information about x 0 whether this proposition is true or false.

直观上,(a) 为假,因为实数 x = 2 不满足 x = x(因为 2 = 4 ≠ 2)。类似地,(b) 为真,因为 x = 1 满足 1 = 1(注意 x = 0 也成立)。陈述 (c) 有点问题:如果变量 x 没有被赋予特定值,而是取所有实数,那么 (c) 没有真值(它不是一个命题)。换句话说,(c) 对于变量 x 的某些值成立,而对于其他值成立,因此陈述 (c) 本身没有确定的真值。相比之下,(d) 中的常量 x 代表一个特定的固定实数。所以 x02=x0 是一个命题,尽管如果没有关于 x 0 的更多信息,我们无法判断这个命题是真还是假。

In general, a statement containing a variable x , such as statement (c) above, is called an open sentence and is denoted by a symbol such as P ( x ), Q ( x ), etc. Open sentences can involve more than one variable; for instance, we could let R ( x , y ) be the formula x 2 + y 2 = 25. An open sentence does not have a truth value; it is not a proposition. One way to turn an open sentence into a proposition is to assign specific values to every variable appearing in it. In our example, R (5, 0) is the true proposition 5 2 + 0 2 = 25; R (2, 4) is the false proposition 2 2 + 4 2 = 25; R (3, − 4) is true; R (7, 10) is false; and so on. R (3, z ) is the open sentence 3 2 + z 2 = 25 involving the new variable z . We can also substitute letters that represent constants into an open sentence. In this text, the letters a , b , c and letters with subscripts (like x 0 ) denote constants. Thus R ( x 0 , b ) is the proposition x 0 2 + b 2 = 25 , but the truth value of this proposition cannot be determined without more information about x 0 and b .

一般来说,包含变量 x 的语句,例如上面的语句 (c),被称为开放语句,并用 P(x)、Q(x) 等符号表示。开放语句可以包含多个变量;例如,我们可以令 R(x, y) 表示公式 x 2 + y 2 = 25。开放语句没有真值;它不是命题。将开放语句转化为命题的一种方法是为其中的每个变量赋予特定的值。在我们的例子中,R(5, 0) 是真命题 5 2 + 0 2 = 25;R(2, 4) 是假命题 2 2 + 4 2 = 25;R(3, -4) 为真; R(7, 10) 为假;以此类推。R(3, z) 是包含新变量 z 的开放语句 3 2 + z 2 = 25。我们还可以将代表常量的字母代入开放语句中。在本例中,字母 a、b、c 以及带下标的字母(例如 x 0 )表示常量。因此,R(x 0 , b) 是命题 x02+b2=25,但如果没有关于 x 0 和 b 的更多信息,则无法确定该命题的真值。

There are two other ways to turn an open sentence P ( x ) into a proposition, which are illustrated in (a) and (b) above: we may quantify the variable x with a phrase such as “for all x ” or “there exists an x .” We use the universal quantifier symbol to abbreviate “for all,” and the existential quantifier symbol to abbreviate “there exists.” These symbols are explained further in the next two definitions.

还有两种方法可以将开放语句 P(x) 转化为命题,如上文 (a) 和 (b) 所示:我们可以用诸如“对于所有 x”或“存在一个 x”之类的短语来量化变量 x。我们使用全称量词符号 ∀ 来表示“对于所有”,使用存在量词符号 ∃ 来表示“存在”。这些符号将在接下来的两个定义中进一步解释。

1.34 Definition of ∀ (Universal Quantifier). Let P ( x ) be an open sentence involving the variable x . The statement “ x , P ( x ) ” is a proposition that is true iff for all objects x 0 , P ( x 0 ) is true. The quantifier x may be translated “for all x ,” “for every x ,” “for each x ,” “for any x ,” among other possibilities. The symbol looks like an upside-down capital A, reminding us of the word ALL.

1.34 ∀(全称量词)的定义。设 P(x) 是一个包含变量 x 的开放语句。命题“∀x,P(x)”为真当且仅当对于所有对象 x 0 ,P(x 0 ) 为真。量词 ∀x 可以翻译为“对于所有 x”、“对于每个 x”、“对于每个 x”、“对于任何 x”等等。符号 ∀ 看起来像一个倒置的大写字母 A,让人联想到单词 ALL(全部)。

1.35 Definition of ∃ (Existential Quantifier). Let P ( x ) be an open sentence involving the variable x . The statement “ x , P ( x ) ” is a proposition that is true iff there is at least one object x 0 making P ( x 0 ) true. The quantifier x may be translated “there is an x ,” “there exists an x ,” “there is at least one x ,” “for some x ,” among other possibilities. The symbol looks like a backwards capital E, reminding us of the word EXISTS.

1.35 存在量词 ∃ 的定义。设 P(x) 是一个包含变量 x 的开放语句。命题“∃x, P(x)”为真当且仅当存在至少一个对象 x 0 使得 P(x 0 ) 为真。量词 ∃x 可以翻译为“存在一个 x”、“存在一个 x”、“至少有一个 x”、“对于某些 x”等等。符号 ∃ 看起来像一个反写的大写字母 E,让人联想到单词 EXISTS(存在)。

Universes; Restricted Quantifiers

宇宙;受限量词

In most discussions, we do not want to be making statements about literally all objects x . Instead, we want to restrict attention to a particular set of objects, such as the set of real numbers, or the set of integers, or the set of continuous real-valued functions. A set U of objects that contains all objects we want to discuss in a particular situation is called a universe or universal set . We write x U to mean that x is a member of the universal set U , and we write x U to mean that x is not in U . (This notation comes from set theory, which we study in Chapter 3 .) By combining this notation with quantifiers, we obtain the restricted quantifiers defined next.

在大多数讨论中,我们并不想对所有对象 x 都做出陈述。相反,我们希望将注意力限制在特定的对象集合上,例如实数集、整数集或连续实值函数集。包含我们在特定情况下想要讨论的所有对象的集合 U 被称为全集或论域。我们用 x ∈ U 表示 x 是全集 U 的成员,用 x ∉ U 表示 x 不属于 U。(这种记号来自集合论,我们将在第三章学习。)通过结合这种记号和量词,我们可以得到接下来定义的受限量词。

1.36. Definition: Restricted Quantifiers. Let U be any set, and let P ( x ) be an open sentence involving x . The statement “ x U , P ( x ) ” means that for every object x 0 in U , P ( x 0 ) is true. The statement “ x U , P ( x ) ” means there exists at least one object x 0 in U for which P ( x 0 ) is true. The symbols “ x U ” and “ x U ” are restricted quantifiers; in contrast, the previous symbols “ x ” and “ x ” are unrestricted quantifiers.

1.36. 定义:受限量词。设 U 为任意集合,P(x) 为包含 x 的开语句。命题“∀x∈U,P(x)”表示对于 U 中的每个对象 x 0 ,P(x 0 ) 为真。命题“∃x∈U,P(x)”表示存在至少一个对象 x 0 属于 U,使得 P(x 0 ) 为真。符号“∀x∈U”和“∃x∈U”是受限量词;与之相对,之前的符号“∀x”和“∃x”是非受限量词。

Next we introduce some notation for certain universes of numbers that arise frequently in mathematics.

接下来,我们将介绍一些在数学中经常出现的某些数论的符号。

1.37. Definition: Notation for Number Systems.

1.37. 定义:数制符号。

(a) Z is the set of all integers , namely Z = { , 3 , 2 , 1 , 0 , 1 , 2 , 3 , } .

(a)Z 是所有整数的集合,即 Z={…,−3,−2,−1,0,1,2,3,…}。

(b) Q is the set of all rational numbers , which are ratios a / b where a is an integer and b is a nonzero integer.

(b)Q 是所有有理数的集合,有理数是比值 a/b,其中 a 是整数,b 是非零整数。

(c) R is the set of all real numbers .

(c)R 是所有实数的集合。

(d) C is the set of all complex numbers , which are expressions a + bi with a , b real, and where i satisfies i 2 = −1.

(d)C 是所有复数的集合,复数是表达式 a + bi,其中 a、b 为实数,且 i 满足 i 2 = −1。

(e) We write Z 0 = { 0 , 1 , 2 , 3 , } for the set of nonnegative integers, and Z > 0 = { 1 , 2 , 3 , } for the set of positive integers. Some authors use N (the set of natural numbers ) to denote the set Z 0 , whereas other authors use N to denote the set Z > 0 . To avoid confusion, we do not use the symbol N except in §8.3 .

(e) 我们用 Z≥0={0,1,2,3,…} 表示非负整数集合,用 Z>0={1,2,3,…} 表示正整数集合。有些作者用 N(自然数集)表示集合 Z≥0,而另一些作者则用 N 表示集合 Z>0。为避免混淆,除 §8.3 外,我们不再使用符号 N。

(f) By analogy with (e), let R > 0 be the set of strictly positive real numbers (not including zero). For a fixed integer b , let Z b be the set of integers that are at least b , and let Z > b be the set of integers strictly greater than b . We similarly define Z b , R > x 0 , R x 0 , and so on.

(f) 类比 (e),令 R>0 为严格正实数(不包括零)的集合。对于固定的整数 b,令 Z≥b 为大于等于 b 的整数的集合,令 Z>b 为大于 b 的整数的集合。类似地,我们定义 Z≠b、R>x0、R≤x0,以此类推。

(g) Given n distinct objects x 1 , x 2 , …, x n , let U = { x 1 , x 2 , …, x n } denote the finite universal set consisting of these n objects.

(g)给定 n 个不同的对象 x 1 , x 2 , …, x n ,令 U = {x 1 , x 2 , …, x n 表示由这 n 个对象组成的有限全集。

This definition is merely setting up notation. In Chapter 8 , we give more precise definitions as part of a formal development of the number systems Z , Q , R , and C .

这个定义仅仅是建立符号体系。在第八章中,我们将给出更精确的定义,作为Z、Q、R和C这几个数系正式发展的一部分。

1.38. Example. True or false?

1.38. 示例。对还是错?

(a) x Z , x + 3 = 2 .

(a) ∃x∈Z,x+3=2。

(b) x Z 0 , x + 3 = 2 .

(b) ∃x∈Z≥0,x+3=2。

(c) x R , x 2 = 1 .

(c) ∃x∈R,x2=−1。

(d) x C , x 2 = 1 .

(d)∃x∈C,x2=−1。

(e) x Z > 0 , x + 5 > 3 .

(e)∀x∈Z>0,x+5>3。

(f) x Q , x + 5 > 3 .

(f)∀x∈Q,x+5>3。

(g) x R , x + 5 > 3 .

(g)∀x∈R,x+5>3。

(h) x Z , x 3 = x .

(h)∀x∈Z,x3=x。

(i) x Z , x 3 = x .

(i)∃x∈Z,x3=x。

(j) x { 1 , 0 , 1 } , x 3 = x .

(j)∀x∈{−1,0,1},x3=x。

Solution. First, (a) is true, since x = −1 is in Z and satisfies x + 3 = 2. In fact, x = −1 is the only real solution of this equation. So (b) is false, since −1 is not in the universe Z 0 . Similarly, (c) is false, since the square of any real number is at least 0, so cannot equal −1. But (d) is true, since x = i is a complex number making the equation true ( x = − i also works). Adding 5 to any positive integer produces an integer 6 or greater, which is also greater than 3, so (e) is true. But (f) is false, since x + 5 > 3 fails to hold for the rational number x = −11/2 (among many other examples). Similarly, (g) is false. Next, (h) is false since 2 Z and 2 3 = 2 is not true. But (i) is true, since 1 is an integer such that 1 3 = 1. Also (j) is true, since each of the three objects b in the finite universe {− 1, 0, 1} does satisfy b 3 = b . These examples show that changing to a different universe can affect the truth of a statement with a restricted quantifier . The type of quantifier ( or ) certainly makes a big difference as well. We also remark that it would not make sense to use unrestricted quantifiers in statements (a) through (j). For, if x is an arbitrary object (not necessarily a number), it is not clear what statements such as x + 3 = 2 and x 3 = x should mean.

解:首先,(a) 为真,因为 x = −1 属于 Z,且满足 x + 3 = 2。事实上,x = −1 是该方程的唯一实数解。因此,(b) 为假,因为 −1 不在 Z≥0 的宇宙中。类似地,(c) 为假,因为任何实数的平方至少为 0,所以不可能等于 −1。但 (d) 为真,因为 x = i 是一个复数,使得该方程成立(x = −i 也成立)。任何正整数加 5 的结果都大于等于 6,而 6 也大于 3,所以 (e) 为真。但 (f) 为假,因为对于有理数 x = −11/2(以及许多其他例子),x + 5 > 3 不成立。类似地,(g) 为假。接下来,(h) 为假,因为 2∈Z 且 2 3 = 2 不成立。但 (i) 为真,因为 1 是一个整数,满足 1 3 = 1。同样,(j) 也为真,因为有限论域 {−1, 0, 1} 中的三个对象 b 都满足 b 3 = b。这些例子表明,转换到不同的论域会影响带有限制量词的陈述的真值。量词的类型(∃ 或 ∀)也会产生很大的影响。我们还要指出,在陈述 (a) 到 (j) 中使用非限制量词是不合理的。因为,如果 x 是一个任意对象(不一定是数字),那么像 x + 3 = 2 和 x 3 = x 这样的陈述究竟是什么意思就不清楚了。

Quantifiers vs. Logical Operators

量词与逻辑运算符

The quantifiers and can be viewed as generalizations of the logical operators (AND) and ∨ (inclusive-OR), respectively. To see why, consider a finite universe U = { z 1 , z 2 , …, z n }. The universal statement “ x U , P ( x ) ” is true iff P ( a ) is true for each fixed object a ∈ { z 1 , …, z n }. In other words, “ x U , P ( x ) ” is true iff P ( z 1 ) is true and P ( z 2 ) is true and … and P ( z n ) is true. Similarly, “ x U , P ( x ) ” is true iff there exists an object b ∈ { z 1 , …, z n } making P ( b ) true, which holds iff P ( z 1 ) is true or P ( z 2 ) is true or … or P ( z n ) is true. In summary,

量词 ∀ 和 ∃ 可以分别视为逻辑运算符 ∧(与)和 ∨(包含或)的推广。为了理解这一点,考虑一个有限论域 U = {z 1 , z 2 , …, z n 。全称命题“∀x∈U,P(x)”为真当且仅当对于每个固定对象 a ∈ {z 1 , …, z n ,P(a) 为真。换句话说,“∀x∈U,P(x)”为真当且仅当 P(z 1 ) 为真且 P(z 2 ) 为真且……且 P(z n ) 为真。类似地,“∃x∈U,P(x)”为真当且仅当存在对象 b ∈ {z 1 , …, z n 使得 P(b) 为真,这成立当且仅当 P(z 1 ) 为真或 P(z 2 ) 为真或 … 或 P(z n ) 为真。总之,

x x { z z 1 z z 2 , , z z n n } , P P ( x x ) P P ( z z 1 ) P P ( z z 2 ) P P ( z z n n ) ;
x x { z z 1 , z z 2 , , z z n n } , P P ( x x ) P P ( z z 1 ) P P ( z z 2 ) P P ( z z n n ) .

In the general case of an infinite universe U , “ x U , P ( x ) ” is a statement that (informally speaking) AND’s together the infinitely many statements P ( a ) as a ranges through all objects in U . Similarly, “ x U , P ( x ) ” is a statement that OR’s together all the statements P ( b ) for b ranging through U . This analogy linking with and with ∨ can help us understand and remember some of the quantifier properties presented later (such as the negation rules in the next section).

对于无限宇宙 U 的一般情况,“∀x∈U,P(x)”表示(通俗地说)将无穷多个命题 P(a) 进行“与”运算,其中 a 取遍 U 中的所有对象。类似地,“∃x∈U,P(x)”表示将所有命题 P(b) 进行“或”运算,其中 b 取遍 U 中的所有对象。这种将 ∀ 与 ∧ 以及 ∃ 与 ∨ 联系起来的类比可以帮助我们理解和记住后面会介绍的一些量词性质(例如下一节中的否定规则)。

Section Summary

章节概要

1. Given any tautology T , any contradiction C , and any propositional form A ,

1.给定任意重言式T、任意矛盾式C和任意命题形式A,

A T A A C ; A T is a tautology; A C is a contradiction. For any forms A and B , A B iff A B is a tautology.

A∧T≡A≡A∨C;A∨T是重言式;A∧C是矛盾式。对于任意形式A和B,A≡B当且仅当A⇔B是重言式。

2. The universal quantifier and the existential quantifier have the following meaning.

2.全称量词∀和存在量词∃具有以下含义。

Symbol

象征

Meaning

意义

x , P ( x )

∀x,P(x)

For all objects x 0 , P ( x 0 ) is true.

对于所有对象 x 0 ,P(x 0 ) 为真。

x U , P ( x )

∀x∈U,P(x)

For all objects x 0 in U , P ( x 0 ) is true.

对于 U 中的所有对象 x 0 ,P(x 0 ) 为真。

x , P ( x )

∃x,P(x)

There exists an object x 0 such that P ( x 0 ) is true.

存在一个对象 x 0 ,使得 P(x 0 ) 为真。

x U , P ( x )

∃x∈U,P(x)

There exists an object x 0 in U such that P ( x 0 ) is true.

U 中存在对象 x 0 ,使得 P(x 0 ) 为真。

Words such as “all,” “any,” “each,” and “every” signal the use of ; phrases such as “there is,” “there exists,” “for some,” and “at least one” signal the use of . Common choices of the universal set U include Z (integers), Q (rational numbers), R (real numbers), Z 0 (nonnegative integers), Z > 0 (positive integers), and finite sets.

诸如“所有”、“任何”、“每个”和“每个”之类的词语表示使用 ∀;诸如“存在”、“有”、“对于某些”和“至少一个”之类的短语表示使用 ∃。全集 U 的常见选择包括 Z(整数)、Q(有理数)、R(实数)、Z≥0(非负整数)、Z>0(正整数)和有限集。

3. For finite universes, quantifiers can be replaced by propositional logic operators as follows:

3.对于有限论域,量词可以用命题逻辑运算符替换,如下所示:

x { z 1 , z 2 , , z n } , P ( x )

∀x∈{z1,z2,…,zn},P(x)

P ( z 1 ) P ( z 2 ) P ( z n ) ;

P(z1)∧P(z2)∧⋯∧P(zn);

x { z 1 , z 2 , , z n } , P ( x )

∃x∈{z1,z2,…,zn},P(x)

P ( z 1 )∨ P ( z 2 )∨ · · · ∨ P ( z n ).

P(z 1 )∨P(z 2 )∨ · · · ∨P(z n ).

Exercises

练习

1. Is each propositional form a tautology, a contradiction, or neither? Explain.

1. 下列各命题形式是重言式、矛盾式,还是两者都不是?请解释。

(a) ( P ( Q R ) ) ( P Q ) ( P ( R ) ) .

(a)(P⇒(Q⇒R))∧(P⇒Q)∧(P∧(∼R))。

(b) (( P R ) ⇔ ( Q R )) ⇒ ( P Q ).

(b)((P∨R) ⇔ (Q∨R)) ⇒ (P ⇔ Q)。

(c) ( P Q ) ⇒ (( R P ) ⇔ ( R Q )).

(c)(P ⇔ Q) ⇒ ((R ⇒ P) ⇔ (R ⇒ Q))。

2. Is each propositional form a tautology, a contradiction, or neither? Explain.

2. 下列各命题形式是重言式、矛盾式,还是两者都不是?请解释。

(a) [ ( P Q ) ( P R ) ( Q R ) ] R .

(a)[(P∨Q)∧(P⇒R)∧(Q⇒R)]⇒R。

(b) [( P Q ) ⇔ R ] ⇔ [ P ⇔ ( Q R )].

(b)[(P ⇔ Q)⇔ R] ⇔ [P ⇔ (Q ⇔ R)].

(c) [ P ⇒ ( Q R )] ⇒ [( P Q ) ⇒ ( P R )].

(c)[P ⇒ (Q ⇒ R)] ⇒ [(P ⇒ Q) ⇒ (P ⇒ R)].

3. Prove Remark 1.32 .

3.证明备注 1.32。

4. Prove: for any propositional forms A and B , if A B is a tautology, then A B is a tautology. Is the converse always true?

4.证明:对于任意命题形式 A 和 B,如果 A⇔B 是重言式,则 A⇒B 也是重言式。其逆命题总是成立吗?

5. Let T be a tautology, C be a contradiction, and A be any propositional form. Prove: (a) A T is a tautology. (b) A C A . (c) A T ( A ) . (d) T C is a contradiction.

5.设 T 为重言式,C 为矛盾式,A 为任意命题形式。证明:(a) A⇒T 是重言式。(b) A⊕C≡A。(c) A⊕T≡(∼A)。(d) T⇔C 是矛盾式。

6. Rewrite each sentence using symbols, not words. (a) The square of any real number is nonnegative. (b) There is an integer whose cube is two. (c) Every positive rational number has the form m / n for some positive integers m and n .

6. 用符号而非文字改写下列句子。(a) 任何实数的平方都是非负数。(b) 存在一个整数,它的立方是 2。(c) 每个正有理数都可以表示为 m/n 的形式,其中 m 和 n 为正整数。

7. Rewrite each statement using words, not symbols.

7.用文字而不是符号重写下列句子。

(a) x Q , x 2 7 . (b) y R , v R , y v = v . (c) a Z , b Z , 25 = a 2 + b 2 .

(a)∀x∈Q,x2≠7。(b)∃y∈R,∀v∈R,yv=v。(c)∃a∈Z,∃b∈Z,25=a2+b2。

8. True or false? (a) x Z , x 2 = 2 . (b) x R , x 2 = 2 . (c) x Q , x 2 > 0 . (d) x Z , x 2 x . (e) x R , x 2 x . (f) x Z , x 2 3 x 10 = 0 .

8.判断正误? (a) ∃x∈Z,x²=2. (b) ∃x∈R,x²=2. (c) ∀x∈Q,x²>0. (d) ∀x∈Z,x²≥x. (e) ∀x∈R,x²≥x. (f) ∃x∈Z,x²−3x−10=0.

9. True or false? Explain.

9. 判断正误并解释原因。

(b) x { 3 , 5 , 7 } , x is an odd integer.

(b)对于所有x∈{3,5,7},x都是奇数。

(b) x , x { 3 , 5 , 7 } x is an odd integer.

(b)∀x,x∈{3,5,7}⇒x是奇数。

(c) x , x { 3 , 5 , 7 } x is an odd integer.

(c)∀x,x∈{3,5,7}∧x是奇数。

(d) x { 3 , 5 , 7 } , x is an even integer.

(d)∃x∈{3,5,7},x 是偶数。

(e) x , x { 3 , 5 , 7 } x is an even integer.

(e) ∃x,x∈{3,5,7}∧x 是偶数。

(e) x , x { 3 , 5 , 7 } x is an even integer.

(e) ∃x,x∈{3,5,7}⇒x 是偶数。

10. True or false? Explain.

10. 判断正误并解释原因。

(a) x Z , 4 x = 7 . (b) x , x Z 4 x = 7 . (c) x , x Z 4 x = 7 . (d) x Z , 4 x = 7 . (e) x , x Z 4 x = 7 . (f) x , x Z 4 x = 7 .

(a) ∀x∈Z,4x=7。(b) ∀x,x∈Z⇒4x=7。(c) ∀x,x∈Z∧4x=7。(d) ∃x∈Z,4x=7。(e) ∃x,x∈Z∧4x=7。(f) ∃x,x∈Z⇒4x=7。

11. For which universes U is the statement x U , y U , x + y = 0 true? (a) Z > 0 (b) Z (c) Q (d) R (e) {0} (f) {− 1, 0, 1}

11.对于哪些宇宙U,命题∀x∈U,∃y∈U,x+y=0为真? (a) Z>0 (b) Z (c) Q (d) R (e) {0} (f) {−1, 0, 1}

12. For which universes U is the statement x U , y U , x y = 1 true? (a) Z > 0 (b) Q > 0 (c) R > 0 (d) Z (e) R (f) {− 1}

12.对于哪些宇宙U,命题∀x∈U,∃y∈U,xy=1为真? (a) Z>0 (b) Q>0 (c) R>0 (d) Z (e) R (f) {− 1}

13. (a) Rewrite the statement x { 1 , 0 , 1 } , x 3 = x without using quantifiers. (b) Rewrite the statement x { 1 , 2 , 3 } , x 2 = 2 x without using quantifiers.

13.(a) 用不含量词的方式重写语句 ∀x∈{−1,0,1},x3=x。(b) 用不含量词的方式重写语句 ∃x∈{1,2,3},x2=2x。

(c) Rewrite the statement “11 is prime and 13 is prime and 17 is prime and 19 is prime” using a quantifier instead of the word AND.

(c)用量词代替“和”来重写语句“11 是质数,13 是质数,17 是质数,19 是质数”。

14. Let x 0 be a fixed real number, and let x be a variable ranging through real numbers. (a) Explain the logical difference between these two sentences: I. “if x 0 > 0, then x 0 2 x 0 .” II. “if x > 0, then x 2 x .” (b) One of the sentences in (a) is not a proposition. Use quantifiers to turn this sentence into a proposition in two different ways, and state whether these propositions are true or false. (c) Do the truth values of the propositions in (b) change if we restrict x to vary through Z instead of R ?

14.设 x 0 为一个固定的实数,x 为一个取值范围为实数的变量。(a) 解释以下两个句子之间的逻辑区别:I.“如果 x 0 > 0,则 x02≥x0。” II.“如果 x > 0,则 x 2 ≥ x。” (b) (a) 中的一个句子不是命题。使用量词以两种不同的方式将该句子转化为命题,并说明这些命题是真还是假。(c) 如果将 x 的取值范围限制在 Z 而不是 R 上,(b) 中命题的真值会改变吗?

15. (a) Express the statement x , P ( x ) using only and propositional logic operators. (b) Express the statement x , Q ( x ) using only and propositional logic operators.

15.(a) 仅使用 ∃ 和命题逻辑运算符表示命题 ∀x,P(x)。(b) 仅使用 ∀ 和命题逻辑运算符表示命题 ∃x,Q(x)。

16. Suppose U and V are sets such that every member of U is also a member of V . Which statements below must be true for every open sentence P ( x )? For those that are true, explain why. For those that are false, illustrate with a specific choice of U , V , and P ( x ).

16. 假设 U 和 V 是两个集合,且 U 中的每个元素也都是 V 中的元素。对于每个开放语句 P(x),下列哪些陈述必然为真?对于为真的陈述,请解释原因。对于为假的陈述,请举例说明 U、V 和 P(x) 的具体组合。

(a) x U , P ( x ) implies x V , P ( x ) .

(a)∀x∈U,P(x)蕴含∀x∈V,P(x)。

(b) x V , P ( x ) implies x U , P ( x ) .

(b)∀x∈V,P(x)蕴含∀x∈U,P(x)。

(c) x U , P ( x ) implies x V , P ( x ) .

(c)∃x∈U,P(x)蕴含∃x∈V,P(x)。

(d) x V , P ( x ) implies x U , P ( x ) .

(d)∃x∈V,P(x)蕴含∃x∈U,P(x)。

17. We know Q ∨(~ Q ) is a tautology. Which statements below must be true for all choices of U and P ( x )? Explain.

17. 我们知道 Q∨(~Q) 是一个重言式。下列哪些陈述对于 U 和 P(x) 的所有取值都必须成立?请解释。

(a) x U , [ P ( x ) ( P ( x ) ) ] .

(a)∀x∈U,[P(x)∨(∼P(x))].

(b) ( x U , P ( x ) ) ( x U , P ( x ) ) .

(b)(∀x∈U,P(x))∨(∀x∈U,∼P(x))。

(c) ( x U , P ( x ) ) ( x U , P ( x ) ) .

(c)(∀x∈U,P(x))∨∼(∀x∈U,P(x))。

18. We know Q ( Q ) is a contradiction. Which statements below must be false for all choices of U and P ( x )? Explain.

18. 我们知道 Q∧(∼Q) 是一个矛盾式。对于 U 和 P(x) 的所有取值,下列哪些陈述必然为假?请解释。

(a) x U , [ P ( x ) ( P ( x ) ) ] .

(a)∃x∈U,[P(x)∧(∼P(x))].

(b) ( x U , P ( x ) ) ( x U , P ( x ) ) .

(b)(∃x∈U,P(x))∧(∃x∈U,∼P(x))。

(c) ( x U , P ( x ) ) ( x U , P ( x ) ) .

(c)(∃x∈U,P(x))∧∼(∃x∈U,P(x))。

19. True or false? Explain each answer.

19.判断正误并解释每个答案。

(a) For all propositional forms A and all contradictions C , C A is a tautology.

(a)对于所有命题形式 A 和所有矛盾 C,C⇒A 是一个重言式。

(b) For all propositional forms A and B , A B is not logically equivalent to B A .

(b)对于所有命题形式 A 和 B,A⇒B 在逻辑上不等价于 B⇒A。

(c) Any two contradictions are logically equivalent.

(c)任意两个矛盾在逻辑上是等价的。

(d) For all logically equivalent forms A and B , ( A B ) ( A B ) .

(d)对于所有逻辑等价的形式 A 和 B,(A⇒B)≡(A⇔B)。

20. Show that any propositional form using only the logical connectives and ∨ is neither a tautology nor a contradiction.

20.证明仅使用逻辑连接词∧和∨的任何命题形式既不是重言式也不是矛盾式。

1.5 Quantifier Properties and Useful Denials

1.5 量词的性质和有用的否定

This section develops more properties of the existential quantifier and the universal quantifier . We look at rules for converting restricted quantifiers to unrestricted quantifiers, translation examples illustrating how English phrases encode quantifiers, and rules for transforming negated quantified statements. We then discuss how to find useful denials of complex statements, which is a crucial skill needed to analyze and create proofs.

本节将进一步探讨存在量词 ∃ 和全称量词 ∀ 的性质。我们将考察将受限量词转换为非受限量词的规则、阐释英语短语如何编码量词的翻译示例,以及转换否定量化语句的规则。随后,我们将讨论如何找到复杂语句的有效否定,这是分析和构建证明的一项关键技能。

Conversion Rules for Restricted Quantifiers

受限量词的转换规则

We start with some rules for eliminating restricted quantifiers. Let U be any set, and consider the statement x U , P ( x ) . This statement is true iff there is an object x 0 in the set U for which P ( x 0 ) is true. Intuitively, this condition holds iff there is an object x 0 (not initially required to be in U ) for which “ x 0 U and P ( x 0 )” is true. Therefore, we obtain the rule

我们首先给出一些消除受限量词的规则。设 U 为任意集合,考虑命题 ∃x∈U,P(x)。该命题为真当且仅当集合 U 中存在对象 x 0 ,使得 P(x 0 ) 为真。直观地说,该条件成立当且仅当存在对象 x 0 (初始时不一定在 U 中),使得“x 0 ∈ U 且 P(x 0 )”为真。因此,我们得到以下规则:

x x U U , P P ( x x ) iff 当……时 x x , ( x x U U P P ( x x ) )

for converting a restricted existential quantifier to an unrestricted existential quantifier. The next example shows that this rule does not work if we blindly replace both symbols by symbols.

用于将受限存在量词转换为非受限存在量词。下一个例子表明,如果我们盲目地将两个 ∃ 符号都替换为 ∀ 符号,则此规则无效。

1.39. Example. Let U be the set of polar bears, and let P ( x ) be the open sentence “ x is white.” The two statements x U , P ( x ) and x , ( x U P ( x ) ) both assert that a white polar bear exists. Similarly, x U , P ( x ) says that all polar bears are white, which (ignoring evolutionary anomalies) is a true proposition. But x , ( x U P ( x ) ) means something entirely different: this statement says that every object x is a white polar bear!

1.39. 示例。设 U 为北极熊的集合,P(x) 为开放语句“x 是白色的”。两个命题 ∃x∈U,P(x) 和 ∃x,(x∈U∧P(x)) 都断言存在一只白色的北极熊。类似地,∀x∈U,P(x) 表示所有北极熊都是白色的,这(忽略进化异常)是一个真命题。但是 ∀x,(x∈U∧P(x)) 的含义完全不同:这个命题表示每个对象 x 都是一只白色的北极熊!

We can fix the example by replacing (AND) by ⇒ (IMPLIES). Consider the statement x , ( x U P ( x ) ) . This says that for all objects x 0 , IF x 0 is a polar bear, THEN x 0 is white. Imagine testing the truth of this statement by considering each object x 0 in the whole universe, one at a time. Some objects x 0 are polar bears, and for these objects the IF-statement is true (having a true hypothesis and true conclusion). All remaining objects x 0 are not polar bears, and for such objects the IF-statement is automatically true (since its hypothesis is false). Thus these objects are essentially irrelevant when determining the truth of the quantified statement.

我们可以通过将 ∧(与)替换为 ⇒(蕴含)来修正这个例子。考虑命题 ∀x,(x∈U⇒P(x))。该命题表示,对于所有对象 x 0 ,如果 x 0 是北极熊,那么 x 0 是白色的。想象一下,要检验这个命题的真假,我们需要逐一考虑整个宇宙中的每个对象 x 0 。有些对象 x 0 是北极熊,对于这些对象,如果命题为真(假设和结论都为真)。所有剩余的对象 x 0 都不是北极熊,对于这些对象,如果命题自动为真(因为其假设为假)。因此,在判断量化命题的真假时,这些对象本质上是无关紧要的。

The rule suggested by the example holds in general:

该示例所揭示的规则具有普遍适用性:

x x U U , P P ( x x ) iff 当……时 x x , ( x x U U P P ( x x ) ) .

Intuitively, if the left side is true, then the implication on the right side must be true for each fixed object x 0 satisfying the hypothesis x 0 U . But the implication is always true for all other objects x 0 (those making the hypothesis x 0 U false), so the right side is true. Similarly, if the right side is true, then applying the right side to each object x 0 U shows that P ( x 0 ) is true for all such objects x 0 . So the left side is also true. Later, after discussing proof methods, we will see that this paragraph is essentially a proof of the conversion rule stated above. We also remark that a major reason for defining the IF truth table as we did was to make sure that this quantifier rule would work (compare to our discussion of the statement x R , x > 2 x 2 > 4 on page 11; this statement is equivalent to x R > 2 , x 2 > 4 ).

直观地说,如果左侧为真,那么对于满足假设 x 0 ∈ U 的每个固定对象 x 0 ,右侧的蕴含式必然为真。但对于所有其他对象 x 0 (即那些使假设 x 0 ∈ U 为假的对象),该蕴含式始终为真,因此右侧为真。类似地,如果右侧为真,那么将右侧应用于每个对象 x 0 ∈ U,可知对于所有这样的对象 x 0 ,P(x 0 ) 都为真。因此左侧也为真。稍后,在讨论证明方法之后,我们将看到本段本质上是对上述转换规则的证明。我们还要指出,我们这样定义 IF 真值表的一个主要原因是确保这个量词规则能够生效(比较我们在第 11 页对语句 ∀x∈R,x>2⇒x2>4 的讨论;该语句等价于 ∀x∈R>2,x2>4)。

Translation Examples

翻译示例

Here are some examples illustrating the process of translating back and forth between English statements and logical symbolism. We use these universes and open sentences: U is the set of roses; V is the set of violets; C is the set of carrots; R ( x ) means x is red; P ( x ) means x is purple; O ( x ) means x is orange; and B ( x ) means x is beautiful. In the first few examples, we encode English statements, first using restricted quantifiers, and then converting to unrestricted quantifiers.

以下是一些示例,说明如何在英语语句和逻辑符号之间进行双向转换。我们使用以下集合和开放语句:U 是玫瑰的集合;V 是紫罗兰的集合;C 是胡萝卜的集合;R(x) 表示 x 是红色的;P(x) 表示 x 是紫色的;O(x) 表示 x 是橙色的;B(x) 表示 x 是美丽的。在前几个示例中,我们首先使用受限量词对英语语句进行编码,然后再转换为非受限量词。

(a) “All violets are purple.” The word “all” signifies a universal statement, so we write x V , P ( x ) . Eliminating the restricted quantifier gives x , ( x V P ( x ) ) .

(a)“所有紫罗兰都是紫色的。” “所有”一词表示全称命题,因此我们写成 ∀x∈V,P(x)。去掉限定量词后得到 ∀x,(x∈V⇒P(x))。

(b) “Some roses are orange.” Here, “some” indicates an existential statement, so we get x U , O ( x ) , which in turn becomes x , ( x U O ( x ) ) . Note that the logical statement would be true even if there were only one orange rose, despite the use of the plural “roses” in the English statement.

(b) “有些玫瑰是橙色的。”这里,“有些”表示存在性命题,因此我们得到 ∃x∈U,O(x),进而得到 ∃x,(x∈U∧O(x))。注意,即使只有一朵橙色的玫瑰,这个逻辑命题也为真,尽管英文命题中使用了复数形式的“玫瑰”。

(c) “Every carrot is orange or purple.” We encode this as x C , ( O ( x ) P ( x ) ) or equivalently x , ( x C ( O ( x ) P ( x ) ) ) .

(c)“每根胡萝卜都是橙色或紫色。”我们将其编码为 ∀x∈C,(O(x)∨P(x)) 或等价地 ∀x,(x∈C⇒(O(x)∨P(x)))。

(d) “There is a purple and orange carrot.” We encode this as x C , ( P ( x ) O ( x ) ) or equivalently x , ( x C P ( x ) O ( x ) ) . On the other hand, “Purple carrots exist, and orange carrots exist” is a different statement, which could be encoded as

(d) “存在紫色和橙色的胡萝卜。” 我们将其编码为 ∃x∈C,(P(x)∧O(x)) 或等价地 ∃x,(x∈C∧P(x)∧O(x))。另一方面,“存在紫色胡萝卜,也存在橙色胡萝卜”是另一个陈述,可以编码为

( x x , ( x x C C P P ( x x ) ) ) ( y , ( y C C O ( y ) ) ) .

(e) “Any red rose is beautiful.” We have not introduced a universe of red roses, but we can use unrestricted quantifiers to write x , ( R ( x ) x U ) B ( x ) . We could also say x U , ( R ( x ) B ( x ) ) .

(e) “任何红玫瑰都很美丽。”我们并没有引入一个由红玫瑰组成的宇宙,但我们可以使用不受限制的量词来写 ∀x,(R(x)∧x∈U)⇒B(x)。我们也可以写成 ∀x∈U,(R(x)⇒B(x))。

(f) “If purple carrots exist, then each white rose is beautiful.” In this sentence, two quantifiers appear within an IF-THEN construction. One possible encoding is

(f) “如果紫色胡萝卜存在,那么每一朵白玫瑰都是美丽的。” 在这句话中,两个量词出现在一个 IF-THEN 结构中。一种可能的编码是

[ x x , ( x x C C P P ( x x ) ) ] [ y , ( ( W W ( y ) y U U ) B B ( y ) ) ] .

Note that adjectives modifying a noun produce symbols in the encoding, whether the noun is existentially or universally quantified.

请注意,修饰名词的形容词在编码中会产生 ∧ 符号,无论该名词是存在量化还是全称量化。

(g) “ x + y = y + x for all real numbers x and y .” Although English grammar allows a quantifier to appear after the variable it quantifies, formal logic requires the quantifier to come first. Thus we write x R , y R , x + y = y + x . Note that the word “and” in this sentence does not translate into the logical operator , but allows the single phrase “for all” to stand for two universal quantifiers.

(g) “对于所有实数 x 和 y,x + y = y + x。” 虽然英语语法允许量词出现在它所量化的变量之后,但形式逻辑要求量词必须位于变量之前。因此,我们写成 ∀x∈R,∀y∈R,x+y=y+x。请注意,这句话中的“and”并不等同于逻辑运算符 ∧,而是允许用“对于所有”这个短语来表示两个全称量词。

(h) “ n = 2 k for some integer k .” Here too, the quantifier at the end of the English sentence must be moved to the front in the logical version, giving k Z , n = 2 k .

(h)“对于某个整数 k,n = 2k。”这里,英文句子末尾的量词在逻辑版本中也必须移到前面,得到 ∃k∈Z,n=2k。

(i) “Roses are red, carrots are orange.” This poetic sentiment reveals a new and unpleasant feature of English: hidden quantifiers and logical operators. The use of the plural “roses” and “carrots” implicitly indicates that these statements are intended to apply to all roses and all carrots. Moreover, the comma joining the two phrases has the same meaning as AND. Thus we write ( x , ( x U R ( x ) ) ( y , ( y C O ( y ) ) . In this case, we could get away with a single quantifier for both clauses, writing

(i) “玫瑰是红色的,胡萝卜是橙色的。” 这句诗意的表达揭示了英语中一个令人不快的全新特征:隐藏的量词和逻辑运算符。复数形式的“玫瑰”和“胡萝卜”隐含地表明,这些陈述旨在适用于所有玫瑰和所有胡萝卜。此外,连接两个短语的逗号与“与”的含义相同。因此,我们可以写成 (∀x,(x∈U⇒R(x))∧(∀y,(y∈C⇒O(y))。在这种情况下,我们可以使用一个量词来表示两个子句,写成:

z z , ( ( z z U U R ( z z ) ) ( z z C C O ( z z ) ) ) .

But if had been ∨ in both statements, the two encodings above would not be equivalent. The first encoding now says “All roses are red OR all carrots are orange,” which is false (as white roses and purple carrots do exist!). But the second encoding (upon eliminating IF) states that every object is not a rose or is red or is not a carrot or is orange, which is true.

但如果两个语句中的 ∧ 都变成了 ∨,那么上面的两种编码就不等价了。第一种编码现在表示“所有玫瑰都是红色的,或者所有胡萝卜都是橙色的”,这是错误的(因为白玫瑰和紫色胡萝卜确实存在!)。而第二种编码(去掉 IF 之后)表示每个物体都不是玫瑰、不是红色、不是胡萝卜或不是橙色,这是正确的。

(j). “Each even integer greater than 2 is the sum of two primes.” This sentence (called Goldbach’s Conjecture ) contains two hidden existential quantifiers signaled by the verb “is.” We can encode it as follows:

(j)“大于 2 的每个偶数都是两个质数之和。”这句话(称为哥德巴赫猜想)包含两个隐藏的存在量词,由动词“是”表示。我们可以将其编码如下:

x x Z Z , ( x x is even 甚至 x x > 2 ) [ y Z Z , z z Z Z , ( y is prime 是主要 z z is prime 是主要 x x = y + z z ) ] .

As this example illustrates, we often allow ourselves to use restricted quantifiers involving a “standard” universe (like Z ), but we apply the conversion rules to move the non-standard restrictions on the variables (like being even, or being prime) into the propositional part of the statement, rather than inventing new universes for the set of even integers larger than 2 or the set of prime integers.

正如这个例子所说明的,我们经常允许自己使用涉及“标准”宇宙(如 Z)的受限量词,但我们会应用转换规则,将变量的非标准限制(如偶数或素数)移到语句的命题部分,而不是为大于 2 的偶数集合或素数集合发明新的宇宙。

Negating Quantified Statements

否定量化陈述

We continue our translation examples, letting U be the universe of roses and letting R ( x ) mean “ x is red.” The next family of examples reveals the subtleties that arise when words with a negating effect are mixed with quantifiers.

我们继续举翻译例子,设 U 为玫瑰花的宇宙,R(x) 表示“x 是红色的”。下一组例子揭示了否定词与量词混合使用时产生的微妙之处。

1.40. Example. Encode each sentence as literally as possible, using only unrestricted quantifiers.

1.40. 示例。尽可能按字面意思对每个句子进行编码,只使用不受限制的量词。

(a) All roses are not red.

(a)并非所有的玫瑰都是红色的。

(b) Not all roses are red.

(b)并非所有的玫瑰都是红色的。

(c) No roses are red.

(c)没有红色的玫瑰。

(d) There does not exist a red rose.

(d)世上没有红玫瑰。

(e) Red roses do not exist.

(e)红玫瑰并不存在。

(f) Some roses are not red.

(f)有些玫瑰不是红色的。

(g) There is a non-red rose.

(g)有一朵非红色的玫瑰。

Solution. (a) We initially say x U , R ( x ) , and convert this to x , ( x U ⇒∼ R ( x ) ) . (b) In this example, the word “not” applies to everything following it, so we get x U , R ( x ) , which converts to x , ( x U R ( x ) ) . When we say “no roses” in part (c), we are making a negative statement about all roses, leading first to x U , R ( x ) and then to x , ( x U ⇒∼ R ( x ) ) . (d) becomes x , ( x U R ( x ) ) ; (e) is encoded in exactly the same way, even though the word “exist” appears at the end. (f) says x , ( x U R ( x ) ) , and (g) is encoded in the same way. Intuitively, each of the four statements (a), (c), (d), and (e) is saying the same thing. Similarly, statements (b), (f), and (g) say the same thing but are not the same as the other four statements. Using restricted quantifiers now, these observations mean that x U , R ( x ) is equivalent to x U , R ( x ) , whereas x U , R ( x ) is equivalent to x U , R ( x ) .

解答。(a) 我们首先写成 ∀x∈U,∼R(x),然后将其转换为 ∀x,(x∈U⇒∼R(x))。(b) 在这个例子中,“非”这个词适用于它后面的所有内容,所以我们得到 ∼∀x∈U,R(x),它转换为 ∼∀x,(x∈U⇒R(x))。当我们在 (c) 部分说“没有玫瑰”时,我们是在对所有玫瑰做出否定陈述,首先得到 ∀x∈U,∼R(x),然后得到 ∀x,(x∈U⇒∼R(x))。(d) 变为 ∼∃x,(x∈U∧R(x));(e) 的编码方式完全相同,即使“存在”这个词出现在末尾。(f) 写成 ∃x,(x∈U∧∼R(x)),(g) 的编码方式也相同。直观上,语句 (a)、(c)、(d) 和 (e) 表达的是同一个意思。类似地,语句 (b)、(f) 和 (g) 表达的是同一个意思,但与其他四个语句并不相同。现在使用限制量词,这些观察结果表明 ∀x∈U,∼R(x) 等价于 ∼∃x∈U,R(x),而 ∃x∈U,∼R(x) 等价于 ∼∀x∈U,R(x)。

The equivalences at the end of the last example hold in general. In other words, for any universe U and any open sentence P ( x ), we have the quantifier negation rules :

上一个例子末尾的等价关系具有普遍性。换句话说,对于任意论域 U 和任意开语句 P(x),我们有量词否定规则:

x x U U , P P ( x x ) iff 当……时 x x U U , P P ( x x ) ;
x x U U , P P ( x x ) iff 当……时 x x U U , P P ( x x ) .

We justify the first rule intuitively, as follows. The left side of this rule is the negation of the statement “There is an x 0 in U making P ( x 0 ) true.” How can this existential statement fail to hold? The statement fails precisely when every single object x 0 in U fails to make P ( x 0 ) true. This is exactly the universal statement on the right side of the rule.

我们可以直观地证明第一条规则,如下所示。这条规则的左侧是对命题“U 中存在一个 x 0 使得 P(x 0 ) 为真”的否定。这个存在命题怎么可能不成立呢?当 U 中的每个对象 x 0 都不能使 P(x 0 ) 为真时,该命题就不成立。这恰好就是规则右侧的全称命题。

Similarly, the left side of the second rule asserts the falsehood of the statement, “For every x 0 in U , P ( x 0 ) is true.” When does this universal statement fail? It fails when, and only when, at least one object x 0 in U fails to make P ( x 0 ) true. This is exactly the existential statement on the right side of the second rule.

类似地,第二条规则的左侧断言了以下命题为假:“对于 U 中的每个对象 x 0 ,P(x 0 ) 为真。” 这个全称命题何时不成立?它不成立,当且仅当 U 中至少存在一个对象 x 0 使得 P(x 0 ) 不成立时。这正是第二条规则右侧的存在性命题。

Negations and Denials

否定和否认

Given any proposition P , recall that the negation of P is the proposition ~ P , which has the opposite truth value as P . We say that a proposition Q is a denial of P iff Q is logically equivalent to ~ P . In general, if P is a complicated statement built from logical operators and quantifiers, then the negation ~ P can be difficult to work with directly. Thus we need to develop techniques for passing from the particular denial ~ P of P to another denial that does not begin with the negation symbol. It turns out that the most useful denials of P are those in which the negation symbol is not applied to any substatement involving a logical operator or quantifier symbol.

给定任意命题 P,回想一下,P 的否定是命题 ~P,其真值与 P 相反。我们称命题 Q 是 P 的否定,当且仅当 Q 在逻辑上等价于 ~P。一般来说,如果 P 是由逻辑运算符和量词构成的复杂语句,那么直接处理其否定 ~P 可能比较困难。因此,我们需要开发一些技巧,以便从 P 的特定否定 ~P 过渡到另一个不以否定符号开头的否定。事实证明,P 最有用的否定是那些否定符号不应用于任何包含逻辑运算符或量词的子语句的否定。

How can we find a useful denial of P ? In this section and previous ones, we have already derived the rules we need to simplify expressions that begin with NOT. These rules are summarized in the following table. The tricky aspect of this table is that we usually need to apply several rules recursively to convert the negation ~ P into the most useful denial in which no subexpression begins with NOT.

如何找到 P 的一个有用的否定?在本节及之前的章节中,我们已经推导出了简化以 NOT 开头的表达式所需的规则。这些规则总结在下表中。该表的难点在于,我们通常需要递归地应用多个规则,才能将否定 ~P 转换为最有用的否定,其中任何子表达式都不以 NOT 开头。

Statement

陈述

Denial of Statement

否认声明

Symbolic Version of Rule

规则的符号版本

A and B

A 和 B

(denial of A) or (denial of B)

(否认A)或(否认B)

( A B ) ( A ) ( B )

∼(A∧B)≡(∼A)∨(∼B)

A or B

A 或 B

(denial of A) and (denial of B)

(否认A)和(否认B)

( A B ) ( A ) ( B )

∼(A∨B)≡(∼A)∧(∼B)

if A, then B

如果 A,则 B

A and (denial of B)

A 和(否认 B)

( A B ) A ( B )

∼(A⇒B)≡A∧(∼B)

not A

不是A

A

一个

~(~ A ) ≡ A

~(~A) ≡ A

For all x , P ( x )

对于所有 x,P(x)

There is x , (denial of P ( x ))

存在 x,(否定 P(x))

x U , P ( x ) iff x U , P ( x )

∼∀x∈U,P(x) 当且仅当 ∃x∈U,∼P(x)

There is x , P ( x )

存在 x,P(x)

For all x , (denial of P ( x ))

对于所有 x,(否定 P(x))

x U , P ( x ) iff x U , P ( x )

∼∃x∈U,P(x) 当且仅当 ∀x∈U,∼P(x)

A iff B

A 当且仅当 B

A or B, not both

A 或 B,两者不可兼得

( A B ) ( A B )

#

∼(A⇔B)≡(A⊕B)

A or B, not both

A 或 B,两者不可兼得

A iff B

A 当且仅当 B

( A B ) ( A B )

∼(A⊕B)≡(A⇔B)

Informally, applying a negation operator converts AND to OR, OR to AND, FOR ALL to EXISTS, EXISTS to FOR ALL, and in these cases we must continue recursively, denying the inputs to these operators. On the other hand, negation converts IFF to XOR and XOR to IFF without needing to negate the inputs (see also Exercises 17 and 18). Applying a negation to something that already begins with NOT removes the NOT. Finally, the most troublesome entry in the table is the denial rule for IF. For emphasis, we repeat the key fact here:

通俗地说,应用否定运算符会将 AND 转换为 OR,OR 转换为 AND,FOR ALL 转换为 EXISTS,EXISTS 转换为 FOR ALL,在这些情况下,我们必须递归地继续否定这些运算符的输入。另一方面,否定运算符会将 IFF 转换为 XOR,XOR 转换为 IFF,而无需否定输入(另见练习 17 和 18)。对已经以 NOT 开头的语句应用否定运算符会移除 NOT。最后,表格中最棘手的条目是 IF 的否定规则。为了强调,我们在此重申关键事实:

The denial of an IF-statement is an AND-statement, not another IF-statement !

否定一个 IF 语句是一个 AND 语句,而不是另一个 IF 语句!

One way to understand this is to recall that “if A then B” is equivalent to “(not A) or B.” Negating the latter expression produces “A and (not B),” as asserted in the table.

理解这一点的一种方法是回想一下,“如果 A 则 B”等价于“(非 A)或 B”。对后一个表达式取反,即可得到“A 且(非 B)”,如表格所示。

1.41. Example. Find useful denials of each statement.

1.41. 示例。找出对每个陈述的有用否定。

(a) 3 is odd or 5 is not prime.

(a)3 是奇数或 5 不是质数。

(b) If a < b < c , then a < c .

(b)如果 a < b < c,则 a < c。

(c) a < c only if a < b < c .

(c)a < c 当且仅当 a < b < c。

Not everyone likes math.

并非人人都喜欢数学。

Solution. A denial of (a) is “3 is not odd and 5 is prime.” Denying the IF-statement in (b) produces “ a < b < c and a c ” (assuming a , b , c are real constants). Statement (c) contains two traps. First, the “only if” construction in (c) is equivalent (before negating) to the statement “If a < c , then a < b < c .” Second, the phrase a < b < c is really an abbreviation for “ a < b and b < c .” Thus, a denial of (c) is “ a < c and ( a b or b c ).” It would be incorrect to replace the parenthesized expression by a b c , which still contains an implicit AND. Note also that (for real numbers) “not ( a < b )” becomes a b , as opposed to a > b .

解答。否定 (a) 可得“3 不是奇数且 5 是质数”。否定 (b) 中的 IF 语句可得“a < b < c 且 a ≥ c”(假设 a、b、c 为实数常数)。语句 (c) 包含两个陷阱。首先,(c) 中的“仅当”结构(在取反之前)等价于语句“如果 a < c,则 a < b < c”。其次,短语 a < b < c 实际上是“a < b 且 b < c”的缩写。因此,否定 (c) 可得“a < c 且 (a ≥ b 或 b ≥ c)”。用 a ≥ b ≥ c 替换括号内的表达式是不正确的,因为后者仍然包含隐含的 AND 运算。另请注意(对于实数),“非 (a < b)”变为 a ≥ b,而不是 a > b。

Finally, the negation of (d) is “It is false that not everyone likes math.” A useful denial of (d) is “Everyone likes math.” (Astute readers may have noticed that (d) and its denials are not really propositions.)

最后,(d) 的否定是“并非每个人都喜欢数学”这句话是错误的。(d) 的一个有用的否定是“每个人都喜欢数学”。(细心的读者可能已经注意到,(d) 及其否定实际上并不是命题。)

Rant on Hidden Quantifiers (Optional)

关于隐藏量词的吐槽(可选)

We have now seen several examples of hidden or implicit quantifiers in English sentences, in which universal or existential quantifiers are suggested by grammatical constructions (like plural subjects) rather than explicitly designated by phrases like “for all” or “there exists.” My personal position is that all mathematical writers should avoid the use of all hidden quantifiers in all circumstances . The reason is that one of the most common sources of error and confusion among beginners (and experts!) in this subject is forgetting to quantify a variable, confusing universal quantifiers with existential quantifiers, changing the meaning of a statement by moving a quantifier to a new location, or misinterpreting the meaning or scope of a quantifier. The correct use of quantifiers is one of the keys to success in reading and writing proofs, and it is one of the hardest skills for newcomers to master. Why, then, should we make it even harder by gratuitously omitting or disguising the quantifiers used in our own writing? I have tried very hard in this text (and elsewhere) to quantify every variable appearing in all proofs and other formal discussions, and I implore you to exert similar efforts in your own writing.

我们已经看到了英语句子中一些隐藏或隐含量词的例子,其中全称量词或存在量词并非像“for all”或“there exists”那样明确地用短语指代,而是通过语法结构(例如复数主语)暗示出来的。我个人认为,所有数学写作者都应该在任何情况下避免使用隐藏量词。原因在于,初学者(以及专家!)在这个领域最常犯的错误和困惑之一就是忘记量化变量、混淆全称量词和存在量词、通过移动量词的位置来改变语句的含义,或者误解量词的含义或范围。正确使用量词是成功阅读和撰写证明的关键之一,也是初学者最难掌握的技能之一。那么,我们为什么要故意省略或掩盖自己写作中使用的量词,从而让学习变得更加困难呢?我在本文(以及其他地方)中尽力量化了所有证明和其他正式讨论中出现的每一个变量,我恳请你们在自己的写作中也付出类似的努力。

Nevertheless, you must be aware that the majority of mathematical authors do not share my view and use hidden quantifiers all over the place. Here is a particularly common convention that you must know: in all theorem statements, any unquantified variables are understood to be universally quantified. The universe of objects often must be inferred from context as well. For example, an algebra text might announce as a theorem the commutative law for addition: x + y = y + x . Filling in quantifiers, the actual theorem being presented is x R , y R , x + y = y + x . However, this convention must be used with care, keeping in mind all the various grammatical constructions that can serve the purpose of quantification. Consider, for instance, the statement: “An even integer n > 2 can be written as the sum of two primes p and q .” The article “an” indicates universal quantification of n , whereas the verb “can be” indicates existential quantification of p and q . To reduce the chance of confusion, and to draw attention to the fact that quantifiers are being used, in this text I prefer the symbols and (or the standard English phrases “for all” and “there exists”) to introduce quantified variables. So I would have written the statement under consideration like this: “For all even integers n > 2, there exist primes p and q such that n = p + q .” One final grammatical point: note the plural verb form EXIST (not EXISTS) agrees with the plural subject PRIMES occurring later in the sentence.

然而,您必须意识到,大多数数学作者并不认同我的观点,他们经常在数学中大量使用隐式量词。这里有一个您必须了解的常见约定:在所有定理陈述中,任何未量化的变量都被理解为全称量词。对象的范围通常也需要从上下文中推断出来。例如,一本代数教材可能会将加法交换律作为定理提出:x + y = y + x。填入量词后,实际的定理是:∀x∈R,∀y∈R,x+y=y+x。但是,必须谨慎使用这种约定,并牢记所有可以用于量化的各种语法结构。例如,考虑以下陈述:“一个大于 2 的偶数 n 可以表示为两个素数 p 和 q 的和。” 定冠词“an”表示对 n 的全称量词,而动词“can be”表示对 p 和 q 的存在量词。为了减少混淆的可能性,并强调量词的使用,本文中我倾向于使用符号 ∀ 和 ∃(或标准的英语短语“for all”和“there exists”)来引入量化变量。因此,我会这样写待讨论的语句:“对于所有大于 2 的偶数 n,存在素数 p 和 q,使得 n = p + q。” 最后一点语法要点:请注意,动词的复数形式 EXIST(而非 EXISTS)与句子后面出现的复数主语 PRIMES 保持一致。

Exercise: Find all the places in this text where, despite my comments here, I inadvertently used a hidden quantifier.

练习:找出本文中所有尽管我在此处进行了评论,但我还是无意中使用了隐藏量词的地方。

Section Summary

章节概要

1. Rules for Converting and Negating Quantifiers.

1.量词转换和否定规则。

x x U U , P P ( x x ) x x , ( x x U U P P ( x x ) ) x x U U , P P ( x x ) x x , ( x x U U P P ( x x ) ) x x U U , P P ( x x ) x x U U , P P ( x x ) x x U U , P P ( x x ) x x U U , P P ( x x )

2. Translation Points. (a) In English sentences, changing the relative positions of quantifiers and logical keywords (especially negative words like “not”) can affect the meaning.

2.翻译要点。(a) 在英语句子中,改变量词和逻辑关键词(尤其是像“not”这样的否定词)的相对位置会影响句子的意思。

(b) In symbolic logic (unlike English), quantifiers must precede the variables they modify.

(b)在符号逻辑中(与英语不同),量词必须位于它们所修饰的变量之前。

(c) Quantifiers can occur within other logical constructions ( example: “if all roses are red, then some violets are blue”). Moving these quantifiers to the front of the symbolic version of the statement can change the meaning.

(c) 量词可以出现在其他逻辑结构中(例如:“如果所有的玫瑰都是红色的,那么有些紫罗兰是蓝色的”)。将这些量词移到语句符号形式的前面可以改变语句的含义。

(d) Quantifiers often occur implicitly in English; for instance, a plural subject may signal a universal quantifier, whereas the verb “is” can disguise one or more existential quantifiers.

(d)量词在英语中经常隐含出现;例如,复数主语可能表示全称量词,而动词“is”可以掩盖一个或多个存在量词。

3. Finding Useful Denials. Applying a negation operator to a complex statement has the following effect on the outermost logical operator: becomes ∨, ∨ becomes , becomes , becomes , ~ cancels out, ⇔ becomes , and becomes ⇔. In the case of , ∨, , and , we must continue by recursively denying the inputs to these operators. Regarding IF, chant this statement aloud five times: THE DENIAL OF A B IS ( B ) . The denial rules are summarized in the table on page 35.

3. 寻找有用的否定。对复杂语句应用否定运算符会对最外层的逻辑运算符产生以下影响:∧ 变为 ∨,∨ 变为 ∧,∀ 变为 ∃,∃ 变为 ∀,~ 被抵消,⇔ 变为 ⊕,⊕ 变为 ⇔。对于 ∧、∨、∀ 和 ∃,我们必须继续递归地否定这些运算符的输入。关于 IF 语句,请大声念诵以下语句五遍:否定 A⇒BIS∧(∼B)。否定规则总结在第 35 页的表格中。

Exercises

练习

1. Convert each statement to an equivalent statement using unrestricted quantifiers.

1.将每个语句转换为使用不受限制的量词的等效语句。

(a) n Z , n is even n is odd . (b) n Z , n is odd n is perfect . (c) x R , x > 3 x + 2 > 5 . (d) q Q > 0 , r Q < 0 , q 3 + r 3 = 1 . (e) x R , n Z , x < n .

(a) ∀n∈Z,n为偶数⊕n为奇数。(b) ∃n∈Z,n为奇数∧n为完全数。(c) ∀x∈R,x>3⇒x+2>5。(d) ∃q∈Q>0,∃r∈Q<0,q³+r³=1。(e) ∀x∈R,∃n∈Z,x<n。

2. Eliminate all propositional operators from the following expressions by converting to restricted quantifiers.

2.通过转换为受限量词,消除下列表达式中的所有命题运算符。

(a) x , x Q x 2 = 3 . (b) x , x Z x 2 Z 0 . (c) x , y , ( x R y R ) x + y R . (d) x , [ x Q ( n , n Z > 0 x n Z ) ] . (e) [ k , k Z n = 2 k ] [ m , m Z n + 3 = 2 m + 1 ] .

(a) ∃x,x∈Q∧x²=3. (b) ∀x,x∈Z⇒x²∈Z≥0. (c) ∀x,∀y,(x∈R∧y∈R)⇒x+y∈R. (d) ∀x,[x∈Q⇒(∃n,n∈Z>0∧xⁿ∈Z)]. (e) [∃k,k∈Z∧n=2k]⇒[∃m,m∈Z∧n+3=2m+1].

3. Let U be the set of people, let T ( x ) mean x is 20 feet tall, and let P be the statement “ x U , T ( x ) .” Write statements equivalent to P using: (a) an unrestricted existential quantifier; (b) a restricted universal quantifier; (c) an unrestricted universal quantifier. Give literal English translations of P and your three answers.

3. 设 U 为人群集合,T(x) 表示 x 的身高为 20 英尺,P 为命题“∼∃x∈U,T(x)”。请分别使用以下三种方式写出与 P 等价的命题:(a) 无限制存在量词;(b) 受限全称量词;(c) 无限制全称量词。请给出 P 的英文直译以及你的三种答案。

4. Let L be the set of lobsters, let R ( x ) mean x is red, and let Q be the statement

4.设 L 为龙虾的集合,R(x) 表示 x 是红色的,Q 为以下陈述:

x L , R ( x ) .” Write statements equivalent to Q using: (a) an unrestricted universal quantifier; (b) a restricted existential quantifier; (c) an unrestricted existential quantifier. Give literal English translations of Q and your three answers.

“∼∀x∈L,R(x).” 请用以下三种方式写出与问题 Q 等价的语句:(a) 无限制的全称量词;(b) 受限的存在量词;(c) 无限制的存在量词。请给出问题 Q 及其三个答案的英文直译。

5. Consider the true sentence “Every human is male or female.” Someone incorrectly encodes this sentence as: ( x H , M ( x ) ) ( x H , F ( x ) ) , where H is the set of humans, M ( x ) means x is male, and F ( x ) means x is female. (a) Translate this encoding back into English (do not use variables or symbols), and explain how the meaning differs from the original sentence. (b) Give a correct encoding of the original sentence.

5. 考虑真命题“每个人都是男性或女性”。有人错误地将此命题编码为:(∀x∈H,M(x))∨(∀x∈H,F(x)),其中 H 代表人类集合,M(x) 表示 x 为男性,F(x) 表示 x 为女性。(a) 将此编码翻译回英文(不使用变量或符号),并解释其含义与原命题有何不同。(b) 给出原命题的正确编码。

6. Give a specific example of a universe U and open sentences P ( x ) and Q ( x ) such that the two statements x U , ( P ( x ) Q ( x ) ) and [ x U , P ( x ) ] [ x U , Q ( x ) ] have different truth values. Explain the difference in meaning between these two statements.

6. 请举一个具体的例子,说明是否存在一个全集 U 和两个开语句 P(x) 和 Q(x),使得两个命题 ∃x∈U,(P(x)∧Q(x)) 和 [∃x∈U,P(x)]∧[∃x∈U,Q(x)] 的真值不同。解释这两个命题含义上的区别。

7. Let H be the set of humans, and let M ( x ) mean x is mortal. Translate each symbolic formula into an English sentence that does not use propositional logic keywords such as AND, IF, etc. Which sentences are true? (a) x , x H M ( x ) . (b) x , x H M ( x ) . (c) x , x H ( M ( x ) ) . (d) x , x H ( M ( x ) ) .

7. 设 H 为人类集合,M(x) 表示 x 是会死的。将下列符号公式翻译成不使用命题逻辑关键词(例如 AND、IF 等)的英文句子。哪些句子是正确的?(a)∀x,x∈H⇒M(x)。(b)∀x,x∈H∧M(x)。(c)∃x,x∈H∧(∼M(x))。(d)∃x,x∈H⇒(∼M(x))。

8. Find useful denials of each statement. All answers should be English sentences. (a) John has brown hair and blue eyes. (b) If x 0 is even then 3 x 0 is odd. (c) All prime numbers are odd. (d) Purple cows do not exist. (e) There are deserts with red sand. (f) The sine function is continuous or the secant function is not bounded.

8.找出对下列陈述的否定。所有答案均应为英文句子。(a) 约翰有棕色的头发和蓝色的眼睛。(b) 如果 x 0 是偶数,那么 3x 0 是奇数。(c) 所有质数都是奇数。(d) 不存在紫色的牛。(e) 存在红色沙子的沙漠。(f) 正弦函数是连续的,或者正割函数无界。

9. Find useful denials of each statement. All answers should be English sentences.

9.找出对每项陈述的有效反驳。所有答案均应为英文句子。

(a) Jamestown was founded in 1607 or Williamsburg is not the capital of North Dakota. (b) 2 + 2 = 4 if 3 + 3 = 7. (c) All rubies are red. (d) Some diamonds are blue. (e) Not all elephants are pink. (f) All carrots are orange iff all rabbits are white. (g) Every integer is either positive or negative, but not both. (h) For all natural numbers x , if x is prime and even, then x = 2.

(a) 詹姆斯敦建于1607年,或者威廉斯堡不是北达科他州的首府。(b) 如果3 + 3 = 7,那么2 + 2 = 4。(c) 所有红宝石都是红色的。(d) 有些钻石是蓝色的。(e) 并非所有大象都是粉色的。(f) 所有胡萝卜都是橙色的当且仅当所有兔子都是白色的。(g) 每个整数要么是正数,要么是负数,但不能同时是正数和负数。(h) 对于所有自然数x,如果x是质数且是偶数,那么x = 2。

10. Convert each proposition to the standard form P Q , and then give a useful denial of the statement.

10.将每个命题转化为标准形式 P ⇒ Q,然后给出对该命题的有用否定。

(a) G has a proper normal subgroup only if G is not simple. (b) A sufficient condition for normality of X is metrizability of X . (c) For f to be Lebesgue integrable, it is necessary that f be measurable. (d) Invertibility of A implies det ( A ) 0 . (e) Whenever n 0 is a power of an odd prime, n 0 has a primitive root. (f) Injectivity of g is sufficient for g to have an inverse.

(a) G 存在真正规子群当且仅当 G 不是单群。(b) X 是正规群的一个充分条件是 X 可度量。(c) f 是勒贝格可积群的必要条件是 f 是可测群。(d) A 的可逆性蕴含 det(A)≠0。(e) 当 n 0 是奇素数的幂时,n 0 存在本原根。(f) g 的单射性是 g 存在逆的充分条件。

11. True or false? Explain.

11. 判断正误并解释原因。

(a) x Z , x 2 = 4 . (b) x , x Z x 2 = 4 . (c) x , x Z x 2 = 4 . (d) x Z , x + 1 = 1 + x . (e) x , x Z x + 1 = 1 + x . (f) x , x Z x + 1 = 1 + x .

(a) ∃x∈Z,x²=−4. (b) ∃x,x∈Z∧x²=−4. (c) ∃x,x∈Z⇒x²=−4. (d) ∀x∈Z,x+1=1+x. (e) ∀x,x∈Z⇒x+1=1+x. (f) ∀x,x∈Z∧x+1=1+x.

12. Encode each English sentence in symbolic form. Only use unrestricted quantifiers in your answers. Let B ( x ) mean x is black, R ( x ) mean x is a raven, and E ( x ) mean x is evil.

12. 将下列英文句子编码成符号形式。答案中只能使用无限制量词。设 B(x) 表示 x 是黑色的,R(x) 表示 x 是乌鸦,E(x) 表示 x 是邪恶的。

Sample Question: All black ravens are evil. Answer: x , [ ( B ( x ) R ( x ) ) E ( x ) ] .

示例问题:所有黑渡鸦都是邪恶的。答案:∀x,[(B(x)∧R(x))⇒E(x)].

(a) Some ravens are evil. (b) Non-black ravens exist. (c) Although every raven is black, not every raven is evil. (d) Each raven is either black or evil, but not both. (e) If some ravens are black, then all ravens are evil. (f) Not all ravens are evil if some ravens are not black.

(a) 有些乌鸦是邪恶的。(b) 存在非黑色的乌鸦。(c) 虽然每只乌鸦都是黑色的,但并非每只乌鸦都是邪恶的。(d) 每只乌鸦要么是黑色的,要么是邪恶的,但不可能两者兼具。(e) 如果有些乌鸦是黑色的,那么所有的乌鸦都是邪恶的。(f) 如果有些乌鸦不是黑色的,那么并非所有乌鸦都是邪恶的。

13. Write useful denials of each statement in the previous exercise; your answers should be English sentences.

13.请用英语句子对上一练习中的每个陈述写出有用的反驳。

14. Let M be the set of melons, let V be the set of vegetables, let G ( x ) mean x is green, let O ( x ) mean x is orange, let T ( x ) mean x is tasty, and let J ( x ) mean x is juicy. Encode each English sentence in symbolic form; your final answers should not use restricted quantifiers. (a) There exists an orange vegetable. (b) Every melon is juicy. (c) No green vegetable is tasty. (d) Some melons are green, some are orange, but all are tasty. (e) If some orange vegetable is juicy, then all green melons are tasty. (f) All tasty melons are orange. (g) Juiciness is a necessary condition for a melon to be tasty. (h) Vegetables are green only if they are not juicy.

14. 设 M 为瓜类的集合,V 为蔬菜的集合,G(x) 表示 x 是绿色的,O(x) 表示 x 是橙色的,T(x) 表示 x 好吃,J(x) 表示 x 多汁。请将下列英文句子编码成符号形式;最终答案中不应使用限制性量词。(a) 存在橙色的蔬菜。(b) 每个瓜都多汁。(c) 没有绿色的蔬菜好吃。(d) 有些瓜是绿色的,有些是橙色的,但都很好吃。(e) 如果有些橙色的蔬菜多汁,那么所有绿色的瓜都很好吃。(f) 所有好吃的瓜都是橙色的。(g) 多汁是瓜好吃的必要条件。(h) 蔬菜是绿色的,当且仅当它们不多汁。

15. Write useful denials of each statement in the previous exercise; your answers should be English sentences.

15.请用英语句子对上一练习中的每个陈述写出有用的反驳。

16. Find all hidden quantifiers in the following sentences, and rewrite the sentences to make all quantifiers explicit.

16.找出下列句子中所有隐藏的量词,并改写句子,使所有量词都明确表达出来。

(a) Eagles have wings, but pigs do not. (b) The square of a real number is nonnegative. (c) Positive integers are expressible as sums of four squares. (d) A cyclic group G can be generated by an element x in G .

(a) 老鹰有翅膀,但猪没有。(b) 实数的平方是非负的。(c) 正整数可以表示为四个平方数的和。(d) 循环群 G 可以由 G 中的元素 x 生成。

17. Use truth tables to show that all of the following propositional forms are possible denials of P Q . (a) P ⇔ (~ Q ) (b) (~ P ) ⇔ Q (c) ( P ) ( Q )

17. 使用真值表证明下列所有命题形式都是 P ⇔ Q 的可能否定。(a) P ⇔ (~Q) (b) (~P) ⇔ Q (c) (~P)⊕(~Q)

18. Which of the following propositional forms are correct denials of P Q ? Explain.

18.下列哪些命题形式是对P⊕Q的正确否定?请解释。

(a) ( P ) ( Q ) (b) ( P ) Q (c) (~ P ) ⇔ (~ Q ) (d) ( P ) ( Q ) (e) P ( Q ) (f) ( P Q )

(a) (∼P)∧(∼Q) (b) (∼P)⊕Q (c) (∼P) ⇔ (∼Q) (d) (∼P)⊕(∼Q) (e) P⊕(∼Q) (f) ∼(P⊕Q)

19. Let F ( x ) mean x is a fish, let A ( x ) mean x is an animal, let S ( x ) mean x can swim, let L ( x ) mean x has lungs, and let W ( x ) mean x is white. Use these open sentences to encode each English sentence in symbolic form. (a) Some fish are white. (b) Not all animals have lungs. (c) Every white fish can swim. (d) Some animals that can swim are not fish, but all fish can swim. (e) Being a fish is a sufficient condition to be able to swim. (f) If no fish have lungs, then some white animals cannot swim.

19. 设 F(x) 表示 x 是鱼,A(x) 表示 x 是动物,S(x) 表示 x 会游泳,L(x) 表示 x 有肺,W(x) 表示 x 是白色的。请使用这些开放式句子将下列英文句子编码为符号形式。(a) 有些鱼是白色的。(b) 并非所有动物都有肺。(c) 每条白色的鱼都会游泳。(d) 有些会游泳的动物不是鱼,但所有鱼都会游泳。(e) 是鱼是能够游泳的充分条件。(f) 如果没有鱼有肺,那么有些白色的动物不会游泳。

20. Write useful denials of each statement in the previous exercise; your answers should be English sentences.

20.请用英语句子对上一练习中的每个陈述写出有用的反驳。

21. Encode each sentence in symbolic form, using only unrestricted quantifiers. Let R ( x ) mean x is a rabbit, U ( x ) mean x is a unicorn, W ( x ) mean x is white, and C ( x ) mean x is cute. (a) All white rabbits are cute. (b) Some rabbits are not cute. (c) All unicorns are white but not cute. (d) There are non-white unicorns. (e) Although not every rabbit is white, every unicorn is cute. (f) All rabbits are cute only if some unicorns are white. (g) Some white objects are unicorns or rabbits, but not both. (h) For all objects, a necessary condition for being a unicorn is being white. (i) The non-cuteness of some rabbits is sufficient for the whiteness of all unicorns.

21. 用符号形式编码下列句子,仅使用无限制量词。设 R(x) 表示 x 是兔子,U(x) 表示 x 是独角兽,W(x) 表示 x 是白色的,C(x) 表示 x 很可爱。(a) 所有白兔都很可爱。(b) 有些兔子不可爱。(c) 所有独角兽都是白色的,但不可爱。(d) 存在非白色的独角兽。(e) 虽然不是每只兔子都是白色的,但每只独角兽都很可爱。(f) 所有兔子都很可爱,当且仅当有些独角兽是白色的。(g) 有些白色的物体是独角兽或兔子,但不是两者都是。(h) 对于所有物体而言,它是独角兽的必要条件是它是白色的。(i) 有些兔子不可爱是所有独角兽都是白色的充分条件。

22. Write useful denials of each statement in the previous exercise; your answers should be English sentences.

22.请对上一练习中的每个陈述写出有用的否定句;你的答案应该用英语句子表达。

23. Find all the hidden quantifiers in the true-false exercises of §1.3 . Consider whether this changes any of your answers to those exercises (e.g., if some whales live in aquariums rather than in the ocean).

23.找出第 1.3 节真假练习中所有隐藏的量词。考虑一下这是否会改变你对这些练习的任何答案(例如,如果有些鲸鱼生活在水族馆而不是海洋中)。

1.6 Denial Practice and Uniqueness Statements

1.6 否认实践和独特性声明

Finding useful denials of complex logical statements is an absolutely fundamental skill needed to understand and create mathematical proofs. The only way to master this skill, like so many others, is practice ! You will practice translating and denying various statements in the first half of this section. Afterward, we consider one of the more subtle ideas in quantified logic — the concept of uniqueness .

找到复杂逻辑命题的有效否定是理解和构建数学证明的一项绝对基础的技能。掌握这项技能的唯一途径,如同掌握其他许多技能一样,就是练习!在本节的前半部分,你将练习翻译和否定各种命题。之后,我们将探讨量化逻辑中一个较为微妙的概念——唯一性。

Practice with Useful Denials

练习有效的否认

In the last section, we gave a table summarizing the rules needed to find useful denials of logical statements. Review that table now, memorizing all the rules, and noting particularly that the denial of an IF-statement is an AND-statement . Next, spend some time working out useful denials of the statements in the next example. Solutions and discussion appear in the next subsection. For complicated sentences, you might begin by converting the sentence (partially or completely) into symbolic form, to see how logical operators nest within the sentence. It can also help to rewrite potentially confusing phrases such as “only if” or “it is necessary” into standard IF-THEN form.

上一节中,我们提供了一个表格,总结了找到逻辑语句有效否定所需的规则。现在请复习该表格,记住所有规则,并特别注意 IF 语句的否定是 AND 语句。接下来,花些时间推导下一个示例中语句的有效否定。答案和讨论将在下一小节中给出。对于复杂的句子,您可以先将句子(部分或全部)转换为符号形式,以了解逻辑运算符在句子中的嵌套方式。将“仅当”或“有必要”等可能令人困惑的短语改写成标准的 IF-THEN 形式也很有帮助。

1.42. Example: Denial Practice. Write a useful denial of each statement.

1.42. 示例:否认练习。针对每项陈述写出一个有效的否认语句。

(a) If a > b , then a 2 > b 2 .

(a)如果 a > b,则 a 2 > b 2

(b) a > 3 iff −2 a < −4.

(b)a > 3 当且仅当 −2a < −4。

(c) All carrots are orange.

(c)所有的胡萝卜都是橙色的。

(d) There is an integer x such that x is odd and x is perfect.

(d)存在整数 x,使得 x 为奇数且 x 为完全数。

(e) All roses are not red.

(e)并非所有玫瑰都是红色的。

(f) Roses are red, yet some violets are not blue.

(f)玫瑰是红色的,但有些紫罗兰不是蓝色的。

(g) 1 + 1 = 2 and 2 × 2 = 4; or 3 2 = 8.

(g)1 + 1 = 2 且 2 × 2 = 4;或 3 2 = 8。

(h) Some cherries are not red, or all apples are red, but not both.

(h)有些樱桃不是红色的,或者所有的苹果都是红色的,但不能两者都是。

(i) Some pigs can fly, if water is wet or the sky is blue.

(i)有些猪会飞,如果水是湿的或者天空是蓝色的。

(j) For a to be prime, it is necessary that a is odd or a > 2.

(j)要使 a 为素数,a 必须是奇数或 a > 2。

(k) If a is prime, then: a is even implies a = 2.

(k)如果 a 是素数,则:a 是偶数意味着 a = 2。

(l) A sufficient condition for the existence of white ravens is the non-existence of black doves.

(l)白渡鸦存在的充分条件是黑鸽不存在。

(m) x Z , y Z , z Z , ( x > 0 and y > 0 and x 3 + y 3 = z 3 ) .

(m) ∃x∈Z,∃y∈Z,∃z∈Z,(x>0且y>0且x3+y3=z3)。

(n) a > b is necessary and sufficient for b −1 > a −1 only if b > 0 and a > 0.

(n)a > b 是 b −1 > a −1 的必要且充分条件,当且仅当 b > 0 且 a > 0。

(o) x , y , z , ( ( x < y + z ) ( z < x y < y ) ) .

(o)∀x,∃y,∀z,((x<y+z)⇒(z<xy<y))。

Solutions to Denial Exercise

否认练习的解决方案

Do not read any further until you have tried all parts of the preceding example yourself!

在你亲自尝试过前面示例中的所有部分之前,不要继续阅读!

(a) A denial of “if P then Q” is “P and not Q.” So one answer is “ a > b and a 2 b 2 .”

(a)否定“如果 P 则 Q”是“P 且非 Q”。因此,一个答案是“a > b 且 a 2 ≤ b 2 ”。

(b) Denying an IF-statement produces an XOR-statement. So one answer is “ a > 3 or −2 a < −4, but not both.”

(b)否定一个 IF 语句会产生一个 XOR 语句。因此,一个答案是“a > 3 或 −2a < −4,但不能同时满足这两个条件”。

(c) Denying a universal statement produces an existential statement. So one answer is “Some carrot is not orange.”

(c) 否定一个普遍命题会产生一个存在命题。因此,一个答案是“有些胡萝卜不是橙色的”。

(d) Denials convert to and to ∨. So one answer is “For all integers x , x is not odd or x is not perfect.”

(d) 否定将 ∃ 转换为 ∀,将 ∧ 转换为 ∨。因此,一个答案是“对于所有整数 x,x 不是奇数或 x 不是完全数。”

(e) Answer: “Some rose is red.” In this example, avoid the temptation to declare that a denial of “All roses are not red” is “All roses are red.” Observe that both of the latter statements are false, so they cannot be denials of each other. The double negation rule can only be used to cancel a NOT operator applying to the entire statement, including quantifiers.

(e) 答案:“有些玫瑰是红色的。” 在这个例子中,不要误以为“并非所有玫瑰都是红色的”的否定就是“所有玫瑰都是红色的”。请注意,后两个陈述都是错误的,因此它们不能互相否定。双重否定规则只能用于否定应用于整个语句(包括量词)的非运算符。

(f) This sentence features an implicit universal quantifier (signaled by the plural subject “roses”), as well as the word YET, which has the same logical meaning as AND. Denying the sentence produces “Some rose is not red, or all violets are blue.”

(f)这句话包含一个隐含的全称量词(由复数主语“玫瑰”表示),以及“然而”一词,其逻辑含义与“与”相同。否定这句话会得出“有些玫瑰不是红色的,或者所有的紫罗兰都是蓝色的”。

(g) Notice the semicolon in the original statement, which serves to group the clauses of the sentence as shown by the parentheses here: “(1 + 1 = 2 and 2 × 2 = 4) or 3 2 = 8.” Using the denial rules for OR and then AND, we get “(1 + 1 ≠ 2 or 2 × 2 ≠ 4), and 3 2 ≠ 8.” To avoid parentheses, we could say: 1 + 1 ≠ 2 or 2 × 2 ≠ 4; and 3 2 ≠ 8.

(g) 注意原语句中的分号,它用于将句子的各个分句分组,如括号所示:“(1 + 1 = 2 且 2 × 2 = 4) 或 3 2 = 8。” 使用 OR 和 AND 的否定规则,我们得到“(1 + 1 ≠ 2 或 2 × 2 ≠ 4),且 3 2 ≠ 8。” 为了避免使用括号,我们可以说:1 + 1 ≠ 2 或 2 × 2 ≠ 4;且 3 2 ≠ 8。

(h) This sentence combines “Some cherries are not red” and “All apples are red” with the exclusive-OR operation. Denying gives “Some cherries are not red iff all apples are red.”

(h)这句话用异或运算将“有些樱桃不是红色的”和“所有苹果都是红色的”结合起来。否定运算的结果为“有些樱桃不是红色的当且仅当所有苹果都是红色的”。

(i) This sentence is an IF-statement where the hypothesis comes at the end. Denying gives “Water is wet or the sky is blue, and all pigs cannot fly.”

(i)这句话是一个“如果”语句,假设放在最后。否定假设会得出“水是湿的,天空是蓝色的,而且所有的猪都不会飞。”

(j) Before denying, the sentence can be rewritten: “If a is prime, then ( a is odd or a > 2).” After denying, we get “ a is prime and a is not odd and a ≤ 2.”

(j)否定之前,该句子可以改写为:“如果 a 是素数,则(a 是奇数或 a > 2)”。否定之后,我们得到“a 是素数且 a 不是奇数且 a ≤ 2”。

(k) The conclusion of this IF-statement is another IF-statement. Denying each IF in turn produces “ a is prime and a is even and a ≠ 2.”

(k)该 IF 语句的结论是另一个 IF 语句。依次否定每个 IF 语句,即可得出“a 是质数且 a 是偶数且 a ≠ 2”。

(l) Before denying, we rewrite the sentence as: “If black doves do not exist, then white ravens exist.” Denying gives “Black doves do not exist and white ravens do not exist.” Since both new clauses begin with NOT, we can further simplify the denial to: “All doves are not black, and all ravens are not white.”

(l)在否定之前,我们将句子改写为:“如果黑鸽不存在,那么白渡鸦就存在。”否定后得到“黑鸽不存在,白渡鸦也不存在。”由于两个新分句都以“NOT”开头,我们可以进一步简化否定语句为:“并非所有鸽子都是黑色的,也并非所有渡鸦都是白色的。”

(m) In statements starting with multiple quantifiers, imagine forming the negation by placing the negation symbol ∼ at the far left. Now move this symbol to the right past each quantifier, flipping quantifiers from to and vice versa. Finally, the negation symbol acts on the ANDs in the middle of the statement, flipping them into ORs. The final denial is

(m) 对于以多个量词开头的语句,想象一下,否定符号 ∼ 放在最左边。现在,将这个符号向右移动,越过每个量词,将量词从 ∃ 变为 ∀,反之亦然。最后,否定符号作用于语句中间的 AND 关系,将它们变为 OR 关系。最终的否定是

x x Z Z , y Z Z , z z Z Z , ( x x 0 y 0 x x 3 + y 3 z z 3 ) .

(n) The statement can be recast as “If ( a > b iff b −1 > a −1 ), then ( b > 0 and a > 0).” Denying the statement produces “( a > b iff b −1 > a −1 ) and ( b ≤ 0 or a ≤ 0).”

(n) 该语句可以改写为“如果 (a > b 当且仅当 b −1 > a −1 ),则 (b > 0 且 a > 0)”。否定该语句则得到“(a > b 当且仅当 b −1 > a −1 ) 且 (b ≤ 0 或 a ≤ 0)”。

(o) Flipping the initial quantifiers and denying the IF, we get “ x , y , z , ( x < y + z ) ( z x y x y y ) .” Note that the original formula z < xy < y contains an implicit AND, which becomes OR in the denial.

(o)将初始量词翻转并否定 IF,我们得到“∃x,∀y,∃z,(x<y+z)∧(z≥xy∨xy≥y)”。注意,原始公式 z < xy < y 包含一个隐含的 AND,在否定中变成了 OR。

Uniqueness

独特性

An object is unique iff it is the only one of its kind. We now introduce a modified version of the existential quantifier that can be used to assert the existence of a unique object satisfying some property.

一个对象是唯一的,当且仅当它是同类对象中唯一的。现在我们引入一个改进的存在量词,它可以用来断言存在一个满足某个性质的唯一对象。

1.43. Definition: Uniqueness Symbol. Let U be a set, and let P ( x ) be an open sentence. The statement ! x U , P ( x ) means that there exists exactly one object x 0 in U for which P ( x 0 ) is true. Equivalently, there exists a unique object x 0 in U making P ( x 0 ) true. The unrestricted quantifier ! x , P ( x ) is defined similarly; this statement means that there exists one and only one object x 0 making P ( x 0 ) true. The exclamation mark following the existential quantifier is called the uniqueness symbol .

1.43. 定义:唯一性符号。设 U 为一个集合,P(x) 为一个开语句。命题 ∃!x∈U,P(x) 表示 U 中存在且仅存在一个对象 x 0 ,使得 P(x 0 ) 为真。等价地,U 中存在唯一对象 x 0 ,使得 P(x 0 ) 为真。无限制量词 ∃!x,P(x) 的定义类似;该命题表示存在且仅存在一个对象 x 0 ,使得 P(x 0 ) 为真。存在量词后的感叹号称为唯一性符号。

1.44. Example. True or false?

1.44. 示例。对还是错?

(a) ! x R , x 2 = x .

(a) ∃!x∈R,x2=x。

(b) ! x Z > 0 , x 2 = x .

(b) ∃!x∈Z>0,x2=x。

(c) ! x R , x 3 = 8 .

(c)∃!x∈R,x3=8。

(d) ! x Z , 3 x = 5 .

(d)∃!x∈Z,3x=5。

(e) ! x Q , 3 x = 5 .

(e)∃!x∈Q,3x=5。

Solution. (a) is false, because there are two real numbers x that satisfy x 2 = x , namely x = 0 and x = 1. On the other hand, (b) is true, since x 2 = x has one and only one solution (namely x = 1) belonging to the given universe Z > 0 . Part (c) is true, since x = 2 is the unique real solution to x 3 = 8. (Note that (− 2) 3 = −8.) Part (c) would be false if we enlarged the universe from R to C . Part (d) is false because there is no integer x for which 3 x = 5. On the other hand, x = 5/3 is in Q and satisfies this equation, so “ x Q , 3 x = 5 ” is true. Moreover, x = 5/3 is the only solution to 3 x = 5 in Q , so that the stronger existence-and-uniqueness statement in (e) is also true. (e) would remain true in the larger universes R or C .

解答:(a) 错误,因为存在两个实数 x 满足 x = x,即 x = 0 和 x = 1。另一方面,(b) 正确,因为 x = x 有且仅有一个解(即 x = 1),该解属于给定的论域 Z>0。(c) 正确,因为 x = 2 是 x = 8 的唯一实数解。(注意 (−2) = −8。)如果我们将论域从 R 扩展到 C,则 (c) 错误。(d) 错误,因为不存在整数 x 使得 3x = 5。另一方面,x = 5/3 属于 Q 且满足该方程,因此“∃x∈Q,3x=5”为真。此外,x = 5/3 是 Q 中方程 3x = 5 的唯一解,因此 (e) 中更强的存在性和唯一性陈述也成立。(e) 在更大的宇宙 R 或 C 中仍然成立。

It is possible to eliminate the uniqueness symbol, replacing ! x , P ( x ) by an equivalent statement using previously introduced logical symbols. This elimination rule gives us insight into the precise meaning of uniqueness, and we often need the rule when giving proofs of uniqueness. To derive the rule, let us consider the logical encoding of several related statements first.

可以通过用前面介绍的逻辑符号表示的等价语句来消除唯一性符号 ∃!x,P(x)。这条消除规则让我们能够更深入地理解唯一性的确切含义,而且在证明唯一性时,我们经常需要用到这条规则。为了推导出这条规则,我们首先考虑几个相关语句的逻辑编码。

Step 1 . How can we encode the following statement? “There exist at least two objects x 0 making P ( x 0 ) true.” A first attempt might be to write x , y , P ( x ) P ( y ) . However, this does not quite work, because we are allowed to pick the same object for x and for y . For instance, we see that x , y , x + 1 = 3 y + 1 = 3 is true by taking x = y = 2. If we intend the variables x and y to represent different (distinct) objects, we must explicitly say so by saying x y . So the given statement can be encoded as follows:

步骤 1. 如何对以下语句进行编码?“至少存在两个对象 x 0 使得 P(x 0 ) 为真。” 一个初步的尝试是写成 ∃x,∃y,P(x)∧P(y)。然而,这种方法并不完全有效,因为我们可以选取同一个对象作为 x 和 y。例如,当 x = y = 2 时,我们发现 ∃x,∃y,x+1=3∧y+1=3 为真。如果我们希望变量 x 和 y 代表不同的对象,则必须明确地声明 x ≠ y。因此,给定的语句可以编码如下:

x x , y , [ ( P P ( x x ) P P ( y ) ) x x y ] .

Step 2. Now consider how to encode “There exists at most one object x 0 making P ( x 0 ) true.” The key is to recognize that the situation described here (having at most one object that works) is the exact logical opposite of the situation in Step 1 (having at least two objects that work). Thus we can obtain the answer by denying the statement from Step 1. One possible denial looks like this:

步骤 2. 现在考虑如何编码“至多存在一个对象 x 0 使得 P(x 0 ) 为真”。关键在于认识到这里描述的情况(至多只有一个对象有效)与步骤 1 中的情况(至少有两个对象有效)在逻辑上正好相反。因此,我们可以通过否定步骤 1 中的语句来获得答案。一种可能的否定方式如下:

x x , y , [ ( P P ( x x ) P P ( y ) ) x x = y ] .

We could continue to simplify the denial by replacing ( P ( x ) P ( y ) ) by (~ P ( x ))∨(~ P ( y )). However, another way to proceed is to remember the equivalence A B ≡ (~ A )∨ B from the Theorem on IF. Using this equivalence in reverse, we obtain the following IF-statement as a possible denial of the statement in Step 1:

我们可以继续简化否定式,将 ∼(P(x)∧P(y)) 替换为 (~P(x))∨(~P(y))。然而,另一种方法是利用 IF 定理中的等价关系 A ⇒ B ≡ (~A)∨B。反向运用此等价关系,我们可以得到以下 IF 语句,作为步骤 1 中语句的一种可能的否定:

x x , y , [ ( P P ( x x ) P P ( y ) ) x x = y ] .

(1.5)

Reading this in English, with some words added for emphasis, (1.5) says that “for all objects x 0 and y 0 , if P ( x 0 ) and P ( y 0 ) both happen to be true, then it must be the case that x 0 actually equals y 0 .” This statement does have the intended effect of preventing more than one object from satisfying the open sentence P ( x ).

用英语阅读这段文字,并添加一些词语以示强调,(1.5) 表示“对于所有对象 x 0 和 y 0 ,如果 P(x 0 ) 和 P(y 0 ) 都为真,那么 x 0 实际上必然等于 y 0 。” 这条语句确实达到了预期的效果,即防止多个对象满足开放语句 P(x)。

Although it already follows from the denial rules, let us confirm directly that (1.5) is true when P ( x ) is satisfied by zero objects, or by exactly one object. First suppose no objects make P ( x ) true. Then for any objects x 0 and y 0 , P ( x 0 ) P ( y 0 ) is false, so the IF-statement is true. Thus, (1.5) is true in this situation. Now suppose exactly one object z 0 makes P ( x ) true. In this case, when x = z 0 and y = z 0 , the IF-statement says ( P ( z 0 ) P ( z 0 ) ) z 0 = z 0 , which is true. On the other hand, for all other choices of x and y , the IF-statement has a false hypothesis, hence is true.

虽然根据否定规则已经可以得出结论,但我们还是直接确认一下,当 P(x) 不被任何对象满足,或者恰好被一个对象满足时,(1.5) 是否成立。首先假设没有任何对象使 P(x) 为真。那么对于任意对象 x 0 和 y 0 ,P(x0)∧P(y0) 为假,因此 IF 语句为真。所以,在这种情况下,(1.5) 为真。现在假设恰好有一个对象 z 0 使 P(x) 为真。在这种情况下,当 x = z 0 且 y = z 0 时,IF 语句表明 (P(z0)∧P(z0))⇒z0=z0,这为真。另一方面,对于 x 和 y 的所有其他选择,IF 语句的假设为假,因此为真。

Step 3. We are now ready to encode the target statement, “There exists exactly one object x 0 making P ( x 0 ) true.” The key is to rewrite this assertion using AND, as follows: “There exists at least one object x 0 making P ( x 0 ) true, AND there exists at most one object x 0 making P ( x 0 ) true.” The first clause can be handled by an ordinary existential quantifier, and we encoded the second clause in Step 2. To summarize, our elimination rule for the uniqueness symbol is:

步骤 3. 现在我们可以对目标语句“恰好存在一个对象 x 0 使得 P(x 0 ) 为真”进行编码了。关键在于使用 AND 重写此断言,如下所示:“至少存在一个对象 x 0 使得 P(x 0 ) 为真,并且至多存在一个对象 x 0 使得 P(x 0 ) 为真。”第一个子句可以用普通的存在量词处理,第二个子句我们在步骤 2 中已经编码过了。总而言之,我们消除唯一性符号的规则是:

! x x , P P ( x x ) ( x x , P P ( x x ) ) ( x x , y , [ ( P P ( x x ) P P ( y ) ) x x = y ] ) .

A similar rule holds for restricted quantifiers. For instance, ! x R , x 3 = 8 can be rewritten as

类似的规则也适用于受限量词。例如,∃!x∈R,x3=8 可以重写为

( x x R , x x 3 = 8 ) ( x x R , y R , ( x x 3 = 8 y 3 = 8 ) x x = y ) .

We can also adapt the reasoning in Steps 1 through 3 to make statements such as “there are at least three objects x 0 making P ( x 0 ) true,” “there are at most two objects x 0 making P ( x 0 ) true,” “there are exactly two objects x 0 making P ( x 0 ) true,” and so on. We also mention that a statement such as, “the solution is unique, if it exists” is asserting the existence of at most one solution, hence could be encoded by the formula (1.5) .

我们还可以运用步骤 1 到 3 中的推理,得出诸如“至少存在三个对象 x 0 使得 P(x 0 ) 为真”、“至多存在两个对象 x 0 使得 P(x 0 ) 为真”、“恰好存在两个对象 x 0 使得 P(x 0 ) 为真”等等的陈述。此外,我们还要指出,诸如“如果解存在,则解是唯一的”之类的陈述断言至多存在一个解,因此可以用公式 (1.5) 来表示。

1.45 Remark. The English word UNIQUE is a special kind of adjective called an absolute (or non-gradable ) adjective. Absolute adjectives cannot be modified by adverbs indicating the degree to which that adjective applies. For example, although it might be very cold today, or somewhat cold, or really cold, or colder than yesterday, we cannot say that an object is very unique, somewhat unique, really unique, more unique, or almost unique. An object either is unique (the one and only one object in a given universe having a specified property), or it is not.

1.45 备注。英语单词 UNIQUE 是一种特殊的形容词,称为绝对形容词(或不可分级形容词)。绝对形容词不能被表示程度的副词修饰。例如,虽然今天可能很冷,或者有点冷,或者非常冷,或者比昨天更冷,但我们不能说一个物体非常独特、有点独特、非常独特、更独特或几乎独特。一个物体要么是独一无二的(在给定的范围内,唯一具有特定属性的物体),要么就不是。

Section Summary

章节概要

1. Denials. Have you fully memorized the table of denial rules on page 35? Keep practicing until the rules become second nature.

1. 否认。你是否已经完全记住了第35页的否认规则表?继续练习,直到这些规则成为你的本能反应。

2. Uniqueness. ! x U , P ( x ) means there exists exactly one object x 0 in U making P ( x 0 ) true. We could say “unique” or “one and only one” instead of “exactly one” here. We can eliminate the uniqueness symbol (the exclamation mark “!”) with this rule:

2.唯一性。∃!x∈U,P(x) 表示在集合 U 中恰好存在一个对象 x 0 ,使得 P(x 0 ) 为真。这里我们可以用“唯一”或“唯一”来代替“恰好一个”。我们可以通过以下规则去掉唯一性符号(感叹号“!”):

! x x U U , P P ( x x ) ( x x U U , P P ( x x ) ) ( x x U U , y U U , [ ( P P ( x x ) P P ( y ) ) x x = y ] ) .

To say there is at most one x 0 in U making P ( x 0 ) true, we write:

为了说明在 U 中至多存在一个 x 0 使得 P(x 0 ) 为真,我们写成:

x x U U , y U U , [ ( P P ( x x ) P P ( y ) ) x x = y ] .

Exercises

练习

1. True or false? Explain.

1. 判断正误并解释原因。

(a) ! x R , x 2 = 2 . (b) ! x R < 0 , x 2 = 2 . (c) ! x Z , x 2 3 x 10 = 0 . (d) ! x Z > 0 , x 2 3 x 10 = 0 . (e) ! x Z 0 , x 2 + 4 x + 4 = 0 . (f) ! x R , x 2 + 4 x + 4 = 0 .

(a) ∃!x∈R,x²=2。(b) ∃!x∈R<0,x²=2。(c) ∃!x∈Z,x²−3x−10=0。(d) ∃!x∈Z>0,x²−3x−10=0。(e) ∃!x∈Z≥0,x²+4x+4=0。(f) ∃!x∈R,x²+4x+4=0。

2. True or false? Explain.

2. 判断正误并解释原因。

(a) ! x Z , 2 < x < 4 . (b) ! x Q , 2 < x < 4 . (c) ! x Z , 2 x 4 . (d) ! y Z , 5 y 7 = 2 . (e) ! y Q , 5 y 7 = 2 . (f) ! y R , 5 y 7 = 2 .

(a) ∃!x∈Z,2<x<4。 (b) ∃!x∈Q,2<x<4。 (c) ∃!x∈Z,2≤x≤4。 (d) ∃!y∈Z,5y−7=2。 (e) ∃!y∈Q,5y−7=2。 (f) ∃!y∈R,5y−7=2。

3. Write a useful denial of each statement. (Assume all unquantified letters are constants.)

3. 请对每个陈述写出一个有效的否定。(假设所有未量化的字母均为常量。)

(a) If ab ≠ 0, then a ≠ 0 and b ≠ 0.

(a)如果 ab ≠ 0,则 a ≠ 0 且 b ≠ 0。

(b) All perfect numbers have last digit 6 or 8.

(b)所有完全数的个位数都是 6 或 8。

(c) ( X is compact) ⇔ ( X is closed and bounded).

(c) (X 是紧集)⇔ (X 是闭集且有界集)。

(d) When X is compact, X is closed and bounded.

(d)当 X 是紧致的,X 是闭的且有界的。

(e) x Z , y Q , x < y < x + 1 .

(e)∀x∈Z,∃y∈Q,x<y<x+1。

(f) If every odd number is prime, then some raven is not black.

(f)如果每个奇数都是质数,那么一定有一只乌鸦不是黑色的。

(g) Not all elephants are pink.

(g)并非所有的大象都是粉红色的。

(g) 2 + 2 = 4 only if 2 × 3 = 9.

(g)2 + 2 = 4 仅当 2 × 3 = 9。

4. Write a useful denial of each statement. (Assume all unquantified letters are constants.)

4. 请对每个陈述写出一个有效的否定。(假设所有未量化的字母均为常量。)

(a) If ab = 0, then a = 0 or b = 0.

(a)如果ab=0,则a=0或b=0。

(a) Some prime numbers p satisfy 10 < p < 20.

(a)有些素数 p 满足 10 < p < 20。

(b) X is compact if X is both complete and totally bounded.

(b)如果 X 既是完备的又是完全有界的,则 X 是紧致的。

(c) X is connected and compact whenever X is a closed interval.

(c)当 X 为闭区间时,X 是连通且紧致的。

(d) If some carrot is not orange, then every even number is prime.

(d)如果某个胡萝卜不是橙色的,那么每个偶数都是质数。

(e) Mauve pigs do not exist.

(e)紫色猪并不存在。

(f) A necessary condition for 1 + 1 = 2 is that 2 × 5 = 11.

(f)1 + 1 = 2 的必要条件是 2 × 5 = 11。

(g) Some violets are blue only if all oranges are green or yellow.

(g)有些紫罗兰只有在所有橙子都是绿色或黄色时才是蓝色的。

(h) ( ( P Q ) R ) ( Q ( R ( P ) ) ) .

(h)((P⊕Q)⇔R)∧(Q⇒(R∨(∼P)))。

5. Write a useful denial of each statement.

5. 写出对每项陈述的有力反驳。

(a) For every positive real number y , there is a real number x with e x = y .

(a)对于每个正实数 y,存在实数 x,使得 e x = y。

(b) For all real x and z , if tan x = tan z , then x = z .

(b)对于所有实数 x 和 z,如果 tan x = tan z,则 x = z。

(c) For all real x and y , y = 3 x + 1 iff x = ( y − 1)/3.

(c)对于所有实数 x 和 y,y = 3x + 1 当且仅当 x = (y − 1)/3。

6. Write a useful denial of each statement.

6. 针对每项陈述写出一个有效的反驳。

(a) A necessary condition for all roses to be red is that some carrots are white or blue.

(a)所有玫瑰都是红色的必要条件是有些胡萝卜是白色或蓝色的。

(b) ϵ R > 0 , δ R > 0 , x R , y R , ( | x y | < δ | sin ( x ) sin ( y ) | < ϵ ) .

(b)∀ϵ∈R>0,∃δ∈R>0,∀x∈R,∀y∈R,(|x−y|<δ⇒|sin⁡(x)−sin⁡(y)|<ϵ)。

(c) For all functions f , if the continuity of f is sufficient for the differentiability of f , then the integrability of f is necessary for the continuity of f .

(c)对于所有函数 f,如果 f 的连续性是 f 可微性的充分条件,那么 f 的可积性是 f 连续性的必要条件。

(d) ( ( P Q ) R ) ( Q ( R ( P ) ) ) .

(d)((P⇒Q)⊕R)∨(Q∧(R⇒(∼P)))。

(e) For some real x , x > 2 implies x 2 > 4 but not conversely.

(e)对于某个实数 x,x > 2 意味着 x 2 > 4,但反之则不然。

7. Write a useful denial of this statement: 0 ∈ H and a , b H , a + b H and c H , c H .

7.写出对以下陈述的有用否定:0 ∈ H 且 ∀a,b∈H,a+b∈H 且 ∀c∈H,−c∈H。

8. For each set H , decide whether the statement in the previous exercise or its denial is true for H . (a) H = Q . (b) H = R 0 . (c) H = Z 0 . (d) H = {− 1, 0, 1}. (e) H is the set of even integers.

8.对于每个集合H,判断上一题中的陈述或其否定是否对H成立。(a) H=Q。(b) H=R≥0。(c) H=Z≠0。(d) H = {−1, 0, 1}。(e) H是偶数的集合。

9. Write a useful denial of each statement.

9. 写出对每项陈述的有力反驳。

(a) e G , x G , e x = x = x e .

(a)∃e∈G,∀x∈G,e⋆x=x=x⋆e。

(b) x P , y P , z P , x y z x z .

(b)∀x∈P,∀y∈P,∀z∈P,x≤y≤z⇒x≤z。

(c) L R , ϵ R > 0 , N Z > 0 , n Z > 0 , n N | a n L | < ϵ .

(c) ∃L∈R,∀ϵ∈R>0,∃N∈Z>0,∀n∈Z>0,n≥N⇒|an−L|<ϵ。

(d) y R , [ ( x S , y x ) ( z R , y < z u S , u < z ) ] .

(d) ∃y∈R,[(∀x∈S,y≤x)∧(∀z∈R,y<z⇒∃u∈S,u<z)].

10. Write a useful denial of each statement.

10. 写出对每项陈述的有力反驳。

(a) M R , x S , y S , d ( x , y ) M .

(a) ∃M∈R,∀x∈S,∀y∈S,d(x,y)≤M。

(b) y G , z G , y z = e = z y .

(b)∀y∈G,∃z∈G,y⋆z=e=z⋆y。

(c) ϵ R > 0 , N Z > 0 , n Z > 0 , n N d ( x n , x ) < ϵ .

(c) ∀ϵ∈R>0,∀N∈Z>0,∃n∈Z>0,n≥N∧d(xn,x)<ϵ.

11. Let x 0 be a fixed object. Encode the following sentence using logical symbols: “ x 0 is the unique object in U making P ( x ) true.”

11.设 x 0 为一个固定对象。使用逻辑符号对以下句子进行编码:“x 0 是 U 中唯一使 P(x) 为真的对象。”

12. Encode the following sentence using logical symbols: “for all but one object x 0 in U , Q ( x 0 ) is true.”

12.使用逻辑符号对以下句子进行编码:“对于 U 中除一个对象 x 0 之外的所有对象,Q(x 0 ) 为真。”

13. Find a useful denial of ! x U , P ( x ) by eliminating the uniqueness symbol and applying the denial rules. Give your answer in symbols and then in English.

13. 通过消除唯一性符号并应用否定规则,找到 ∃!x∈U,P(x) 的一个有效否定。请用符号和英文分别给出答案。

14. Consider this statement: “If ! x , P ( x ) is true, then x , P ( x ) must be true.”

14.考虑以下陈述:“如果 ∃!x,P(x) 为真,那么 ∃x,P(x) 也必须为真。”

(a) Is this statement true for all open sentences P ( x )? Explain.

(a)该陈述对所有开语句 P(x) 都成立吗?请解释。

(b) Repeat (a) for the converse of the given statement.

(b)对给定命题的逆命题重复(a)。

(c) Repeat (a) for the contrapositive of the given statement.

(c)对给定命题的逆否命题重复(a)。

15. Consider this statement: “If x U , P ( x ) is true, then x , P ( x ) must be true.”

15.考虑以下陈述:“如果 ∃x∈U,P(x) 为真,则 ∃x,P(x) 必定为真。”

(a) Is this statement true for all choices of U and P ( x )? Explain.

(a)对于 U 和 P(x) 的所有选择,该陈述是否都成立?请解释。

(b) Repeat (a) for the converse of the given statement.

(b)对给定命题的逆命题重复(a)。

16. Consider this statement: “If ! x U , P ( x ) is true, then ! x , P ( x ) must be true.”

16.考虑以下陈述:“如果 ∃!x∈U,P(x) 为真,那么 ∃!x,P(x) 也必须为真。”

(a) Is this statement true for all choices of U and P ( x )? Explain.

(a)对于 U 和 P(x) 的所有选择,该陈述是否都成立?请解释。

(b) Repeat (a) for the converse of the given statement.

(b)对给定命题的逆命题重复(a)。

17. Let P ( x ) be a fixed open sentence. Find symbolic versions of each statement below that do not use the uniqueness symbol.

17.设 P(x) 为一个固定开语句。求下列各语句的符号形式,使其不使用唯一性符号。

(a) There exist at least three objects x 0 making P ( x 0 ) true.

(a)至少存在三个对象 x 0 使得 P(x 0 ) 为真。

(b) There exist at most two objects x 0 making P ( x 0 ) true.

(b)至多存在两个对象 x 0 使得 P(x 0 ) 为真。

(c) There exist exactly two objects x 0 making P ( x 0 ) true.

(c)恰好存在两个对象 x 0 使得 P(x 0 ) 为真。

18. Encode the following sentence using logical symbols: “There exist exactly four objects x 0 making P ( x 0 ) true.”

18.使用逻辑符号对以下句子进行编码:“恰好存在四个对象 x 0 使得 P(x 0 ) 为真。”

1 Some exercises and examples in Chapter assume familiarity with even integers …, − 4, − 2, 0, 2, 4, … and odd integers …, − 5, − 3, − 1, 1, 3, 5, …. Formal definitions of even and odd appear in §2.1 .

1 本章中的一些练习和示例假定读者熟悉偶数…, −4, −2, 0, 2, 4, …和奇数…, −5, −3, −1, 1, 3, 5, …。偶数和奇数的正式定义见第2.1节。

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2

Proofs

证明

2.1  Definitions, Axioms, Theorems, and Proofs

2.1 定义、公理、定理和证明

In most branches of science, such as physics or chemistry, we develop and test theories based on experiments. Mathematics, on the other hand, is a deductive science, in which we use logical rules to derive a large body of truths from a small set of initial statements, called axioms. This section introduces the main ingredients in mathematical theories: undefined terms, definitions, axioms, theorems, inference rules, and proofs.

在大多数科学分支中,例如物理学或化学,我们基于实验来发展和检验理论。而数学则是一门演绎科学,它运用逻辑规则从少量初始陈述(称为公理)中推导出大量真理。本节将介绍数学理论的主要组成部分:未定义术语、定义、公理、定理、推理规则和证明。

Undefined Terms and Definitions

未定义术语和定义

In mathematics, we would like to give a precise and rigorous meaning to every word and symbol that we use. This is typically done by giving formal definitions for various mathematical terms. However, all words and symbols are defined by means of other words and symbols, so it is not possible to define everything! The solution is to begin our theory with a collection of undefined terms , such as “true,” “false,” “for all,” “set,” and “membership in a set.” We can often convey an intuitive idea of what an undefined term is supposed to mean by giving synonyms for the term or specific examples. For instance, we might say that a set is a collection of objects, or we might give examples of sets such as {1, 2, 3} or the set of all cows. But, these informal explanations are not mathematical definitions. Ultimately, the meaning of the undefined terms is captured by the axioms we use to describe their properties.

在数学中,我们希望为使用的每一个词语和符号赋予精确而严谨的含义。通常的做法是为各种数学术语给出正式的定义。然而,所有词语和符号都是通过其他词语和符号来定义的,因此不可能定义一切!解决办法是从一系列未定义术语开始构建我们的理论,例如“真”、“假”、“对于所有”、“集合”和“属于某个集合”。我们通常可以通过给出术语的同义词或具体例子来传达未定义术语的直观含义。例如,我们可以说集合是对象的集合,或者我们可以给出集合的例子,例如{1, 2, 3}或所有奶牛的集合。但是,这些非正式的解释并非数学定义。最终,未定义术语的含义是由我们用来描述其性质的公理所概括的。

Most of the words and symbols we use are, in fact, defined terms . For instance, we have already defined the precise meaning of words such as AND, OR, and IF by means of truth tables that show how these words are related to the undefined concepts of “true” and “false.” Typically, a mathematical definition describes a new object, relationship, or symbol via an expression consisting of previously defined symbols. Almost every definition contains quantifiers and logical keywords. For example, the following IFF-statement can be viewed as a definition of the uniqueness symbol :

我们使用的大部分词语和符号实际上都是已定义的术语。例如,我们已经通过真值表定义了“与”、“或”和“如果”等词语的精确含义,真值表展示了这些词语与“真”和“假”这两个未定义概念之间的关系。通常,数学定义通过由先前定义的符号组成的表达式来描述新的对象、关系或符号。几乎每个定义都包含量词和逻辑关键字。例如,以下“如果成立则成立”语句可以被视为唯一性符号的定义:

! x x U U , P P ( x x ) ( x x U U , P P ( x x ) ) ( x x U U , y U U , ( [ P P ( x x ) P P ( y ) ] x x = y ) .

In the next few sections, we will be practicing proofs using four basic concepts that you may have already seen informally: even numbers, odd numbers, divisibility, and rational numbers. We now give formal definitions of these concepts.

在接下来的几节中,我们将练习运用四个基本概念进行证明,这些概念你可能已经非正式地接触过:偶数、奇数、整除性和有理数。现在,我们将给出这些概念的正式定义。

2.1. Definition of EVEN, ODD, DIVIDES, and RATIONAL

2.1. 偶数、奇数、除法和有理数的定义

(a) For all n , n is e v e n means: k Z , n = 2 k .

(a)对于所有 n,niseven 表示:∃k∈Z,n=2k。

(b) For all n , n is o d d means: k Z , n = 2 k + 1.

(b)对于所有 n,nisodd 表示:∃k∈Z,n=2k+1。

(c) For all integers a and b , a d i v i d e s b means: c Z , b = a c .

(c)对于所有整数 a 和 b,a 整除 b 表示:∃c∈Z,b=ac。

(d) For all x , x is rational means: m Z , n Z , ( n 0 x = m / n ) .

(d)对于所有 x,x 是有理数意味着:∃m∈Z,∃n∈Z,(n≠0∧x=m/n)。

Before continuing, we offer some advice and remarks about definitions. When studying any mathematical subject, you must memorize all definitions perfectly as soon as they occur. Do so now for the four definitions listed above. Note that each definition consists of the defined term (appearing in the box on the left) followed by the definition text (appearing in the box on the right). The overall definition asserts the logical equivalence of the defined term and the definition text, for all choices of the variables in the defined term.

在继续之前,我们先就定义提供一些建议和说明。学习任何数学科目时,你必须在遇到定义时立即将其牢记于心。现在就请牢记上面列出的四个定义。请注意,每个定义都由被定义项(出现在左侧方框中)和定义文本(出现在右侧方框中)组成。整个定义断言,对于被定义项中所有变量的选择,被定义项和定义文本在逻辑上是等价的。

The quantifiers and universes appearing on the right side of each definition are a crucial part of the definition; do not forget them! For example, when asked what “ n is odd” means, the answer “ k , n = 2 k + 1 ” is incorrect (the universe is omitted), and the answer “ n = 2 k + 1” is even worse (the quantifier on k is omitted).

每个定义右侧出现的量词和论域是定义的关键部分;千万不要忘记它们!例如,当被问及“n 是奇数”是什么意思时,答案“∃k,n=2k+1”是不正确的(省略了论域),而答案“n = 2k + 1”则更糟糕(省略了对 k 的量词)。

The only unimportant feature of the definitions are the letters used on each side. Different letters may be used instead, as long as there is no conflict with previously introduced letters. Sometimes we must use a different letter on the right side, if the letters in the original definition already have a different meaning. Furthermore, expressions can be substituted for the letters on the left side of a definition. These observations are basic, but they are so crucial that we give an example.

定义中唯一无关紧要的特征是两侧使用的字母。可以使用不同的字母,只要不与之前引入的字母冲突即可。有时,如果原始定义中的字母已有不同的含义,我们就必须在右侧使用不同的字母。此外,定义左侧的字母可以用表达式代替。这些观察虽然基础,但却至关重要,因此我们举个例子。

2.2. Example. The phrase “ m is odd” means k Z , m = 2 k + 1 . We could also have said z Z , m = 2 z + 1 or q Z , m = 2 q + 1 as the definition of this phrase. The phrase “ k is even” means u Z , k = 2 u ; here, we needed to change the quantified variable from k to some other letter (in this case, u ) because k already had a different meaning. The phrase “5 x + 7 is odd” means k Z , 5 x + 7 = 2 k + 1 . The phrase “ c divides c 3 ” means d Z , c 3 = c d (we had to change the quantified variable from c to d here). The phrase “ m 2 + n 2 is rational” means a Z , b Z , ( b 0 m 2 + n 2 = a / b ) .

2.2. 示例。“m 是奇数”表示 ∃k∈Z,m=2k+1。我们也可以用 ∃z∈Z,m=2z+1 或 ∃q∈Z,m=2q+1 来定义这个短语。“k 是偶数”表示 ∃u∈Z,k=2u;这里,我们需要将量化变量从 k 改为其他字母(在本例中为 u),因为 k 已经有了不同的含义。“5x + 7 是奇数”表示 ∃k∈Z,5x+7=2k+1。“c 整除 c 3 ”表示 ∃d∈Z,c3=cd(这里我们需要将量化变量从 c 改为 d)。短语“m 2 + n 2 是有理数”意味着 ∃a∈Z,∃b∈Z,(b≠0∧m2+n2=a/b)。

We reiterate that formal mathematical definitions must be meticulously memorized, including all quantifiers and universes . It is fine, and indeed helpful, to have intuitive descriptions of what a formal definition is saying. For example, the intuition for the definition of “rational” is that a rational number is the ratio of two integers. However, this phrase does not capture the full content of the definition (the existential quantifiers should be explicit, and the denominator must be nonzero).

我们重申,必须一丝不苟地记住所有正式的数学定义,包括所有量词和论域。对正式定义的内容进行直观描述固然很好,也很有帮助。例如,“有理数”的定义直观地表明,有理数是两个整数的比值。然而,这种表述并没有涵盖定义的全部内容(存在量词必须明确指出,且分母必须非零)。

Axioms

公理

In developing a mathematical theory, we would ideally like to derive every true statement as a logical consequence of previously known statements. However, as in the case of definitions and undefined terms, it is impossible to prove literally every statement in this way. For, at the very beginning, we have not derived any true statements, so we have nothing to work with. The solution is to begin our theory with a small collection of initial statements, called axioms or postulates , that are assumed to be true without proof. Each axiom is a proposition that may use quantifiers, logical symbols, undefined terms, and defined terms.

在构建数学理论时,理想情况下,我们希望所有真命题都能作为先前已知命题的逻辑推论而得出。然而,就像定义和未定义项的情况一样,我们不可能逐条证明所有命题。因为在理论构建之初,我们并没有推导出任何真命题,也就无从下手。解决办法是从一组初始命题(称为公理或假设)开始构建理论,这些命题无需证明即可被视为真。每条公理都是一个命题,其中可以使用量词、逻辑符号、未定义项和已定义项。

What are some examples of axioms? We obtain a large supply of axioms by agreeing that any instance of a tautology is an axiom . Recall that a tautology is a propositional form, such as P ∨(¬ P ), that is true for all values of the propositional variables appearing in the form. An instance of a tautology is any statement obtained from the propositional form by replacing all propositional variables by specific propositions. For example, here are some instances of the tautology P ∨(∼ P ): “1 + 1 = 2 or 1 + 1 ≠ 2;” “All roses are red or not all roses are red;” “The sine function is continuous or the sine function is not continuous;” “(0 = 1 and some pigs can fly) or it is false that (0 = 1 and some pigs can fly).” All of these statements are axioms. Instances of tautologies are particularly trustworthy axioms, since the truth of these statements is virtually forced upon us by the way we have agreed to use the logical connectives such as NOT, AND, and OR.

公理有哪些例子?我们可以通过约定任何重言式都是公理来获得大量的公理。回想一下,重言式是一种命题形式,例如 P∨(¬P),它对于形式中出现的所有命题变量的值都为真。重言式的实例是指通过将命题形式中的所有命题变量替换为特定命题而得到的任何语句。例如,以下是重言式 P∨(¬P) 的一些实例:“1 + 1 = 2 或 1 + 1 ≠ 2”;“所有玫瑰都是红色的或并非所有玫瑰都是红色的”;“正弦函数是连续的或正弦函数是不连续的”;“(0 = 1 且有些猪会飞) 或 (0 = 1 且有些猪会飞) 为假”。所有这些语句都是公理。同义反复的例子是特别值得信赖的公理,因为这些陈述的真值实际上是由我们使用诸如 NOT、AND 和 OR 之类的逻辑连接词的方式强加给我们的。

Every new definition can be viewed as a special kind of axiom, called a definitional axiom. For example, we can rewrite the definition of EVEN (given above) as the following axiom:

每个新的定义都可以看作是一种特殊的公理,称为定义公理。例如,我们可以将上面给出的“偶数”的定义重写为以下公理:

n n , [ n n is even 甚至 k k Z Z , n n = 2 k k ] .

On one hand, the forward implication of this IFF-statement lets us eliminate the defined term “ n is even” and replace it with the definition text “ k Z , n = 2 k ” (where k is a new variable). On the other hand, the converse implication lets us go back from the definition text to the defined term, when convenient. The definitions of ODD, DIVIDES, and RATIONAL can similarly be recast as axioms involving universally quantified IFF-statements. For instance:

一方面,该逆正则表达式的前向蕴含允许我们消除定义项“n 为偶数”,并将其替换为定义文本“∃k∈Z,n=2k”(其中 k 为新变量)。另一方面,逆向蕴含允许我们在方便时从定义文本返回到定义项。类似地,奇数、除数和有理数的定义也可以重写为包含全称量化的逆正则表达式的公理。例如:

a 一个 Z Z , b b Z Z , [ a 一个 divides 分割 b b c c Z Z , b b = a 一个 c c ] .

In what follows, we state all new definitions as definitional axioms in this way.

在下文中,我们将所有新定义都以这种方式表述为定义公理。

2.3. Remark. Some mathematical texts present new definitions as IF-statements rather than IFF-statements. For instance, a text might say, “We define an integer n to be even IF n = 2 k for some integer k .” This practice, while very common, is logically wrong! Each definition needs to be an “if and only if” statement telling us precisely when the newly defined term is to be considered true.

2.3 备注。一些数学文献将新的定义表述为“如果”语句,而非“当且仅当”语句。例如,某篇文献可能会这样写:“我们定义整数 n 为偶数,当且仅当存在某个整数 k,使得 n = 2k。”这种做法虽然很常见,但在逻辑上是错误的!每个定义都应该是一个“当且仅当”语句,明确地告诉我们新定义的术语何时为真。

All mathematical theories share a certain common core of logical axioms (e.g., instances of tautologies, and axioms concerning quantifiers and the equality symbol). Each particular theory, such as the theory of sets or the theory of integers or the theory of Euclidean geometry, contains additional axioms giving specific properties of the objects under investigation. Some of the axioms in the theory of sets will be introduced later when we study set theory. In the theory of integers, here is a sample of some of the possible axioms we might start with. (We assume Z , +, ·, 0, and 1 are undefined terms .)

所有数学理论都共享一些共同的核心逻辑公理(例如,重言式、量词和等号相关的公理)。每一种具体的理论,例如集合论、整数论或欧几里得几何,都包含一些额外的公理,这些公理描述了所研究对象的具体性质。集合论中的一些公理将在我们学习集合论时介绍。在整数论中,以下是一些我们可能使用的初始公理示例。(我们假设 Z、+、·、0 和 1 是未定义的项。)

2.4. Some Axioms for Z

2.4. Z 的一些公理

(a) Closure under Addition: x Z , y Z , x + y Z .

(a)加法封闭性:∀x∈Z,∀y∈Z,x+y∈Z。

(b) Commutativity of Addition: x Z , y Z , x + y = y + x .

(b)加法交换律:∀x∈Z,∀y∈Z,x+y=y+x。

(c) Associativity of Addition: x Z , y Z , z Z , ( x + y ) + z = x + ( y + z ) .

(c)加法结合律:∀x∈Z,∀y∈Z,∀z∈Z,(x+y)+z=x+(y+z)。

(d) Additive Identity: x Z , x + 0 = x .

(d)加法单位元:∀x∈Z,x+0=x。

(e) Additive Inverses: x Z , y Z , x + y = 0 .

(e)加法逆元:∀x∈Z,∃y∈Z,x+y=0。

(f) Closure under Multiplication: x Z , y Z , x y Z .

(f)乘法封闭性:∀x∈Z,∀y∈Z,x⋅y∈Z。

(g) Commutativity of Multiplication: x Z , y Z , x y = y x .

(g)乘法交换律:∀x∈Z,∀y∈Z,x⋅y=y⋅x。

(h) Associativity of Multiplication: x Z , y Z , z Z , ( x y ) z = x ( y z ) .

(h)乘法结合律:∀x∈Z,∀y∈Z,∀z∈Z,(x⋅y)⋅z=x⋅(y⋅z)。

(i) Multiplicative Identity: x Z , x 1 = x .

(i)乘法单位元:∀x∈Z,x⋅1=x。

(j) Distributive Law: x Z , y Z , z Z , x ( y + z ) = ( x y ) + ( x z ) .

(j)分配律:∀x∈Z,∀y∈Z,∀z∈Z,x⋅(y+z)=(x⋅y)+(x⋅z)。

(k) No Zero Divisors: x Z , y Z , ( x 0 y 0 ) x y 0 .

(k)无零因子:∀x∈Z,∀y∈Z,(x≠0∧y≠0)⇒x⋅y≠0。

These axioms describe some of the basic properties of integer addition and integer multiplication. We could list more axioms giving properties of equalities and inequalities among integers. In fact, all the axioms we have listed can be proved from even more basic axioms that we do not state here. However, for our initial practice with proofs, we assume that all of the above statements and similar basic algebraic facts are already known. This includes arithmetical facts such as 2 + 2 = 4, ( 1 / 2 Z ) , etc., as well as properties of inequalities such as:

这些公理描述了整数加法和整数乘法的一些基本性质。我们还可以列出更多描述整数之间等式和不等式性质的公理。事实上,我们列出的所有公理都可以从更基本的公理(此处未列出)证明。然而,为了便于我们进行证明的初步练习,我们假设以上所有陈述以及类似的代数基本事实都已被掌握。这包括诸如 2 + 2 = 4、∼(1/2∈Z) 等算术事实,以及诸如以下不等式性质:

x x Z Z , y Z Z , z z Z Z , if 如果 x x < y and z z > 0 then 然后 x x z z < y z z .

We also assume that analogous facts about the real number system are already known. However, unless otherwise noted, we do not allow ourselves to use any fact that specifically concerns the concepts of even numbers, odd numbers, divisibility, and rational numbers. In Chapter 8 , we give a detailed development of the real number system starting from a list of 19 axioms. This development makes extensive use of the proof techniques to be presented in the coming sections.

我们还假设关于实数系统的类似事实已被人们所知。然而,除非另有说明,我们不允许使用任何专门涉及偶数、奇数、整除性和有理数概念的事实。在第8章中,我们将从19条公理出发,详细阐述实数系统。这一阐述过程将大量运用后续章节中介绍的证明技巧。

Theorems, Inference Rules, and Proofs

定理、推理规则和证明

Starting from the axioms, we use rules of logic to deduce new true statements called theorems . Each axiom is also considered a theorem. The sequence of steps leading from the axioms to a given theorem is called a proof of that theorem. An individual step in a proof uses an inference rule to combine one or more previous theorems to obtain another theorem. Our first inference rule involves the logical keyword IF.

从公理出发,我们运用逻辑规则推导出新的真命题,这些命题被称为定理。每条公理本身也被视为一个定理。从公理推导出某个定理的一系列步骤,称为该定理的证明。证明中的每一步都运用推理规则,将一个或多个先前的定理结合起来,从而得到另一个定理。我们的第一个推理规则涉及逻辑关键字“如果”。

2.5. Inference Rule for IF. Suppose P Q and P are already known to be theorems. Then we may deduce Q as a new theorem.

2.5. IF 的推理规则。假设 P ⇒ Q 且 P 已知是定理。那么我们可以推导出 Q 为一个新的定理。

We can informally justify this rule using the truth table for IF, shown here:

我们可以使用如下所示的 IF 真值表来非正式地证明这条规则:

P

P

Q

P Q

P ⇒ Q

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

T

T

T

F

F

F

F

T

T

In the situation described in the rule, P is a known theorem, and hence is a true proposition. Thus we must be in one of the first two rows of the truth table. But P Q is also known to be true, forcing us to be in the first row of the truth table. In this row, Q is a true proposition, so we may safely conclude that Q is a theorem. Similar reasoning with the truth table for IF establishes the following inference rule.

在规则描述的情况下,P 是一个已知的定理,因此是一个真命题。所以我们必定位于真值表的前两行之一。但是 P ⇒ Q 也已知为真,这迫使我们位于真值表的第一行。在这一行中,Q 是一个真命题,因此我们可以得出结论:Q 是一个定理。对 IF 的真值表进行类似的推理,可以建立以下推理规则。

2.6. Contrapositive Inference Rule for IF. Suppose P Q and ∼ Q are already known to be theorems. Then we may deduce ∼ P as a new theorem.

2.6. IF 的逆否命题推理规则。假设 P ⇒ Q 和 ∼Q 已知为定理。那么我们可以推导出 ∼P 作为新定理。

We can also justify the Contrapositive Inference Rule as follows. Given that P Q is a known theorem, the logically equivalent statement (∼ Q ) ⇒ (∼ P ) must also be true. Given that ∼ Q is also known to be true, we can deduce the truth of ∼ P by applying the Inference Rule for IF to the theorems (∼ Q ) ⇒ (∼ P ) and ∼ Q .

我们还可以通过以下方式证明逆否命题推理规则。已知 P ⇒ Q 是一个已知定理,那么逻辑等价的语句 (∼Q) ⇒ (∼P) 也必然为真。已知 ∼Q 也为真,我们可以通过将 IF 推理规则应用于定理 (∼Q) ⇒ (∼P) 和 ∼Q 来推导出 ∼P 的真值。

On the other hand, suppose P Q and Q are known theorems. Is it always safe to deduce P as a new theorem? Consulting the truth table for IF, we see that the answer is no! More specifically, in row 3 of the truth table, P Q and Q are both true, yet P is false. The incorrect deduction of P from known theorems P Q and Q is sometimes called the converse error .

另一方面,假设 P ⇒ Q 和 Q 是已知定理。那么,推导出 P 作为新定理总是安全的吗?查阅 IF 的真值表,我们发现答案是否定的!更具体地说,在真值表的第 3 行,P ⇒ Q 和 Q 都为真,但 P 为假。从已知定理 P ⇒ Q 和 Q 错误地推导出 P 的情况有时被称为逆错误。

Some inference rules involve quantifiers, as we see in the next rule. This rule formally recasts our earlier intuitive description of what a universal quantifier means.

有些推理规则涉及量词,如下一条规则所示。这条规则正式地重述了我们之前对全称量词含义的直观描述。

2.7. Inference Rule for ALL. Suppose x U , P ( x ) is a known theorem and c U is a known theorem, where c is a variable or expression denoting a particular object. Then we may deduce the new theorem P ( c ).

2.7. 所有情况的推理规则。假设对于所有 x∈U,P(x) 是一个已知定理,c∈U 也是一个已知定理,其中 c 是表示特定对象的变量或表达式。那么我们可以推导出新的定理 P(c)。

Intuitively, if we already know that property P is true for all objects x in U , and if we also know that the expression c represents an object in U , then we can conclude that property P is true for the particular object c .

直观地说,如果我们已经知道性质 P 对 U 中的所有对象 x 都成立,并且我们也知道表达式 c 表示 U 中的一个对象,那么我们可以得出结论,性质 P 对特定对象 c 也成立。

2.8. Example. Suppose we have already proved the theorem x R , x 2 0 , and suppose a and b are fixed positive real numbers. Then a and b are also positive real numbers, and hence c = a b is in R because R is closed under subtraction. Now we can apply the Inference Rule for ALL, replacing the quantified variable x in “ x 2 ≥ 0” by the expression c . We deduce the new theorem ( a b ) 2 0 . Now we manipulate this inequality using known algebraic facts. Expanding the square, we get a 2 a b + b 0 . Rearranging terms, we get a + b 2 a b and then ( a + b ) / 2 a b . This inequality says that the arithmetic mean of two positive numbers is greater than or equal to the geometric mean of those numbers.

2.8 示例。假设我们已经证明了定理 ∀x∈R,x²≥0,并且假设 a 和 b 是固定的正实数。那么 a 和 b 也是正实数,因此 c=a−b 属于 R,因为 R 对减法封闭。现在我们可以应用 ALL 推理规则,将“x 2 ≥ 0”中的量化变量 x 替换为表达式 c。我们推导出新的定理 (a−b)²≥0。现在我们利用已知的代数性质来处理这个不等式。展开平方,我们得到 a−2ab+b≥0。重新排列项,我们得到 a+b≥2ab,进而得到 (a+b)/2≥ab。这个不等式表明两个正数的算术平均值大于或等于这两个数的几何平均值。

We can obtain more inference rules for AND, OR, IFF, etc., by analyzing truth tables. Some examples of these rules are considered in the exercises. But these new inference rules turn out to be redundant, because they already follow by combining tautology-based axioms with the Inference Rule for IF. For instance, consider this inference rule for AND: given that P Q is a known theorem, we can deduce Q as a new theorem. To see why this rule is redundant, note that ( P Q ) Q is a tautology and hence an axiom. Suppose that we have proved P Q already. Using the Inference Rule for IF, we see that Q is a new theorem. Similarly, the Contrapositive Inference Rule for IF follows from the Inference Rule for IF and an appropriate tautology.

我们可以通过分析真值表,得到更多关于与、或、因果关系等的推理规则。练习中会给出一些例子。但这些新的推理规则实际上是冗余的,因为它们已经可以通过结合基于重言式的公理和因果关系的推理规则得出。例如,考虑以下与的推理规则:已知 P∧Q 是一个定理,我们可以推导出 Q 是一个新定理。为了理解这条规则的冗余性,请注意 (P∧Q)⇒Q 是一个重言式,因此也是一个公理。假设我们已经证明了 P∧Q。利用因果关系的推理规则,我们发现 Q 是一个新定理。类似地,因果关系的逆否命题推理规则也可以由因果关系的推理规则和一个合适的重言式得出。

2.9. Remark. In many logic texts, the Inference Rule for IF is called the modus ponens rule, abbreviated MP. “Modus ponens” is a Latin phrase meaning “method of affirming.” The Contrapositive Inference Rule for IF is called the modus tollens rule (MT), which means “method of raising.” The Inference Rule for ALL is called universal instantiation (UI). We shall not use these terms.

2.9. 备注。在许多逻辑教材中,如果成立的推理规则被称为肯定前件式(modus ponens),简称MP。“肯定前件式”是拉丁语,意为“肯定的方法”。如果成立的逆否命题推理规则被称为否定前件式(modus tollens),简称MT,意为“否定的方法”。所有成立的推理规则被称为全称实例化(universal instantiation,简称UI)。本文将不使用这些术语。

More on Mathematical Theories (Optional)

更多数学理论相关内容(可选)

This optional subsection gives a bit more detail on how logicians use axiom systems and inference rules to develop formal mathematical theories. The problem of not being able to define the initial words and symbols is dealt with in the following way. First, we invent a formal language using a highly restricted set of symbols (such as ∼, ⇒, , =, ∈, and letters for variables and constants). There are concrete rules determining which strings of symbols in this language constitute terms and formulas . Certain strings of symbols in the language are designated as axioms , and certain inference rules are given. The key is that each inference rule can be executed by performing very basic mechanical manipulations on strings of symbols, without needing to know what any of the symbols mean. For instance, one inference rule (corresponding to the Inference Rule for IF discussed earlier) says: if the string of symbols consisting of a formula P followed by the symbol ⇒ followed by a formula Q is a known theorem, and if the string of symbols P is a known theorem, then the string of symbols Q is a new theorem. The inference rule for universal quantifiers says (roughly): if one of our known theorems consists of the symbol , then a variable symbol x , then a formula P , and if t is any term, then we get a new theorem by writing the string P with all occurrences of the symbol x replaced by the string of symbols t . We see that axioms, proofs, and theorems are all syntactic constructions — properties of the formal language that make no specific reference to the underlying meaning of the symbols.

本可选小节将更详细地介绍逻辑学家如何使用公理系统和推理规则来构建形式数学理论。无法定义初始词和符号的问题可以通过以下方式解决。首先,我们发明一种形式语言,该语言使用一组高度受限的符号(例如∼、⇒、∀、=、∈,以及用于表示变量和常量的字母)。存在一些具体的规则来确定该语言中的哪些符号串构成项和公式。语言中的某些符号串被指定为公理,并给出了一些推理规则。关键在于,每条推理规则都可以通过对符号串执行非常基本的机械操作来执行,而无需知道任何符号的含义。例如,一条推理规则(对应于前面讨论的IF推理规则)指出:如果由公式P、符号⇒和公式Q组成的符号串是一个已知定理,并且如果符号串P是一个已知定理,那么符号串Q就是一个新的定理。全称量词的推理规则(大致)指出:如果我们已知的某个定理由符号∀、变量符号x和公式P组成,并且如果t是任意项,那么我们可以通过将符号串P中所有出现的符号x替换为符号串t来得到一个新的定理。由此可见,公理、证明和定理都是句法构造——形式语言的属性,它们并不具体提及符号的底层含义。

Of course, the only reason we care about proving theorems is because of what they mean ! Thus, every formal language has an associated semantic component that attempts to define the meaning of the terms, formulas, and theorems in the language. The idea is to create an interpretation for the formal theory by introducing a specific universe (set) of objects for the variables to range through, and stipulating the meaning (within this specific model) of each symbol in the language. For instance, the meaning of the logical connectives is defined using truth tables, as before. A given formal theory can have many different interpretations, so the meaning of each formula in the theory depends on which interpretation is used. In some interpretations, all of the axioms of the theory become true statements; such interpretations are called models of the theory. (Actually, this is not quite true: some theories are inconsistent , and have no models at all.)

当然,我们之所以关心定理的证明,唯一的原因就是它们的意义!因此,每一种形式语言都对应着一个语义组件,试图定义该语言中术语、公式和定理的含义。其思路是通过引入一个特定的对象集合(即变量取值范围),并规定语言中每个符号(在这个特定模型中)的含义,来为形式理论创建一个解释。例如,逻辑连接词的含义仍然使用真值表来定义。一个给定的形式理论可以有多种不同的解释,因此该理论中每个公式的含义取决于所使用的解释。在某些解释中,该理论的所有公理都成为真命题;这样的解释被称为该理论的模型。(实际上,这种说法并不完全正确:有些理论是不一致的,根本没有模型。)

It can be shown that all theorems of a formal language (not just the axioms) become true statements in any model of that theory. A celebrated result of mathematical logic ( Gödel’s Completeness Theorem ) asserts that, conversely, any formula of a theory that is true in all models of that theory is provable from the axioms of that theory. This result is absolutely amazing, because it means that the apparently mindless symbol-manipulation rules of a formal theory are powerful enough to discover all true statements expressible within that theory! Put another way, the concrete syntactic machinery of proofs and formal theorems delivers abstract semantic results applicable to a wide variety of situations. Many theorems of modern mathematical logic have a similar flavor, establishing the equivalence of a syntactic concept and a semantic concept. For instance, another version of the Completeness Theorem states that a theory has no models iff a contradictory formula of the form P ( P ) is a formal theorem within that theory.

可以证明,形式语言的所有定理(而不仅仅是公理)在该理论的任何模型中都为真。数学逻辑的一个著名结果(哥德尔完备性定理)断言,反过来,任何在该理论的所有模型中都为真的公式都可以从该理论的公理中证明。这个结果令人惊叹,因为它意味着形式理论中看似简单的符号操作规则,竟然强大到足以发现该理论中所有可表达的真命题!换句话说,证明和形式定理的具体句法机制能够得出适用于各种情况的抽象语义结果。现代数学逻辑的许多定理都具有类似的特征,它们建立了句法概念和语义概念的等价性。例如,完备性定理的另一个版本指出,一个理论没有模型,当且仅当形如 P∧(∼P) 的矛盾公式是该理论中的形式定理。

One practical benefit of organizing mathematical theories using abstract axiom systems is economy of thought. By selecting appropriate general axioms at the outset, we can simultaneously prove facts about many different situations all at once. As an example of this, we now describe the axioms used to define an abstract algebraic structure called a group . The undefined symbols are G (representing the set of objects in the group), (representing an algebraic operation that combines two elements of G to produce a new element of G ), and e (representing an identity element for the operation). The axioms are listed next .

使用抽象公理系统组织数学理论的一个实际好处是思维的简洁性。通过在一开始就选择合适的通用公理,我们可以同时证明许多不同情况下的事实。例如,我们现在描述用于定义称为群的抽象代数结构的公理。未定义的符号是 G(表示群中的对象集合)、⋆(表示将 G 中的两个元素组合成 G 的一个新元素的代数运算)和 e(表示该运算的单位元)。公理列表如下。

2.10. Group Axioms.

2.10. 群公理。

(a) Closure: x G , y G , x y G .

(a)闭包:∀x∈G,∀y∈G,x⋆y∈G。

(b) Associativity: x G , y G , z G , x ( y z ) = ( x y ) z .

(b)结合律:∀x∈G,∀y∈G,∀z∈G,x⋆(y⋆z)=(x⋆y)⋆z。

(c) Identity: x G , x e = x e x = x .

(c)恒等式:∀x∈G,x⋆e=x∧e⋆x=x。

(d) Inverses: x G , y G , x y = e y x = e .

(d) 逆:∀x∈G,∃y∈G,x⋆y=e∧y⋆x=e。

Starting from these axioms, we can prove many theorems about abstract groups. One such theorem is the Left Cancellation Law:

从这些公理出发,我们可以证明许多关于抽象群的定理。其中一个定理就是左消去律:

a 一个 G G , x x G G , y G G , ( a 一个 x x = a 一个 y ) x x = y .

The advantage of the axiomatic setup is that all theorems proved for abstract groups from the group axioms are automatically available for any particular concrete group we may happen to be working with. For example, it can be shown that the integers under addition, the nonzero real numbers under multiplication, and invertible 3 × 3 matrices under matrix multiplication all satisfy the group axioms. Each of these mathematical structures therefore has a left cancellation law. Rather than reproving this law (and all the other theorems of group theory) each time a new system comes along, we need only verify that the new system satisfies the four initial axioms of a group, and then we know that all the theorems are true for this system.

公理化框架的优势在于,所有基于群公理证明的抽象群定理都自动适用于我们可能正在研究的任何具体群。例如,可以证明整数在加法运算下、非零实数在乘法运算下以及可逆的3×3矩阵在矩阵乘法运算下都满足群公理。因此,这些数学结构都具有左消去律。我们无需每次遇到新系统时都重新证明该律(以及群论中的所有其他定理),只需验证新系统是否满足群的四个初始公理,即可知道所有定理对该系统都成立。

Section Summary

章节概要

  1. Elements of Mathematical Theories. A mathematical theory starts with a small collection of undefined terms and introduces definitions for new concepts that combine logical operators, undefined terms, and previous definitions. Similarly, the theory starts with axioms (statements assumed to be true without proof) and uses chains of inference rules called proofs to deduce new theorems. When learning a new theory, it is critical to memorize definitions, axioms, and theorem statements as soon as they arise.
    数学理论的要素。一个数学理论从少量未定义的术语开始,并引入新概念的定义,这些定义结合了逻辑运算符、未定义术语和先前的定义。类似地,该理论从公理(无需证明即可假定为真的陈述)开始,并使用称为证明的推理规则链来推导出新的定理。学习新理论时,至关重要的是要尽快记住定义、公理和定理陈述。
  2. Definitions of Even, Odd, Divides, Rational. Memorize these definitional axioms, which will be the subject of practice proofs in the next few sections. (a) For all n , n is even iff k Z , n = 2 k . (b) For all n , n is odd iff k Z , n = 2 k + 1 . (c) For all a , b Z , a divides b iff c Z , b = a c . (d) For all x , x is rational iff m Z , n Z , ( n 0 x = m / n ) .

    Note that quantifiers and universes are critical features of the definition, whereas the letters used are not important. Expressions can be substituted for the letters in the defined term, and letters in the definition text must be changed if they already have a different meaning.


    偶数、奇数、整除、有理数的定义。请记住这些定义公理,它们将在接下来的几节中进行练习证明。 (a) 对于所有 n,n 为偶数当且仅当 ∃k∈Z,n=2k。 (b) 对于所有 n,n 为奇数当且仅当 ∃k∈Z,n=2k+1。 (c) 对于所有 a,b∈Z,a 整除 b 当且仅当 ∃c∈Z,b=ac。 (d) 对于所有 x,x 为有理数当且仅当 ∃m∈Z,∃n∈Z,(n≠0∧x=m/n)。 注意,量词和论域是定义的关键特征,而使用的字母并不重要。可以用表达式代替定义术语中的字母,如果定义文本中的字母已经有不同的含义,则必须更改这些字母。
  3. Axioms. Every instance of a tautology is an axiom. Every new definition can be viewed as an axiom. Some axioms for Z are listed in item 2.4. The closure axioms state that the sum or product of any two integers is also an integer. There are axioms asserting the commutative, associative, distributive, and identity laws for addition and multiplication. Another axiom says that the product of any two nonzero integers is always nonzero.
    公理。每个重言式都是一条公理。每个新的定义都可以被视为一条公理。Z 的一些公理列于第 2.4 项。封闭性公理指出,任意两个整数的和或积仍然是整数。还有一些公理断言加法和乘法满足交换律、结合律、分配律和恒等律。另一条公理指出,任意两个非零整数的乘积始终非零。
  4. Inference Rules. The Inference Rule for IF says that if P Q and P are already known theorems, we may deduce Q as a new theorem. The Inference Rule for ALL says that if x U , P ( x ) and c U are already known theorems, we may deduce P ( c ) as a new theorem. Many more inference rules follow from the rule for IF using appropriate tautologies.
    推理规则。IF 推理规则指出,如果 P ⇒ Q 且 P 是已知定理,我们可以推导出 Q 作为新定理。ALL 推理规则指出,如果 ∀x∈U,P(x) 和 c ∈ U 是已知定理,我们可以推导出 P(c) 作为新定理。许多其他推理规则可以通过适当的重言式从 IF 规则推导而来。

Exercises

练习

  1. Rewrite each statement by expanding definitions. For statements with negative logic, give a useful denial of the expanded definition. (Do not prove anything here.) [ Sample Question: u 2 is not odd. Answer: k Z , u 2 2 k + 1 .] (a) 37 is odd. (b) k 2 + k is even. (c) m 3 − 5 is rational. (d) 7 divides c . (e) 3 does not divide 5. (f) 2 is not rational.
    通过扩展定义重写每个语句。对于包含否定逻辑的语句,给出扩展定义的有效否定。(此处无需证明任何内容。)[示例问题:u 2 不是奇数。答案:∀k∈Z,u2≠2k+1。] (a) 37 是奇数。 (b) k 2 + k 是偶数。 (c) m 3 − 5 是有理数。 (d) 7 能整除 c。 (e) 3 不能整除 5。 (f) 2 不是有理数。
  2. Rewrite each statement by expanding definitions. For statements with negative logic, give a useful denial of the expanded definition. (a) −18 is even. (b) ℓ 2 + 3 k + 1 is odd. (c) m 2 + n 2 is rational. (d) c + 1 divides c 2 − 1. (e) ab does not divide 8 b + c . (f) e + π is not rational.
    通过扩展定义重写每个语句。对于包含否定逻辑的语句,请给出扩展定义的有效否定。 (a) −18 是偶数。 (b) ℓ 2 + 3k + 1 是奇数。 (c) m2+n2 是有理数。 (d) c + 1 能整除 c 2 − 1。 (e) ab 不能整除 8b + c。 (f) e + π 不是有理数。
  3. Informally justify each inference rule by analyzing a truth table, imitating our discussion of the Inference Rule for IF. (a) Given known theorems P Q and ∼ Q , deduce the new theorem ∼ P . (b) Given known theorems A and B , deduce the new theorem A B . (c) Given the known theorem A B , deduce the new theorem B . (d) Given the known theorem ( Q ) ( R ( R ) ) , deduce the new theorem Q .
    通过分析真值表,非正式地证明每个推理规则,模仿我们对 IF 推理规则的讨论。 (a) 已知定理 P ⇒ Q 和 ∼Q,推导出新定理 ∼P。 (b) 已知定理 A 和 B,推导出新定理 A∧B。 (c) 已知定理 A∧B,推导出新定理 B。 (d) 已知定理 (∼Q)⇒(R∧(∼R)),推导出新定理 Q。
  4. Show that each proposed inference rule is not correct by analyzing truth tables. (a) Given the known theorem A B , deduce the new theorem A . (b) Given known theorems P Q and ∼ P , deduce the new theorem ∼ Q . (c) Given the known theorem P Q , deduce the new theorem P Q .
    通过分析真值表证明每个提出的推理规则都是不正确的。 (a) 已知定理 A∨B,推导出新定理 A。 (b) 已知定理 P ⇒ Q 和 ∼P,推导出新定理 ∼Q。 (c) 已知定理 P ⇒ Q,推导出新定理 P ⇔ Q。
  5. Decide, with explanation, whether each proposed inference rule is always valid. (a) Given the known theorem Q , deduce the new theorem P Q . (b) Given the known theorem Q , deduce the new theorem P Q . (c) Given the known theorem P Q , deduce the new theorem P Q . (d) Given known theorems P Q and ∼ P , deduce the new theorem ∼ Q . (e) Given the known theorem P ⇒ ( Q R ), deduce the new theorem ( P Q ) ⇒ R . (f) Given the known theorem ( P Q ) ⇒ R , deduce the new theorem P ⇒ ( Q R ).
    判断并解释每个提出的推理规则是否总是有效的。 (a) 已知定理 Q,推导出新定理 P∨Q。 (b) 已知定理 Q,推导出新定理 P⊕Q。 (c) 已知定理 P ⇔ Q,推导出新定理 P ⇒ Q。 (d) 已知定理 P ⇔ Q 和 ∼P,推导出新定理 ∼Q。 (e) 已知定理 P ⇒ (Q ⇒ R),推导出新定理 (P ⇒ Q) ⇒ R。 (f) 已知定理 (P ⇒ Q) ⇒ R,推导出新定理 P ⇒ (Q ⇒ R)。
  6. Show that the following inference rule is valid by a truth table analysis: given known theorems P Q R , P S , Q S , and R S , deduce the new theorem S .
    通过真值表分析证明以下推理规则有效:给定已知定理 P∨Q∨R、P ⇒ S、Q ⇒ S 和 R ⇒ S,推导出新定理 S。
  7. (a) Write the contrapositive of the IF-statement in Axiom 2.4 (k). (b) Write the converse of your answer to part (a). Is this converse true for all x , y Z ?
    (a) 写出公理 2.4(k) 中 IF 语句的逆否命题。(b) 写出 (a) 部分答案的逆命题。该逆命题对所有 x,y∈Z 都成立吗?
  8. Suppose “ x Z , 4 x is even ” is a known theorem, and a and b are fixed positive integers. Which of the following conclusions can be drawn from this theorem using the Inference Rule for ALL? Explain. (a) 8 is even. (b) 6 is even. (c) −100 is even. (d) 0.4 is even (taking x = 1/10). (e) 4 b is even. (f) 4( a + 3 b ) is even. (g) 4( a / b ) is even.
    假设“∀x∈Z,4x 是偶数”是一个已知定理,且 a 和 b 是固定的正整数。利用“所有”推理规则,可以从该定理得出以下哪些结论?请解释。 (a) 8 是偶数。 (b) 6 是偶数。 (c) −100 是偶数。 (d) 0.4 是偶数(取 x = 1/10)。 (e) 4b 是偶数。 (f) 4(a + 3b) 是偶数。 (g) 4(a/b) 是偶数。
  9. Suppose “ k Z , k 2 k is even” is a known theorem, and a and b are fixed positive integers. Which conclusions can be drawn from this theorem using the Inference Rule for ALL? Explain. (a) 20 is even. (b) 6 is even. (c) 8 is even. (d) −30 is even. (e) −1/4 is even (taking k = 1/2). (f) ( a + b ) 2 − ( a + b ) is even. (g) ( a / b ) 2 − ( a / b ) is even. [ Sample Answer: (a) CAN be deduced by taking k = 5 in the known theorem, since k 2 k = 25 − 5 = 20.]
    假设“∀k∈Z,k²−k 为偶数”是一个已知定理,且 a 和 b 是固定的正整数。利用全集推理规则,可以从该定理得出哪些结论?请解释。 (a) 20 为偶数。 (b) 6 为偶数。 (c) 8 为偶数。 (d) −30 为偶数。 (e) −1/4 为偶数(取 k = 1/2)。 (f) (a + b) 2 − (a + b) 为偶数。 (g) (a/b) 2 − (a/b) 为偶数。 [示例答案:(a) 可以通过在已知定理中取 k = 5 推导得出,因为 k 2 − k = 25 − 5 = 20。]
  10. Suppose ≤, =, and the logical operators are undefined concepts in a mathematical theory of real numbers. Use these symbols to give formal definitions of: (a) x < y ; (b) x > y ; (c) x y ; (d) z is strictly between x and y .
    假设在实数数学理论中,≤、= 和逻辑运算符是未定义的概念。使用这些符号给出下列各式的正式定义: (a) x < y; (b) x > y; (c) x ≥ y; (d) z 严格介于 x 和 y 之间。
  11. Using only arithmetic operators (+, ·, <) and logical symbols, give definitions of the following concepts, assuming m , n Z and x R : (a) m is a perfect square (examples: 0, 1, 4, 9, 100); (b) n is a multiple of m ; (c) x is a half-integer (examples: 5/2, −22.5, 100/2); (d) n is prime (examples: 2, 3, 5, 7, 11, 101).
    仅使用算术运算符(+、·、<)和逻辑符号,给出下列概念的定义,假设 m、n∈Z 且 x∈R: (a) m 是完全平方数(例如:0、1、4、9、100); (b) n 是 m 的倍数; (c) x 是半整数(例如:5/2、-22.5、100/2); (d) n 是素数(例如:2、3、5、7、11、101)。
  12. Consider this proposed inference rule involving propositional forms P , Q , R , and S : “given known theorems P , Q , and R , deduce the new theorem S .” (a) Assume ( P Q R ) S is a tautology. Argue informally that the proposed inference rule is valid. (b) Now assume that the proposed inference rule always works. Argue informally that ( P Q R ) S is a tautology.
    考虑以下涉及命题形式 P、Q、R 和 S 的推理规则:“已知定理 P、Q 和 R,推导出新定理 S。” (a) 假设 (P∧Q∧R)⇒S 是一个重言式。非正式地论证该推理规则的有效性。 (b) 现在假设该推理规则总是有效的。非正式地论证 (P∧Q∧R)⇒S 是一个重言式。
  13. Suppose the words IF and NOT are undefined concepts in a certain logical theory. Using only these words, complete the following definitions of other logical keywords. (a) P OR Q means… (b) P AND Q means… (c) P IFF Q means… (d) P XOR Q means…
    假设在某个逻辑理论中,“如果”和“非”是未定义的概念。仅使用这两个词,完成以下其他逻辑关键字的定义。 (a) P OR Q 的含义是…… (b) P AND Q 的含义是…… (c) P IFF Q 的含义是…… (d) P XOR Q 的含义是……
  14. Suppose the symbols , ∼, and are undefined symbols in a certain logical theory. Using only these symbols and parentheses, complete the following definitions of other logical symbols. (a) P Q means… (b) P Q means… (c) P Q means… (d) x , P ( x ) means…
    假设符号 ∧、∼ 和 ∃ 在某个逻辑理论中是未定义的符号。仅使用这些符号和括号,完成以下其他逻辑符号的定义。 (a) P ∨ Q 表示… (b) P ⇒ Q 表示… (c) P ⇔ Q 表示… (d) ∀x,P(x) 表示…
  15. Let x , y , a , b be fixed real numbers. Show how to deduce the new theorem ( ax + by ) 2 ≤ ( a 2 + b 2 )( x 2 + y 2 ) from the known theorem z R , 0 z 2 using the Inference Rule for ALL and algebraic manipulations.
    设 x、y、a、b 为固定的实数。利用所有推理规则和代数运算,从已知定理 ∀z∈R,0≤z2 推导出新定理 (ax + by) 2 ≤ (a 2 + b 2 )(x 2 + y 2 )。
  16. Which structures ( G , , e ) satisfy all the group axioms? Explain. (a) G = { − 1, 0, 1}, a b means a + b , e is 0. (b) G = { − 1, 1}, a b means a · b , e is 1. (c) G = Z , a b is always 4, e is 4. (d) G = {0, 1, 2}, a 0 = a = 0 a for all a in G , a b = 0 for all nonzero a , b G , e is 0. (e) G = {0, 1, 2, …}, a b is the maximum of a and b , e is 0. (f) G = Z , a b = a + b 2 , e is 2.
    哪些结构 (G,⋆,e) 满足群的所有公理?解释。 (a) G = { − 1, 0, 1},a⋆b 表示 a + b,e 为 0。 (b) G = { − 1, 1},a⋆b 表示 a · b,e 为 1。 (c) G=Z,a⋆b 始终为 4,e 为 4。 (d) G = {0, 1, 2},对于 G 中的所有 a,a⋆0=a=0⋆a;对于所有非零的 a, b ∈ G,a⋆b=0;e 为 0。 (e) G = {0, 1, 2, …},a⋆b 是 a 和 b 中的最大值,e 为 0。 (f) G=Z, a⋆b=a+b−2,e 为 2。
  17. Assume ( G , , e ) is a group and a , b , c are fixed elements of G such that a b = e = b c . Use the group axioms and the Inference Rule for ALL to prove a = c .
    假设 (G,⋆,e) 是一个群,a、b、c 是 G 的固定元素,且 a⋆b=e=b⋆c。利用群的公理和所有推理规则证明 a = c。
  18. In a certain logical theory, suppose we adopt the following axioms for equality: (i) x , x = x ; (ii) a , b , c , ( a = c b = c ) a = b . (a) Let r and s be fixed objects. Prove r = s s = r , justifying every step using axiom (i), axiom (ii), an instance of a tautology, the Inference Rule for IF, or the Inference Rule for ALL. [ Hint: P [ ( ( P Q ) R ) ( Q R ) ] is a tautology.] (b) Let r , s , t be fixed objects. Prove ( r = s s = t ) r = t .
    在某个逻辑理论中,假设我们采用以下等式公理: (i) ∀x, x=x; (ii) ∀a, ∀b, ∀c, (a=c∧b=c)⇒a=b。 (a) 设 r 和 s 为固定对象。证明 r = s ⇒ s = r,并用公理 (i)、公理 (ii)、重言式实例、IF 推理规则或 ALL 推理规则来证明每一步的合理性。[提示:P⇒[((P∧Q)⇒R)⇒(Q⇒R)] 是一个重言式。] (b) 设 r、s、t 为固定对象。证明 (r=s∧s=t)⇒r=t。
  19. Let n be a fixed integer. Use the axioms for Z given in the section to prove 0 · n = 0. Which axioms are used in the proof?
    设 n 为一个固定的整数。利用本节给出的 Z 的公理证明 0 · n = 0。证明过程中用到了哪些公理?
  20. Suppose, instead of using all tautologies, we only allow instances of the following three tautologies as axioms for our theory: (i) P ⇒ ( Q P ); (ii) ( P ⇒ ( Q R )) ⇒ (( P Q ) ⇒ ( P R )); (iii) ((∼ Q ) ⇒ (∼ P )) ⇒ ( P Q ). Let A and B be specific, fixed propositions. Using only instances of (i), (ii), and (iii) and the Inference Rule for ⇒, prove: (a) A A ; (b) (∼ B ) ⇒ ( B A ). [Remarkably, it can be shown that all tautologies are derivable in this way from (i), (ii), and (iii).]
    假设我们不使用所有重言式,而只允许以下三个重言式的实例作为我们理论的公理: (i) P ⇒ (Q ⇒ P); (ii) (P ⇒ (Q ⇒ R)) ⇒ ((P ⇒ Q) ⇒ (P ⇒ R)); (iii) ((∼Q) ⇒ (∼P)) ⇒ (P ⇒ Q). 设 A 和 B 为特定的固定命题。仅使用 (i)、(ii) 和 (iii) 的实例以及 ⇒ 的推理规则,证明:(a) A ⇒ A;(b) (∼B) ⇒ (B ⇒ A)。[值得注意的是,可以证明所有重言式都可以通过这种方式从 (i)、(ii) 和 (iii) 推导出来。]

2.2 Proving Existence Statements and IF Statements

2.2 证明存在性语句和 IF 语句

Our goal in the next several sections is to learn how to write precise, formal, mathematical proofs. The cornerstone of our exposition is the idea that the logical structure of a statement dictates the structure of the proof of that statement . We will give a series of rules, called proof templates , for automatically constructing the structural outline of a proof of a statement based on the main logical operator appearing in that statement. For instance, one of the proof templates says that to prove an IF-statement P Q , we may write a proof that starts: “Assume P is true. Prove Q is true.” We can then continue to use proof templates recursively to generate a proof of the new goal, Q , aided by the extra information that P is true. We introduce proof templates for existential statements and IF-statements in this section; other logical operators are covered later.

接下来几节的目标是学习如何编写精确、正式的数学证明。我们阐述的核心思想是:一个命题的逻辑结构决定了该命题证明的结构。我们将给出一系列规则,称为证明模板,用于根据命题中出现的主要逻辑运算符自动构建该命题证明的结构框架。例如,其中一个证明模板指出,要证明一个 IF 语句 P ⇒ Q,我们可以编写一个以“假设 P 为真。证明 Q 为真。”开头的证明。然后,我们可以递归地使用证明模板,利用 P 为真的额外信息,生成对新目标 Q 的证明。本节将介绍存在命题和 IF 语句的证明模板;其他逻辑运算符将在后续章节中讲解。

Initial Advice on Proofs

关于校样的初步建议

To gain facility at writing proofs, it is crucial to memorize the proof templates , and practice using the proof templates . Equally crucial is to memorize all definitions (did I say that already?), since a very common step in proofs is to replace a defined concept with its definition. You cannot carry out this step if you have not memorized the relevant definition.

为了熟练掌握证明的写作,记住证明模板并练习使用它们至关重要。同样重要的是记住所有定义(我之前是不是已经说过?),因为证明中一个非常常见的步骤是用定义替换已定义的概念。如果你没有记住相关的定义,就无法完成这一步骤。

An essential feature of proofs is that every formula, statement, and letter in a proof has a logical status that needs to be explicitly indicated with appropriate words and phrases. For instance, the logical status of a given statement might be a known result, something to be proved, or a temporary assumption. The logical status of a letter in a proof might be a fixed (constant) object not chosen by us, a specific (constant) object that we choose, or a quantified variable. Here are some examples of sentences in which the logical status words have been written in bold: “ Assume n is even.” “ We must prove n + 3 is odd.” “ We know n ≥ 0 or n < 0.” “Let x 0 be a fixed, but arbitrary object.” The proof templates often direct us to write certain sentences that contain logical status words — do not omit these words! Note, for instance, that the IF-template mentioned above contains the logical status words “ assume ” and “ prove .” In our initial examples of proofs below, we stress the use of logical status words by placing them in bold type .

证明的一个关键特征是,证明中的每个公式、语句和字母都具有逻辑状态,需要用适当的词语和短语明确标明。例如,某个语句的逻辑状态可能是已知结果、待证明的内容或临时假设。证明中某个字母的逻辑状态可能是我们未选择的固定(常量)对象、我们选择的特定(常量)对象或量化变量。以下是一些用粗体标出逻辑状态词的句子示例:“假设 n 为偶数。”“我们必须证明 n + 3 为奇数。”“我们知道 n ≥ 0 或 n < 0。”“设 x 0 为一个固定但任意的对象。”证明模板通常会指导我们编写包含逻辑状态词的特定句子——切勿省略这些词!例如,请注意上面提到的 IF 模板包含逻辑状态词“假设”和“证明”。在以下证明的初始示例中,我们通过用粗体字强调逻辑状态词的使用。

At any point in a proof, we can replace any statement by an equivalent statement, which has the same logical status as the original statement. For instance, suppose A B is already known (in the most common case, this might be a definitional axiom in which a new term appearing in A is defined via the statement B ). If we have assumed A in a proof, then we can continue the proof by saying that we have assumed B . If we already know A , then we can continue by saying that we know B . Finally, if we are trying to prove A , then we can instead try to prove B . These replacement rules are used constantly in proofs, e.g., every time we expand a definition.

在证明的任何阶段,我们都可以将其替换为与原语句逻辑地位相同的等价语句。例如,假设 A ⇔ B 已知(最常见的情况是,这可能是一条定义公理,其中 A 中出现的新项通过语句 B 来定义)。如果我们在证明中假设了 A,那么我们可以继续证明,并声明我们假设了 B。如果我们已经知道 A,那么我们可以继续证明,并声明我们知道 B。最后,如果我们试图证明 A,那么我们可以转而尝试证明 B。这些替换规则在证明中经常用到,例如,每次我们展开定义时都会用到。

In many places in a proof, we may need to introduce new variables and constants. Whenever this occurs, we must take care not to reuse a letter that has already been given a meaning. For instance, suppose we have just written a line in a proof that says: “ Assume n is even and m is even.” To continue the proof, we might try expanding the definition of “even” in both places. An incorrect way to do this would be to say: “ We have assumed k Z , n = 2 k and k Z , m = 2 k .” The error arises because there is no guarantee that the same integer k will work for both n and m (indeed, this would only be true if n = m ). One possible correct way to proceed would be to say: “ We have assumed k Z , n = 2 k and j Z , m = 2 j ,” provided that k and j have not already been given a meaning.

在证明过程中,我们经常需要引入新的变量和常量。每当出现这种情况时,我们必须注意不要重复使用已经赋予特定含义的字母。例如,假设我们刚刚在证明中写了一行:“假设 n 为偶数且 m 为偶数。” 为了继续证明,我们可能会尝试扩展“偶数”的定义。一种错误的做法是:“我们假设 ∃k∈Z,n=2k 且 ∃k∈Z,m=2k。” 之所以出现错误,是因为我们无法保证同一个整数 k 可以同时适用于 n 和 m(实际上,只有当 n = m 时才成立)。一种正确的做法是:“我们假设 ∃k∈Z,n=2k 且 ∃j∈Z,m=2j”,前提是 k 和 j 尚未被赋予特定含义。

Proving Existence Statements: Proof By Example

证明存在性命题:举例证明

Here is our first official proof template, which tells us that we can prove existential statements by constructing a specific example of an object that works .

这是我们的第一个正式证明模板,它告诉我们,我们可以通过构造一个有效对象的具体例子来证明存在性陈述。

2.11. Proof Template for Proving x U , P ( x ) .

2.11.证明 ∃x∈U,P(x) 的证明模板。

Choose a specific object x 0 that you think satisfies the statement. (This x 0 may depend on constants that have already appeared in the proof.)

选择一个你认为满足该陈述的特定对象 x 0 。(此 x 0 可能依赖于证明中已出现的常量。)

Prove that x 0 is in the universe U .

证明 x 0 属于宇宙 U。

Prove that P ( x 0 ) is true.

证明 P(x 0 ) 为真。

The proof template does not tell you how to choose the object x 0 — that is up to you! Often, the hardest part of applying this template is deciding on an appropriate x 0 to use. This proof method is called proof by construction or proof by example . Let us turn to some illustrations of this proof template, which use the definitions given in the previous section. Note that the statements we prove in our initial examples are very basic and may seem obvious. We have deliberately done this to allow the reader to focus on each new proof template and the structural features of proofs, without getting sidetracked by complex mathematical content. That will come soon enough!

证明模板不会告诉你如何选择对象 x 0 ——这完全取决于你!通常,应用此模板最难的部分在于决定使用哪个合适的 x 0 。这种证明方法被称为构造式证明或示例式证明。让我们来看一些使用此证明模板的示例,这些示例使用了上一节中给出的定义。请注意,我们在初始示例中证明的语句非常基础,可能看起来显而易见。我们特意这样做是为了让读者专注于每个新的证明模板和证明的结构特征,而不会被复杂的数学内容分散注意力。这些内容很快就会出现!

2.12. Example. Prove that 52 is even.

2.12. 例题。证明 52 是偶数。

Proof. By definition of “even,” we must prove that k Z , 52 = 2 k . Choose k = 26. We know that 26 is an integer, and 52 = 2 × 26 by arithmetic.

证明。根据“偶数”的定义,我们必须证明存在 k∈Z,52=2k。选择 k = 26。我们知道 26 是一个整数,并且根据算术可知 52 = 2 × 26。

Comments: The first step in this proof is replacing a defined term by the text of its definition. We do this step all the time in proofs. (Did you memorize the definitions?) This produced an existence statement that needed to be proved. Here we could immediately see which specific k would work, but it was still necessary to point out that k was in the universe Z . If you think this is not really necessary, try imitating this proof to prove that 52 is odd — what goes wrong?

注释:此证明的第一步是用定义文本替换已定义的术语。我们在证明中经常会用到这一步。(你记住这些定义了吗?)这产生了一个需要证明的存在性陈述。在这里,我们可以立即看出哪个特定的 k 值成立,但仍然需要指出 k 属于宇宙 Z。如果你认为这并非必要,请尝试模仿此证明来证明 52 是奇数——哪里出错了?

2.13. Example. Prove that 0.36 is rational.

2.13. 例题。证明 0.36 是有理数。

Proof. By definition of “rational,” we must prove m Z , n Z , n 0 0.36 = m / n . Choose m = 36 and n = 100. We know 36 is an integer, 100 is an integer, 100 ≠ 0, and 0.36 = 36/100.

证明。根据“有理数”的定义,我们必须证明 ∃m∈Z,∃n∈Z,n≠0∧0.36=m/n。选择 m = 36 和 n = 100。我们知道 36 是整数,100 是整数,100 ≠ 0,且 0.36 = 36/100。

Comments: Other choices of m and n also work (for instance, m = 9 and n = 25), but the proof template only requires us to find a single valid example.

评论:m 和 n 的其他选择也有效(例如,m = 9 和 n = 25),但证明模板只要求我们找到一个有效的例子。

2.14. Example. Prove that 7 divides 0.

2.14. 例题。证明 7 能整除 0。

Proof. By definition of “divides,” we must prove w Z , 0 = 7 w . Choose w = 0. Note that 0 Z and 0 = 7 · 0.

证明。根据“整除”的定义,我们必须证明存在 w∈Z,0=7w。选择 w = 0。注意 0∈Z 且 0 = 7 · 0。

Comments: Was it obvious to you before reading the proof that 7 would divide 0? Also observe that logical status can be indicated in many ways; here we said “note that” to introduce some known arithmetical facts.

评论:在阅读证明之前,你是否已经明显看出 7 能整除 0?另外,请注意,逻辑状态可以用多种方式表示;这里我们用“注意”来引入一些已知的算术事实。

In the next example, and throughout this text, we shall sometimes include commentary within a proof indicating what we are thinking and how we are generating the next step of the proof. This commentary appears in square brackets, and is not part of the proof itself.

在接下来的例子中,以及本文后续部分,我们有时会在证明过程中加入注释,说明我们的思路以及下一步证明的推导过程。这些注释用方括号括起来,并非证明本身的一部分。

2.15. Example. Suppose a is a fixed (constant) integer. Prove that 4 a 2 + 11 is odd.

2.15. 例。假设 a 是一个固定的(常数)整数。证明 4a 2 + 11 是奇数。

#

Proof. By definition of “odd,” we must prove k Z , 4 a 2 + 11 = 2 k + 1 . [It is probably not evident at this stage what k to choose. So we instead continue by manipulating 4 a 2 + 11 as follows.] By algebra, we know that 4 a 2 + 11 = 4 a 2 + 10 + 1 = 2(2 a 2 + 5) + 1. Choose k = 2 a 2 + 5; the previous equation shows that 4 a 2 + 11 = 2 k + 1 is true. Moreover, since a and 2 and 5 are all integers, a 2 and 2 a 2 and k = 2 a 2 + 5 are also integers because Z is closed under addition and multiplication.

证明。根据“奇数”的定义,我们必须证明存在 k∈Z,4a²+11=2k+1。[此时可能还不清楚应该选择哪个 k。因此,我们继续对 4a²+11 进行如下运算。] 通过代数运算,我们知道 4a²+11 = 4a²+10 + 1 = 2(2a²+5) + 1。选择 k = 2a²+5;前面的等式表明 4a²+11 = 2k + 1 成立。此外,由于 a、2 和 5 都是整数,所以 a 2 和 2a 2 以及 k = 2a 2 + 5 也是整数,因为 Z 对加法和乘法是封闭的。

Comments: We frequently need the closure properties of Z to see that the objects we choose in proofs do belong to the universe Z as needed. Also note that the integer k that we chose in the proof was allowed to be an expression involving the previously introduced constant a . This is what we mean in the proof template when we say that the chosen object x 0 is allowed to “depend on” previous constants.

注释:我们经常需要利用 Z 的封闭性来验证我们在证明中选择的对象是否确实属于 Z 域。另请注意,我们在证明中选择的整数 k 可以是一个包含先前引入的常数 a 的表达式。这就是我们在证明模板中所说的,所选对象 x 0 可以“依赖于”先前的常数的含义。

2.16. Example. Let c be a fixed integer. Prove that c divides 5 c 3 .

2.16. 例。设 c 为一个固定的整数。证明 c 能整除 5c 3

Proof. We must prove m Z , 5 c 3 = c m . Choose m = 5 c 2 , which is an integer since it is the product of the integers 5, c , and c . By algebra , cm = c (5 c 2 ) = 5 c 3 , as needed.

证明。我们必须证明 ∃m∈Z,5c3=cm。选择 m = 5c 2 ,这是一个整数,因为它是整数 5、c 和 c 的乘积。通过代数运算,cm = c(5c 2 ) = 5c 3 ,满足要求。

Comments: When we expanded the definition of “divides,” we used m (not c ) as the quantified variable since c already had a meaning. As in the last proof, our choice of m in this proof is allowed to depend on the previously introduced constant c .

注释:当我们扩展“除法”的定义时,我们使用 m(而非 c)作为量化变量,因为 c 已有含义。与上一个证明一样,本证明中 m 的选择可以依赖于先前引入的常数 c。

Direct Proof of an IF-Statement

直接证明 IF 语句

Theorems involving IF-statements are ubiquitous in mathematics, and we will develop several different methods for proving IF-statements. The most basic method, called direct proof , is presented in the following fundamental proof template .

涉及条件语句的定理在数学中无处不在,我们将介绍几种不同的证明条件语句的方法。最基本的方法称为直接证明,其基本证明模板如下所示。

2.17. Proof Template for a Direct Proof of P Q .

2.17. P ⇒ Q 的直接证明的证明模板。

Assume P is true. Prove Q is true. [To prove Q , use the assumed statement P and other known facts.]

假设 P 为真。证明 Q 为真。[为了证明 Q,请使用假设的命题 P 和其他已知事实。]

This proof template features a new idea — the notion of temporarily assuming a statement that may or may not actually be true. We can explain informally why the proof template is valid by considering the truth table for IF one more time:

这个证明模板引入了一个新概念——暂时假设一个可能为真也可能为假的陈述。我们可以再次查看 IF 语句的真值表,用通俗易懂的方式解释这个证明模板的有效性:

P

P

Q

P Q

P ⇒ Q

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

T

T

T

F

F

F

F

T

T

When using the direct proof template, our goal is to show that P Q is true, but we do not yet know that this statement is true. We do know that P , being a proposition, must be either true or false. Now, in the event that P is false, we see from the truth table that P Q is automatically true; we do not need to prove anything in this case. So we may as well assume we are in the other case, where the hypothesis P is true. If we can then deduce that Q must be true, we must be in the first row of the truth table, and in that row, P Q is true. Note carefully that, at the end of the proof, we have not proved that Q itself must be true. Our proof of Q relies on the assumption that P is true, which may not actually be correct. The final conclusion of the proof, i.e., that P Q is true, does not rely on the temporary assumption that P is true.

在使用直接证明模板时,我们的目标是证明 P ⇒ Q 为真,但我们并不知道这个命题是否为真。我们知道 P 作为命题,其结果要么为真,要么为假。如果 P 为假,我们可以从真值表中看出 P ⇒ Q 自动为真;在这种情况下,我们无需证明任何内容。因此,我们可以假设我们处于另一种情况,即假设 P 为真。如果我们能够由此推断出 Q 必定为真,那么我们必定处于真值表的第一行,而在这一行中,P ⇒ Q 为真。请注意,在证明的最后,我们并没有证明 Q 本身必定为真。我们对 Q 的证明依赖于 P 为真的假设,而这个假设实际上可能并不正确。证明的最终结论,即 P ⇒ Q 为真,并不依赖于 P 为真的这个暂时假设。

Let us start with an example using defined terms you probably have not seen, to show how a proof template can guide you in an unfamiliar setting.

让我们从一个例子开始,使用一些你可能从未见过的定义术语,来展示证明模板如何在陌生的环境中指导你。

2.18. Example. Outline the proof of this statement: “If f is continuous, then f is measurable.” Here, f is a fixed function. Even without knowing the meaning of “continuous” or “measurable,” we can still begin the proof by saying: Assume that f is continuous. We must prove that f is measurable. To proceed, we would have to know the definitions of continuity and measurability.

2.18. 示例。概述以下命题的证明:“如果 f 是连续的,那么 f 是可测的。”这里,f 是一个固定的函数。即使不知道“连续”或“可测”的含义,我们仍然可以这样开始证明:假设 f 是连续的。我们必须证明 f 是可测的。为了继续进行证明,我们需要知道连续性和可测性的定义。

2.19. Example. Let x 0 be a fixed integer. Prove: if x 0 is odd, then x 0 + 7 is even.

2.19. 例。设 x 0 为一个固定的整数。证明:如果 x 0 是奇数,则 x 0 + 7 是偶数。

Proof. Assume x 0 is odd. We will prove x 0 + 7 is even. [How can we continue? Our only real option is to start expanding definitions; note how we carry along logical status words.] We have assumed there is an integer k with x 0 = 2 k + 1. We must prove m Z , x 0 + 7 = 2 m . From our assumption , we see that x 0 + 7 = 2 k + 1 + 7 = 2 k + 8 = 2( k + 4). Choose m = k + 4 , which is an integer since k is an integer. Our calculation shows that x 0 + 7 = 2 m , as needed.

证明。假设 x 0 为奇数。我们将证明 x 0 + 7 为偶数。[我们该如何继续?我们唯一的选择是扩展定义;注意我们是如何引入逻辑状态词的。] 我们假设存在一个整数 k,使得 x 0 = 2k + 1。我们必须证明 ∃m∈Z, x0+7=2m。根据我们的假设,我们看到 x 0 + 7 = 2k + 1 + 7 = 2k + 8 = 2(k + 4)。选择 m = k + 4,因为 k 是整数,所以 m 也是整数。我们的计算表明 x 0 + 7 = 2m,符合要求。

Comments: We did not get to choose k , which came from an assumed existential statement. But, we did get to use k when we chose m in the existential statement to be proved . In general, statements that are already known (or assumed) are treated entirely differently from statements that have yet to be proved. This is one reason it is crucial to indicate the logical status of every statement in a proof.

注释:我们无法选择 k,因为 k 源自一个假定的存在命题。但是,当我们在待证明的存在命题中选择 m 时,我们确实使用了 k。一般来说,已知(或假定)的命题与尚未证明的命题的处理方式截然不同。这正是为什么在证明中明确指出每个命题的逻辑状态至关重要的原因之一。

2.20. Example. Let a , b , c be fixed (constant) integers. Prove: if a divides b and b divides c , then a divides c .

2.20. 例。设 a、b、c 为固定的(常数)整数。证明:如果 a 整除 b 且 b 整除 c,则 a 整除 c。

Proof. Assume a divides b and b divides c . We must prove a divides c . Writing out the definition three times, we have assumed that for some integer x , b = ax ; and for some integer y , c = by ; we must prove w Z , c = a w . Combining our assumptions , we see that c = by = ( ax ) y = a ( xy ). So, choosing w = xy , c = aw holds. Note that w Z , since w is the product of the two integers x and y .

证明。假设 a 整除 b 且 b 整除 c。我们需要证明 a 整除 c。将定义写三遍,我们假设对于某个整数 x,b = ax;对于某个整数 y,c = by;我们需要证明存在 w∈Z,c=aw。结合我们的假设,我们看到 c = by = (ax)y = a(xy)。因此,选择 w = xy,c = aw 成立。注意 w∈Z,因为 w 是两个整数 x 和 y 的乘积。

Comments: When we wrote out the definition of “divides” in three places, note that we used three different letters ( x , y , and w ) for the quantified variable. The original definition used c for this variable; but that letter already has a meaning in this setting, so we must avoid it.

注释:我们在三处写出“除法”的定义时,请注意我们使用了三个不同的字母(x、y 和 w)来表示量化变量。原始定义中该变量使用 c;但 c 在此语境下已有其他含义,因此我们必须避免使用它。

The Inference Rule for EXISTS (Optional)

EXISTS 的推理规则(可选)

Recall that the Inference Rule for ALL (sometimes called universal instantiation or UI) allows us to pass from a known universal statement x U , P ( x ) to particular instances of that statement. Specifically, we are allowed to choose any object c known to be in U and deduce that P ( c ) is a true proposition. We now discuss a related rule, the Inference Rule for EXISTS, which is sometimes called existential instantiation or EI. This rule lets us extract information from a known existential statement. Suppose we know (or have assumed) the statement x U , P ( x ) . The EI rule allows us to invent a new constant symbol, say x 0 , and assert that x 0 U and P ( x 0 ) are true. It is crucial to note that we do not have the freedom to make a specific choice of x 0 (in contrast to the UI rule). In other words, we know that x 0 exists, but we do not have any control over what it is (other than knowing it is in U and makes P ( x ) true). The symbol x 0 created by the EI rule must be a new constant, which does not already have some other meaning assigned to it.

回想一下,全称推理规则(有时称为全称实例化或UI)允许我们从已知的全称命题∀x∈U,P(x)推导出该命题的特定实例。具体来说,我们可以选择任何已知属于U的对象c,并推断出P(c)为真。现在我们讨论一个相关的规则,存在推理规则(有时称为存在实例化或EI)。该规则允许我们从已知的存在命题中提取信息。假设我们知道(或假设)命题∃x∈U,P(x)。EI规则允许我们构造一个新的常量符号,例如x 0 ,并断言x 0 ∈ U且P(x 0 )为真。需要特别注意的是,我们不能自由地选择x 0 (与全称推理规则不同)。换句话说,我们知道 x 0 存在,但我们无法控制它的具体值(除了知道它在 U 中且使 P(x) 为真)。由 EI 规则生成的符号 x 0 必须是一个新的常量,它不能被赋予任何其他含义。

In the main text, we tacitly use the EI rule without explicit mention, since the rule often confuses beginners. To see how this rule works, let us return to the first few lines of the proof in Example 2.19 . We initially assumed x 0 is odd, and stated the new goal of proving x 0 + 7 is even. Invoking the definition of ODD, we see that we have assumed an existential statement, namely k Z , x 0 = 2 k + 1 . To proceed with the proof, we must process the existential quantifier using the Inference Rule for EXISTS (EI). That rule provides us with a new constant k 0 that we cannot choose, but which we know satisfies k 0 Z and x 0 = 2 k 0 + 1. Here k (with no subscript) is a quantified variable , whereas k 0 is a fixed, but unknown constant . Technically, x 0 = 2 k + 1 is not a proposition (it is an open sentence), but x 0 = 2 k 0 + 1 is a proposition. The proof given earlier in this section did not draw attention to this logical nuance, instead passing directly from the assumption “ x 0 is odd” to the existence of a constant integer (called k ) satisfying x 0 = 2 k + 1. Here is a version of the proof where EI is invoked explicitly, and where subscripts are used throughout to distinguish quantified variables from constants.

在正文中,我们默认使用了 EI 规则,而没有明确提及,因为该规则常常会让初学者感到困惑。为了理解该规则的运作方式,让我们回到例 2.19 证明的前几行。我们最初假设 x 0 为奇数,并提出了证明 x 0 + 7 为偶数的新目标。根据奇数的定义,我们发现我们假设了一个存在性命题,即 ∃k∈Z,x0=2k+1。为了继续证明,我们必须使用存在性推理规则 (EI) 来处理这个存在量词。该规则为我们提供了一个新的常数 k 0 ,我们无法选择它,但我们知道它满足 k0∈Z 且 x 0 = 2k 0 + 1。这里 k(无下标)是一个量化变量,而 k 0 是一个固定的但未知的常数。严格来说,x 0 = 2k + 1 不是一个命题(它是一个开放语句),但 x 0 = 2k 0 + 1 是一个命题。本节前面给出的证明没有注意到这一逻辑细微差别,而是直接从假设“x 0 为奇数”推导出存在一个常数整数(称为 k)满足 x 0 = 2k + 1。以下是显式调用 EI 的证明版本,其中通篇使用下标来区分量化变量和常量。

Proof. Assume x 0 is odd; we must prove x 0 + 7 is even. We assumed k Z , x 0 = 2 k + 1 . By EI, let k 0 be a fixed integer satisfying x 0 = 2 k 0 + 1. We must prove m Z , x 0 + 7 = 2 m . Choose m 0 = k 0 + 4, which is in Z since it is the sum of two integers. Now calculate x 0 + 7 = 2 k 0 + 1 + 7 = 2 k 0 + 8 = 2( k 0 + 4) = 2 m 0 . We have now proved m Z , x 0 + 7 = 2 m , so x 0 + 7 is indeed even.

证明。假设 x 0 为奇数;我们需要证明 x 0 + 7 为偶数。我们假设存在 k∈Z,x0=2k+1。根据 EI,设 k 0 为满足 x 0 = 2k 0 + 1 的固定整数。我们需要证明存在 m∈Z,x0+7=2m。选择 m 0 = k 0 + 4,由于它是两个整数之和,因此它属于 Z。现在计算 x 0 + 7 = 2k 0 + 1 + 7 = 2k 0 + 8 = 2(k 0 + 4) = 2m 0 。我们已经证明存在 m∈Z,x0+7=2m,所以 x 0 + 7 确实是偶数。

Formal Justification of Proof by Example (Optional)

实例证明的形式化论证(可选)

We hope the proof template for proving an existential statement x , P ( x ) is very plausible at the intuitive level: if you are able to choose a specific object x 0 and prove P ( x 0 ) is true, then surely x , P ( x ) is true! However, we can also give a more formal justification of this proof technique based on an unexpected connection to the Inference Rule for ALL. One way to formulate this rule is to introduce axioms of the form ( x , P ( x ) ) P ( x 0 ) . There is one such axiom for each open sentence P ( x ), variable x , and constant x 0 . We can replace the Inference Rule for ALL by these new axioms, since the Inference Rule for IF allows us to combine the axiom ( x , P ( x ) ) P ( x 0 ) and a known statement x , P ( x ) to deduce the new theorem P ( x 0 ). Now recall that every IF-statement is logically equivalent to its contrapositive. Apply this comment to the new axiom ( x , P ( x ) ) P ( x 0 ) to see that ( P ( x 0 ) ) ⇒∼ x , P ( x ) has the logical status of an axiom. Finally, use denial rules to simplify this statement into the form P ( x 0 ) x , P ( x ) . We can now justify “proof by example” as follows: once we have proved P ( x 0 ) is true, we use the previous formula and the Inference Rule for IF to obtain the new theorem x , P ( x ) . For simplicity, we gave this discussion for unrestricted quantifiers, but the same argument works for restricted quantifiers.

我们希望证明存在命题 ∃x,P(x) 的证明模板在直觉层面上非常合理:如果你能够选择一个特定的对象 x 0 并证明 P(x 0 ) 为真,那么 ∃x,P(x) 也必然为真!然而,我们还可以基于与“全集推理规则”之间意想不到的联系,对这种证明技巧给出更正式的论证。表述该规则的一种方法是引入形如 (∀x,P(x))⇒P(x0) 的公理。对于每个开放语句 P(x)、变量 x 和常量 x 0 ,都存在一个这样的公理。我们可以用这些新公理替换所有命题的推理规则,因为如果命题的推理规则允许我们将公理 (∀x,P(x))⇒P(x0) 与已知命题 ∀x,P(x) 结合起来,从而推导出新定理 P(x 0 )。现在回想一下,每个如果命题的语句在逻辑上都等价于它的逆否命题。将此结论应用于新公理 (∀x,∼P(x))⇒∼P(x0),可以看出 (∼∼P(x0))⇒∼∀x,∼P(x) 具有公理的逻辑地位。最后,使用否定规则将此命题简化为 P(x0)⇒∃x,P(x) 的形式。现在我们可以这样证明“举例证明”的合理性:一旦我们证明了 P(x 0 ) 为真,我们就可以使用之前的公式和 IF 推理规则来得到新的定理 ∃x,P(x)。为简便起见,我们这里仅针对无限制量词进行讨论,但同样的论证也适用于有限制量词。

Section Summary

章节概要

  1. Proof Templates. The logical structure of a statement dictates the structure of the proof of that statement. For each logical operator discussed earlier (like ⇒, , etc.), we will give a proof template for proving statements built from that operator. Statements containing several operators are proved by applying several proof templates recursively.
    证明模板。语句的逻辑结构决定了该语句证明的结构。对于前面讨论过的每个逻辑运算符(例如 ⇒、∃ 等),我们将给出一个证明模板,用于证明由该运算符构成的语句。包含多个运算符的语句需要递归地应用多个证明模板来证明。
  2. General Advice for Proofs. Memorize all definitions! Memorize and practice the proof templates! Do not forget to introduce every statement, formula, and letter in a proof with words indicating their logical status! Do not reuse letters that already have a meaning! Do not overuse exclamation marks! You can always replace a statement by an equivalent statement, which has the same logical status as the original statement. We use this step all the time when expanding definitions.
    证明的一般建议:记住所有定义!记住并练习证明模板!不要忘记在证明中用文字说明每个语句、公式和字母的逻辑地位!不要重复使用已有含义的字母!不要过度使用感叹号!你可以用一个与原语句逻辑地位相同的等价语句来替换原语句。我们在扩展定义时经常会用到这一步骤。
  3. Proof by Example. To prove x U , P ( x ) : choose a specific object x 0 (which can depend on previously introduced constants); prove x 0 U ; and prove P ( x 0 ). We often need closure properties of U to see that the chosen object does belong to U .
    示例证明。要证明 ∃x∈U,P(x):选择一个特定的对象 x 0 (它可以依赖于前面引入的常量);证明 x 0 ∈ U;并证明 P(x 0 )。我们通常需要利用 U 的封闭性来判断所选对象是否属于 U。
  4. Direct Proof. To prove P Q : assume P ; prove Q . After writing these two lines, we often continue by expanding definitions appearing in P and Q , or using another proof template to begin proving Q .
    直接证明。要证明 P ⇒ Q:假设 P 成立;证明 Q。写完这两行之后,我们通常会继续展开 P 和 Q 中出现的定义,或者使用其他证明模板来开始证明 Q。

Exercises

练习

  1. Let a and b be fixed integers. Give careful proofs of each statement, making sure to use appropriate logical status words. (a) −53 is odd. (b) 2.25 is rational. (c) 1 divides a . (d) 8 a − 6 b is even. (e) b is rational.
    设 a 和 b 为固定整数。请仔细证明下列各命题,并确保使用恰当的逻辑状态词。(a) −53 是奇数。(b) 2.25 是有理数。(c) 1 能整除 a。(d) 8a − 6b 是偶数。(e) b 是有理数。
  2. Let r and s be fixed integers. Prove the following statements. (a) 4298 is even. (b) 1.3/5.2 is rational. (c) − s divides s . (d) 6 r + 4 s − 3 is odd. (e) r s divides r 3 s 3 .
    设 r 和 s 为固定整数。证明下列命题。(a) 4298 是偶数。(b) 1.3/5.2 是有理数。(c) −s 能整除 s。(d) 6r + 4s − 3 是奇数。(e) r − s 能整除 r − s。
  3. Let x , y be fixed integers. Prove the following statements. (a) 169 is rational. (b) 10 7 − 1 is odd. (c) x divides x 3 . (d) If x is even, then 3 x 2 is even. (e) If x is odd and y is even, then y x is odd.
    设 x 和 y 为固定整数。证明下列命题。(a) 169 是有理数。(b) 10 − 1 是奇数。(c) x 能整除 x¹。(d) 若 x 为偶数,则 3x² 为偶数。(e) 若 x 为奇数且 y 为偶数,则 y − x 为奇数。
  4. Let m and n be fixed integers, and let x be a fixed real number. Prove: (a) If m is odd, then 5 m − 3 is even. (b) If 5 divides n , then 5 divides mn . (c) If x is rational, then mx + n is rational. (d) m /( m 2 + n 2 + 1) is rational.
    设 m 和 n 为固定整数,x 为固定实数。证明:(a) 若 m 为奇数,则 5m − 3 为偶数。(b) 若 5 能整除 n,则 5 能整除 mn。(c) 若 x 为有理数,则 mx + n 为有理数。(d) m/(m + n + 1) 为有理数。
  5. Let a , b , c be fixed integers. Prove: (a) If a is even, then 3 a + 5 is odd. (b) If a divides b and b divides c , then ab divides c 2 . (c) If a b , then ( a + b )/( a b ) is rational.
    设 a、b、c 为固定整数。证明:(a) 若 a 为偶数,则 3a + 5 为奇数。(b) 若 a 整除 b 且 b 整除 c,则 ab 整除 c。(c) 若 a ≠ b,则 (a + b)/(a − b) 为有理数。
  6. Let a , b , c , r , s be fixed integers. Prove: if c divides r and c divides s , then c divides ar + bs .
    设 a、b、c、r、s 为固定整数。证明:如果 c 整除 r 且 c 整除 s,则 c 整除 ar + bs。
  7. Let y , z be fixed real numbers. Prove: if y and z are both rational, then y + z is rational.
    设 y、z 为固定的实数。证明:如果 y 和 z 都是有理数,那么 y + z 也是有理数。
  8. Prove the following existential statements. (a) m Z , n Z , 30 m + 7 n = 1 . (b) x Z , y Z , 13 x + 5 y = 1 . (c) a Z , b Z , 100 a + 39 b = 1 .
    证明下列存在命题。(a) ∃m∈Z,∃n∈Z,30m+7n=1。(b) ∃x∈Z,∃y∈Z,13x+5y=1。(c) ∃a∈Z,∃b∈Z,100a+39b=1。
  9. Prove the following statements. (a) a Z > 0 , b Z > 0 , a 2 + b 2 = 100 . (b) x Z , y Z , z Z , x 5 + y 5 = z 5 . (c) a Z , b Z , c Z , d Z , a 2 + b 2 + c 2 + d 2 = 1001 .
    证明下列命题。(a) ∃a∈Z>0,∃b∈Z>0,a²+b²=100。(b) ∃x∈Z,∃y∈Z,∃z∈Z,x⁵+y⁵=z⁵。(c) ∃a∈Z,∃b∈Z,∃c∈Z,∃d∈Z,a²+b²+c²+d²=1001。
  10. Let x be a fixed real number. Use facts from algebra to prove: (a) If 3 x + 7 = 13, then x = 2. (b) If x = 2, then x 3 − 6 x 2 + 11 x = 6. (c) If x 2 + 2 x − 15 = 0, then x = 3 or x = −5. (d) If x > 3, then x 2 + 5 x − 4 > 20.
    设 x 为一个固定的实数。利用代数知识证明:(a) 若 3x + 7 = 13,则 x = 2。(b) 若 x = 2,则 x − 6x + 11x = 6。(c) 若 x + 2x − 15 = 0,则 x = 3 或 x = −5。(d) 若 x > 3,则 x + 5x − 4 > 20。
  11. Write the converse of each statement in the previous exercise. Explain informally whether each converse is true or false.
    写出上一题中每个命题的逆命题。并用通俗易懂的方式解释每个逆命题的真假。
  12. Write just the outline of a proof of each statement (imitate Example 2.18 ). Do not expand unfamiliar definitions. (a) If a group G has 49 elements, then G is commutative. (b) If a metric space X is complete and totally bounded, then X is compact. (c) If the series n = 1 a n converges, then lim n →∞ a n = 0. (d) If a function f has a right inverse, then f is surjective.
    仅写出每个命题的证明概要(模仿例 2.18)。不要展开不熟悉的定义。 (a) 如果群 G 有 49 个元素,则 G 是交换群。 (b) 如果度量空间 X 是完备的且完全有界的,则 X 是紧致的。 (c) 如果级数 ∑n=1∞an 收敛,则 lim n →∞ a n = 0。 (d) 如果函数 f 有右逆函数,则 f 是满射。
  13. Outline the proof of each statement using the templates for IF and . (a) If a group G has 700 elements, then x G , x has order 7. (b) If p is prime and k Z , p = 4 k + 1 , then y Z , z Z , p = y 2 + z 2 . (c) If g is a polynomial of odd degree, then for some real number r , g ( r ) = 0.
    使用 IF 和 ∃ 的模板概述每个陈述的证明。 (a) 如果群 G 有 700 个元素,则 ∃x∈G,x 的阶为 7。 (b) 如果 p 是素数且 ∃k∈Z,p=4k+1,则 ∃y∈Z,∃z∈Z,p=y2+z2。 (c) 如果 g 是奇数次多项式,则对于某个实数 r,g(r) = 0。
  14. Let m and n be fixed positive integers. Carefully explain the error in the following incorrect proof of the statement: “if m is even, then m / n is even.” Proof. Assume m is even. We must prove m / n is even. We have assumed there exists an integer k with m = 2 k . We must prove p Z , m / n = 2 p . Dividing the assumed equation m = 2 k by n (which is not zero) produces m / n = (2 k )/ n = 2( k / n ). Hence, choosing p = k / n , we do indeed have m / n = 2 p . So m / n is even.
    设 m 和 n 为固定的正整数。请仔细解释以下证明中的错误:“如果 m 是偶数,则 m/n 是偶数。” 证明:假设 m 是偶数。我们需要证明 m/n 是偶数。我们假设存在一个整数 k,使得 m = 2k。我们需要证明存在 p∈Z,使得 m/n = 2p。将假设的等式 m = 2k 除以 n(n 不为零),得到 m/n = (2k)/n = 2(k/n)。因此,取 p = k/n,我们确实有 m/n = 2p。所以 m/n 是偶数。
  15. Let a , b , and c be fixed positive integers. We are trying to prove: “if a divides b and a divides c , then a divides b + c .” Find and correct the error in the proof below. Proof. Assume a divides b and a divides c ; we must prove a divides b + c . We have assumed there is an integer k with b = ak . We have also assumed there is an integer k with c = ak . We must prove m Z , b + c = a m . Choose m = 2 k , which is an integer. Use the assumptions to compute b + c = ak + ak = a (2 k ) = am , as needed.
    设 a、b 和 c 为固定的正整数。我们要证明:“如果 a 整除 b 且 a 整除 c,则 a 整除 b + c。”找出并改正以下证明中的错误。证明:假设 a 整除 b 且 a 整除 c;我们需要证明 a 整除 b + c。我们假设存在整数 k 使得 b = ak。我们也假设存在整数 k 使得 c = ak。我们需要证明存在 m∈Z,b+c=am。选择 m = 2k,这是一个整数。根据需要,利用假设计算 b + c = ak + ak = a(2k) = am。
  16. Let n be a fixed integer. Find the error in the following proof that n 2 + n is even. Proof. We must prove there exists k such that n 2 + n = 2 k . Choose k = ( n 2 + n )/2. Multiplying both sides by 2, we see that n 2 + n = 2 k does hold.
    设 n 为一个固定的整数。找出以下证明 n 2 + n 为偶数的错误之处。证明:我们需要证明存在 k 使得 n 2 + n = 2k。取 k = (n 2 + n)/2。两边同乘以 2,我们发现 n 2 + n = 2k 成立。
  17. Let x be a fixed integer. Criticize the following proof of the statement “if x is even then x 3 − 1 is odd.” Proof. x = 2 k , k Z , x 3 − 1 = (2 k ) 3 − 1 = 8 k 3 − 1 = 2(4 k 3 − 1) + 1, 4 k 3 1 Z .
    设 x 为一个固定的整数。批判以下关于“如果 x 是偶数,则 x 3 − 1 是奇数”的证明。证明:x = 2k,k∈Z,x 3 − 1 = (2k) 3 − 1 = 8k 3 − 1 = 2(4k 3 − 1) + 1,4k3−1∈Z。
  18. Let c be a fixed integer. Criticize the following proof of the statement “if 10 divides c + 4, then 10 divides c − 6.” Assume 10 divides c − 6, which means a Z , c 6 = 10 a . Prove 10 divides c + 4, which means b Z , c + 4 = 10 b . Choose b = a + 1, which is in Z by closure. Use algebra to compute 10 b = 10( a + 1) = 10 a + 10 = c − 6 + 10 = c + 4.
    设 c 为一个固定的整数。请批判以下关于“如果 10 能整除 c + 4,那么 10 也能整除 c − 6”的证明。假设 10 能整除 c − 6,这意味着 ∃a∈Z, c−6=10a。证明 10 能整除 c + 4,这意味着 ∃b∈Z, c+4=10b。取 b = a + 1,根据封闭性,b 属于 Z。利用代数方法计算 10b = 10(a + 1) = 10a + 10 = c − 6 + 10 = c + 4。
  19. Let r and s be fixed real numbers. Find the error in the following proof that if r and s are rational and s ≠ 0, then r 2 / s 2 is rational. Proof. Assume r and s are rational and s ≠ 0. Prove r 2 / s 2 is rational. We must prove m Z , n Z , n 0 and r 2 / s 2 = m / n . Choose m = r 2 and n = s 2 ; note that n ≠ 0 because s ≠ 0.
    设 r 和 s 为固定的实数。找出以下证明中的错误:如果 r 和 s 是有理数且 s ≠ 0,则 r/s 是有理数。证明:假设 r 和 s 是有理数且 s ≠ 0。证明 r/s 是有理数。我们需要证明存在 m∈Z,n∈Z,n≠0 且 r/s = m/n。选择 m = r/s,n = s/s;注意 n ≠ 0,因为 s ≠ 0。
  20. Give a correct proof of the statement in the previous problem.
    请给出上一题中命题的正确证明。
  21. Let a and b be fixed nonzero integers. Find the error in the following proof of the statement “if a divides b and b divides a then b = ± a .” Proof. Assume a divides b and b divides a . Prove b = ± a . By assumption, there is an integer c with b = ca and a = cb . By substitution, a = cb = c ( ca ) = c 2 a . Since a ≠ 0, we can divide by a to get c 2 = 1. Then c = 1 or c = −1, hence b = a or b = − a .
    设 a 和 b 为固定的非零整数。找出以下证明“如果 a 整除 b 且 b 整除 a,则 b = ±a”的错误。证明:假设 a 整除 b 且 b 整除 a。证明 b = ±a。根据假设,存在整数 c,使得 b = ca 且 a = cb。代入得 a = cb = c(ca) = ca。由于 a ≠ 0,我们可以两边同除以 a,得到 c = 1。因此 c = 1 或 c = -1,所以 b = a 或 b = -a。
  22. Define a rational number r to be dyadic iff m Z , k Z 0 , r = m / 2 k . Let x and y be fixed rational numbers. Prove: (a) If x and y are dyadic, then xy is dyadic. (b) If x and y are dyadic, then x + y is dyadic.
    定义有理数 r 为二重数当且仅当存在 m∈Z,k∈Z≥0,r=m/2k。设 x 和 y 为固定的有理数。证明:(a) 若 x 和 y 是二重数,则 xy 也是二重数。(b) 若 x 和 y 是二重数,则 x + y 也是二重数。
  23. Use the group axioms in § 2.1 to prove this theorem about fixed elements a , b , c in a group ( G , , e ) : if a b = a c , then b = c .
    利用 §2.1 中的群公理来证明关于群 (G,⋆,e) 中固定元素 a、b、c 的这个定理:如果 a⋆b=a⋆c,则 b = c。
  24. Define an integer m to be bisquare iff a Z , b Z , m = a 2 + b 2 . Let x , y be fixed integers. Prove: (a) 13 is bisquare. (b) 200 is bisquare. (c) 610 is bisquare. (d) x 2 is bisquare. (e) If x is bisquare, then xy 2 is bisquare. (f) Not every integer is bisquare.
    定义整数 m 为双平方数当且仅当存在 a∈Z, b∈Z,使得 m=a²+b²。设 x, y 为固定整数。证明:(a) 13 是双平方数。(b) 200 是双平方数。(c) 610 是双平方数。(d) x 2 是双平方数。(e) 若 x 是双平方数,则 xy 2 也是双平方数。(f) 并非每个整数都是双平方数。
  25. Let r and s be fixed integers. Prove: if r and s are bisquare, then rs is bisquare.
    设 r 和 s 为固定整数。证明:如果 r 和 s 都是双平方数,那么 rs 也是双平方数。

2.3 Contrapositive Proofs and IFF Proofs

2.3 逆否命题证明和因果函数证明

Recall our direct proof template for proving P Q : “ Assume P . Prove Q .” Let us see what happens when we use this template in the following example.

回想一下我们证明 P ⇒ Q 的直接证明模板:“假设 P,证明 Q。” 让我们看看在下面的例子中使用这个模板会发生什么。

2.21. Example. Let x be a fixed integer. Prove: if x 2 is even, then x is even.

2.21. 例。设 x 为一个固定的整数。证明:如果 x 2 是偶数,则 x 也是偶数。

Proof. Assume x 2 is even; prove x is even. We have assumed that for some integer k , x 2 = 2 k . We must prove m Z , x = 2 m . Taking the square root of the known equation x 2 = 2 k , we deduce

证明。假设 x 2 为偶数;证明 x 为偶数。我们假设对于某个整数 k,x 2 = 2k。我们必须证明存在 m∈Z,x=2m。对已知方程 x 2 = 2k 开平方根,我们推导出

x x = 2 k k = 4 ( k k / 2 ) = 2 k k / 2 .

So choosing m = k / 2 , we have x = 2 m , as needed.

因此选择 m=k/2,我们得到 x = 2m,符合要求。

This proof has a minor flaw and a major flaw — can you spot them? The minor flaw is that x might be negative, but 2 k / 2 is always nonnegative. This can be corrected by using the negative of the square root when x < 0. The major flaw is that the proof did not check that m was in Z , as required by the definition. In general, dividing an integer k by 2 and then taking a square root produces a real number that may not be an integer. There is no straightforward way to correct this error. We need a new approach, called the contrapositive proof method.

这个证明存在一个小缺陷和一个大缺陷——你能发现它们吗?小缺陷在于 x 可能为负数,但 2k/2 始终为非负数。当 x < 0 时,可以使用平方根的负数来修正这个问题。大缺陷在于证明没有验证 m 是否属于整数集 Z,而这是定义所要求的。一般来说,将整数 k 除以 2 再开平方根得到的实数可能不是整数。目前没有直接的方法可以修正这个错误。我们需要一种新的方法,称为逆否命题证明法。

Contrapositive Proofs

逆否命题证明

Recall that to prove any statement A , we can instead prove a new statement B that is known to be equivalent to A . In particular, taking A to be the IF-statement P Q , we know that the contrapositive IF-statement (∼ Q ) ⇒ (∼ P ) is equivalent to A . Using the direct proof template to prove this new statement, we arrive at the following new proof template for proving the original IF-statement, called proof by contrapositive .

回想一下,为了证明任何命题 A,我们可以转而证明一个已知与 A 等价的新命题 B。特别地,如果 A 是 IF 命题 P ⇒ Q,我们知道其逆否命题 (∼Q) ⇒ (∼P) 与 A 等价。使用直接证明模板来证明这个新命题,我们得到了以下用于证明原 IF 命题的新证明模板,称为逆否命题证明法。

2.22. Contrapositive Proof Template to Prove P Q .

2.22. 证明 P ⇒ Q 的逆否命题证明模板。

Assume Q . Prove P .

假设 ∼Q。证明 ∼P。

[We often replace ∼ Q and ∼ P by useful denials of Q and P , respectively. These are found using the denial rules.]

[我们通常用对 Q 和 P 的有效否定分别替换 ∼Q 和 ∼P。这些否定可以通过否定规则找到。]

We can repair the direct proof in Example 2.21 by doing a contrapositive proof instead. For this new proof, and throughout our initial discussion of proofs, we assume that the following fact about odd and even numbers is already known:

我们可以通过反证法来修正例 2.21 中的直接证明。对于这个新的证明,以及我们之前对证明的讨论,我们假设关于奇数和偶数的以下事实是已知的:

n n Z Z , n n is not even 甚至都不是 n n is odd. 很奇怪。

(2.1)

(Though you are probably aware of this fact, it is not so easy to prove! We provide a proof of this fact later.) Since (∼ P ) ⇔ Q is equivalent to P ⇔ (∼ Q ), the fact also says that for all integers n , n is even iff n is not odd. Since (∼ P ) ⇔ Q is also equivalent to P Q , yet another way to rephrase the fact is to say that every integer is even or odd, but not both. With this fact in hand, let us return to our initial example.

(虽然你可能已经知道这个事实,但证明起来并不容易!我们稍后会给出证明。)由于 (∼P) ⇔ Q 等价于 P ⇔ (∼Q),这个事实也表明,对于所有整数 n,n 为偶数当且仅当 n 不是奇数。由于 (∼P) ⇔ Q 也等价于 P⊕Q,这个事实还可以表述为:每个整数要么是偶数,要么是奇数,但不能既是偶数又是奇数。有了这个事实,让我们回到最初的例子。

2.23. Example. Let x be a fixed integer. Prove: if x 2 is even, then x is even.

2.23. 例。设 x 为一个固定的整数。证明:如果 x 2 是偶数,则 x 也是偶数。

Proof by Contrapositive. Assume x is not even. Prove x 2 is not even. Using the known fact, we have assumed x is odd, which means there is an integer k with x = 2 k + 1. Similarly, we must prove x 2 is odd, which means m Z , x 2 = 2 m + 1 . Squaring the known equation x = 2 k + 1 gives x 2 = 4 k 2 + 4 k + 1 = 2(2 k 2 + 2 k ) + 1. So choosing m = 2 k 2 + 2 k , which is an integer by closure properties of Z , we have x 2 = 2 m + 1 as needed.

逆否命题证明。假设 x 不是偶数。证明 x 2 不是偶数。利用已知事实,我们假设 x 是奇数,这意味着存在整数 k 使得 x = 2k + 1。类似地,我们必须证明 x 2 是奇数,这意味着存在 m∈Z,x2=2m+1。将已知方程 x = 2k + 1 平方,得到 x 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1。因此,选择 m = 2k 2 + 2k,根据 Z 的封闭性质,m 是一个整数,我们得到 x 2 = 2m + 1,满足要求。

Comments: We can see retroactively why the contrapositive proof works and the direct proof fails. Namely, proving the contrapositive allows us to pass from assumed information about x to needed information about x 2 by squaring an equation. This step succeeds because Z is closed under addition and multiplication. On the other hand, in the direct proof we tried to go from assumptions about x 2 to needed information about x by taking square roots. Here we did not have closure properties to guarantee that the integer m (chosen at the end of the proof) must be an integer.

评论:我们可以事后看出为什么逆否命题证明有效而直接证明无效。具体来说,证明逆否命题允许我们通过对等式进行平方运算,从关于 x 的假设信息推导出关于 x 的所需信息。这一步之所以成功,是因为 Z 对加法和乘法封闭。另一方面,在直接证明中,我们试图通过开平方根运算,从关于 x 的假设推导出关于 x 的所需信息。这里我们没有封闭性质来保证整数 m(在证明的最后选择)必须是整数。

2.24. Example. Let x and y be fixed real numbers. Prove: if x 2 y + xy 2 < 30, then x < 2 or y < 3. Proof by Contrapositive. Assume x ≥ 2 and y ≥ 3. Prove x 2 y + xy 2 ≥ 30. To proceed, we need to use two known theorems from algebra regarding inequalities:

2.24. 例。设 x 和 y 为固定的实数。证明:若 xy + xy<sub>1</sub> < 30,则 x < 2 或 y < 3。用逆否命题证明。假设 x ≥ 2 且 y ≥ 3。证明 xy<sub>2</sub> + xy<sub>3</sub> ≥ 30。为了继续,我们需要用到代数中关于不等式的两个已知定理:

a 一个 , b b , c c , d d R , ( a 一个 b b c c d d ) a 一个 + c c b b + d d ;

(2.2)

a 一个 , b b , c c , d d R , ( a 一个 b b 0 c c d d 0 ) a 一个 c c b b d d 0.

(2.3)

Taking a = x , b = 2, c = y , and d = 3 in the first theorem, we deduce that x + y ≥ 2 + 3, i.e., x + y ≥ 5. [We have just used the Inference Rules for ALL and IF.] Using the same values in the second theorem, we deduce that xy ≥ 2 · 3 = 6. Now, since x + y ≥ 5 ≥ 0 and xy ≥ 6 ≥ 0, we can use the second theorem again to deduce that ( x + y ) xy ≥ 30. By algebra, we know ( x + y ) xy = x 2 y + xy 2 . So x 2 y + xy 2 ≥ 30, completing the proof. Comments: How did we know to try a contrapositive proof here, rather than a direct proof? Perhaps we started a direct proof by saying: “Assume x 2 y + xy 2 < 30. Prove x < 2 or y < 3.” At this point, we probably got stuck because it was unclear how to move forward from the rather complicated assumption. Doing a contrapositive proof instead provides the much more tractable assumption “ x ≥ 2 and y ≥ 3.”

在第一个定理中,令 a = x,b = 2,c = y,d = 3,我们推导出 x + y ≥ 2 + 3,即 x + y ≥ 5。[我们刚刚使用了 ALL 和 IF 的推理规则。] 在第二个定理中,使用相同的值,我们推导出 xy ≥ 2 · 3 = 6。现在,由于 x + y ≥ 5 ≥ 0 且 xy ≥ 6 ≥ 0,我们可以再次使用第二个定理推导出 (x + y)xy ≥ 30。根据代数运算,我们知道 (x + y)xy = xy + xy。因此,xy + xy ≥ 30,证明完毕。评论:我们怎么知道这里应该用逆否命题证明,而不是直接证明呢?或许我们一开始就直接证明:“假设 xy + xy < 30。证明 x < 2 或 y < 3。” 这时我们可能就卡住了,因为不清楚如何从这个相当复杂的假设出发继续证明。而使用逆否命题证明则能得到更容易处理的假设:“x ≥ 2 且 y ≥ 3”。

2.25. Example. Let x and y be fixed integers. Prove: if xy is irrational, then x is irrational or y is irrational.

2.25. 例。设 x 和 y 为固定整数。证明:如果 xy 是无理数,则 x 是无理数或 y 是无理数。

Proof. [The given statement has a lot of negative logic in it (note “irrational” means “not rational”), so we try simplifying it with a contrapositive proof. Denying the conclusion, we begin with the line:] Assume x is rational AND y is rational. [Denying the hypothesis, we continue by saying:] Prove xy is rational. [We continue by expanding definitions:] We have assumed x = m / n and y = p / q for some integers m , n , p , q with n and q nonzero. We must prove a Z , b Z , b 0 ( x y = a / b ) . Multiplying together our assumptions , we deduce that xy = ( mp )/( nq ). Choose a = mp and b = nq ; note a and b are integers by closure of Z , and b ≠ 0 since b is the product of two nonzero integers. Also xy = a / b by the above calculation. [We have proved our goal, so the proof is now done.]

证明。[给定的陈述包含很多否定逻辑(注意“无理数”意味着“非理性”),因此我们尝试用逆否命题证明来简化它。否定结论,我们从以下语句开始:] 假设 x 是有理数且 y 是有理数。[否定假设,我们继续说:] 证明 xy 是有理数。[我们继续展开定义:] 我们假设 x = m/n 且 y = p/q,其中 m、n、p、q 为整数,且 n 和 q 非零。我们必须证明 ∃a∈Z,∃b∈Z,b≠0∧(xy=a/b)。将我们的假设相乘,我们推导出 xy = (mp)/(nq)。选择 a = mp 和 b = nq;注意根据 Z 的封闭性,a 和 b 都是整数,并且 b ≠ 0,因为 b 是两个非零整数的乘积。此外,根据上述计算,xy = a/b。 [我们已经证明了我们的目标,所以证明工作到此结束。]

Proving an AND Statement

证明 AND 语句

Let us take a break from the logical subtleties of IF-statements to introduce one of the more straightforward proof templates. This template tells us how to prove an AND-statement .

让我们暂时放下 IF 语句的逻辑细节,来介绍一个更直接的证明模板。这个模板告诉我们如何证明 AND 语句。

2.26. Proof Template to Prove P Q .

2.26.证明 P∧Q 的证明模板。

Part 1. Prove P .

第一部分。证明 P。

Part 2. Prove Q .

第二部分。证明问题。

I hope this proof template is fully believable to you at the intuitive level. But we can still justify it by appealing to the truth table defining , as follows. Consulting the truth table, we see that the composite statement P Q is true if and only if the two separate statements P (on the one hand) and Q (on the other hand) are both true. Thus to prove the composite statement, it is enough to prove P by itself and also to prove Q by itself. These proofs often proceed by recursively using more proof templates. More generally, to prove an AND-statement of the form P 1 P 2 P n , we must carry out n independent subproofs: prove P 1 ; next, prove P 2 ; … finally, prove P n .

我希望这个证明模板在直觉层面上完全可信。但我们仍然可以通过参考定义“∧”的真值表来进一步证明它,如下所示。查阅真值表可知,复合命题 P∧Q 为真当且仅当两个单独的命题 P(一方面)和 Q(另一方面)都为真。因此,要证明复合命题,只需分别证明 P 和 Q 即可。这些证明通常会递归地使用更多的证明模板。更一般地,要证明形如 P1∧P2∧⋯∧Pn 的“与”命题,我们必须执行 n 个独立的子证明:证明 P 1 ;接下来,证明 P 2 ;……最后,证明 P n

2.27. Example. Let b be a fixed integer. Prove 1 divides b and b divides b .

2.27. 例。设 b 为一个固定的整数。证明 1 能整除 b,且 b 也能整除 b。

Proof. First, we prove 1 divides b . By definition, we must prove c Z , b = 1 c . Choose c = b ; note c is an integer and 1 c = 1 b = b . Second, we prove b divides b . We must prove d Z , b = b d . Choose d = 1; this is an integer, and bd = b 1 = b .

证明。首先,我们证明 1 整除 b。根据定义,我们必须证明存在 c∈Z,b=1c。选择 c = b;注意 c 是整数,且 1c = 1b = b。其次,我们证明 b 整除 b。我们必须证明存在 d∈Z,b=bd。选择 d = 1;这是一个整数,且 bd = b1 = b。

2.28. Example. Let x and y be fixed real numbers. Prove: if x < y then x < ( x + y )/2 < y .

2.28. 例。设 x 和 y 为固定的实数。证明:如果 x < y,则 x < (x + y)/2 < y。

Proof. Assume x < y . Prove x < ( x + y )/2 < y . The goal contains a hidden AND; so we must prove x < ( x + y )/2 AND ( x + y )/2 < y .

证明。假设 x < y。证明 x < (x + y)/2 < y。目标中包含一个隐藏的“与”关系;因此,我们必须证明 x < (x + y)/2 且 (x + y)/2 < y。

Part 1. Prove x < ( x + y )/2. Adding x to both sides of the assumption x < y , we deduce 2 x < x + y . Multiplying this by the positive number 1/2, we get x < ( x + y )/2.

第一部分。证明 x < (x + y)/2。在假设 x < y 的两边同时加上 x,我们得出 2x < x + y。将此式乘以正数 1/2,我们得到 x < (x + y)/2。

Part 2. Prove ( x + y )/2 < y . Adding y to both sides of the assumption x < y , we deduce x + y < 2 y . Multiplying this by the positive number 1/2, we get ( x + y )/2 < y .

第二部分。证明 (x + y)/2 < y。在假设 x < y 的两边同时加上 y,我们得出 x + y < 2y。将此式乘以正数 1/2,我们得到 (x + y)/2 < y。

Proving IFF Statements

证明 IFF 语句

Our next proof template tells us how to prove an “if and only if” statement P Q . The template arises by recalling that P Q is equivalent to ( P Q ) ( Q P ) . We already know how to prove this new statement, using the templates for and ⇒. Writing this out, we get the following .

我们的下一个证明模板告诉我们如何证明“当且仅当”命题 P ⇔ Q。该模板源于我们回顾 P ⇔ Q 等价于 (P⇒Q)∧(Q⇒P)。我们已经知道如何使用 ∧ 和 ⇒ 的模板来证明这个新命题。写出来,我们得到以下结果。

2.29. Proof Template for Proving P Q .

2.29. 证明 P ⇔ Q 的证明模板。

Part 1. Prove P Q (by any method).

第一部分。证明 P ⇒ Q(用任何方法)。

Part 2. Prove Q P (by any method).

第二部分。证明 Q ⇒ P(用任意方法)。

We should make a few comments about this template. The proof method is called a two-part proof of the IFF-statement, since the proof requires us to prove two separate IF-statements. Part 1 is often called the forward direction or the direct implication . Part 2 is often called the backward direction or the converse implication . If we are proving “ P is necessary and sufficient for Q ,” part 1 can be called the proof of sufficiency , and part 2 can be called the proof of necessity .

关于这个模板,我们需要做一些说明。这种证明方法被称为IFF语句的两部分证明,因为证明过程需要我们证明两个独立的IF语句。第一部分通常被称为正向推导或直接蕴含。第二部分通常被称为反向推导或逆向蕴含。如果我们证明的是“P是Q的充要条件”,那么第一部分可以称为充分性证明,第二部分可以称为必要性证明。

Beginners sometimes forget to write part 2 by the time they finish writing part 1 of the proof. Thus, they have omitted half of the proof! To avoid this, I recommend setting up the entire proof structure shown above, by writing “ Part 1 . Prove P Q ,” leaving lots of space, writing “ Part 2 . Prove Q P ,” and then filling in both parts. Another remark is that there are multiple ways of completing part 1 — we could use direct proof, contrapositive proof, or proof by contradiction (to be discussed later). Independently of part 1, we have similar choices for how to prove part 2. The natural choice is to try a direct proof for each part. Often, however, one direction of the proof may be easier using a contrapositive proof (or perhaps a contradiction proof). Remember that these options are available. One final point: any temporary assumptions made in part 1 (e.g., “assume P ”) are no longer available when we reach part 2. Similarly, things proved in part 1 using the assumptions in part 1 can no longer be used in part 2 (unless we can reprove them using the assumptions in part 2). On the other hand, variables and constants introduced in part 1 of the proof are allowed to be reused (with fresh meanings) in part 2 — although it may be preferable to use new letters instead to avoid confusion.

初学者有时会在完成证明的第一部分后忘记写第二部分。这样一来,他们就漏掉了证明的一半!为了避免这种情况,我建议先搭建好上面所示的完整证明结构:写上“第一部分:证明 P ⇒ Q”,留出足够的空间,再写上“第二部分:证明 Q ⇒ P”,然后分别填写两部分。另外需要注意的是,完成第一部分的方法有很多种——我们可以使用直接证明、逆否命题证明或反证法(稍后会讨论)。与第一部分无关,我们也有类似的方法来证明第二部分。最自然的选择是尝试对每一部分都进行直接证明。然而,通常情况下,使用逆否命题证明(或者反证法)可能更容易证明某个方向的结论。请记住,这些都是可行的方法。最后一点:第一部分中做出的任何临时假设(例如,“假设 P”)在第二部分中都不再适用。同样,第一部分中基于这些假设证明的结论也不能在第二部分中使用(除非我们能在第二部分中使用同样的假设重新证明它们)。另一方面,证明的第一部分中引入的变量和常量可以在第二部分中重复使用(但需要赋予新的含义)——尽管为了避免混淆,最好使用新的字母来表示它们。

2.30. Example. Let x be a fixed integer. Prove x is even iff x 2 is even.

2.30. 例。设 x 为一个固定的整数。证明 x 为偶数当且仅当 x 2 为偶数。

Proof. Part 1. Assume x is even; prove x 2 is even. We have assumed there is an integer k with x = 2 k . We must prove m Z , x 2 = 2 m . Squaring the assumption gives x 2 = 4 k 2 = 2(2 k 2 ). Choose m = 2 k 2 ; then m Z and x 2 = 2 m , as needed.

证明。第一部分。假设 x 为偶数;证明 x 2 为偶数。我们假设存在整数 k 使得 x = 2k。我们需要证明存在 m∈Z,x2=2m。假设成立后,可得 x 2 = 4k 2 = 2(2k 2 )。选择 m = 2k 2 ;则 m∈Z 且 x 2 = 2m,满足要求。

Part 2. We must prove: “if x 2 is even, then x is even.” We did this earlier in the section (using a contrapositive proof), so we do not repeat the proof here.

第二部分。我们必须证明:“如果 x 2 是偶数,那么 x 也是偶数。”我们在本节前面已经证明了这一点(使用逆否命题证明),所以这里不再赘述。

2.31. Example. Let r be a fixed real number. Prove r is rational iff − r is rational.

2.31. 例。设 r 为一个固定的实数。证明 r 是有理数当且仅当 −r 是有理数。

Proof. Part 1. Assume r is rational. Prove r is rational. We have assumed there are integers p , q with q ≠ 0 and r = p / q . We must prove m , n Z , n 0 r = m / n . Since − r = ( − p )/ q , we can choose m = − p and n = q ≠ 0; these are integers with − r = m / n .

证明。第一部分。假设 r 是有理数。证明 −r 是有理数。我们假设存在整数 p 和 q,其中 q ≠ 0,且 r = p/q。我们需要证明存在 m, n∈Z, n≠0 ∧ −r = m/n。由于 −r = ( − p)/q,我们可以选择 m = −p 和 n = q ≠ 0;这些整数满足 −r = m/n。

Part 2. Assume r is rational. Prove r is rational. We have assumed there are integers p , q with q ≠ 0 and − r = p / q . We must prove m , n Z , n 0 r = m / n . Since r = −( − r ) = ( − p )/ q , we can choose m = − p and n = q ≠ 0; these are integers with r = m / n .

第二部分:假设 −r 是有理数。证明 r 是有理数。我们假设存在整数 p 和 q,其中 q ≠ 0,且 −r = p/q。我们需要证明存在 m, n∈Z, n≠0 ∧ r = m/n。由于 r = −( − r) = ( − p)/q,我们可以选择 m = −p 和 n = q ≠ 0;这些是满足 r = m/n 的整数。

Comments: Part 2 is totally separate from part 1, so we are allowed to reuse the letters p , q , m , n in part 2 (which no longer have the meanings they did in part 1). Despite the extreme similarity of the two parts of this proof, both parts are needed to have a complete proof of the IFF-statement. On the other hand, we will see later that part 1 of the proof really proved a universal statement: r R , ( r Q r Q ) . This leads to a shorter proof of part 2 that relies on the result proved in part 1. Using the Inference Rule for ALL, we can replace r by − r in the universal statement written above, to see that ( r ) Q ( r ) Q . Since −( − r ) = r by algebra, we get ( r ) Q r Q , which proves part 2 without expanding any definitions.

注释:第二部分与第一部分完全独立,因此我们可以在第二部分中重复使用字母 p、q、m、n(它们不再具有第一部分中的含义)。尽管此证明的两部分极其相似,但两部分对于完整证明 IFF 语句都是必需的。另一方面,我们稍后会看到,证明的第一部分实际上证明了一个全称命题:∀r∈R,(r∈Q⇒−r∈Q)。这使得第二部分的证明更加简洁,并且依赖于第一部分中证明的结果。利用全称命题的推理规则,我们可以将上述全称命题中的 r 替换为 −r,从而得到 (−r)∈Q⇒−(−r)∈Q。由于 −( − r) = r(代数关系),我们得到 (−r)∈Q⇒r∈Q,这证明了第二部分,而无需展开任何定义。

2.32. Example. Outline the proof of this statement:

2.32. 示例。简述以下命题的证明:

“if X R k , then: X is compact iff X is closed and bounded.”

“如果 X⊆Rk,则:X 是紧致的当且仅当 X 是闭集且有界的。”

Solution. We can use the proof templates to generate the structural skeleton of the proof without knowing the definitions of “compact,” “closed,” and “bounded.” Here is the proof outline:

解决方案。我们可以使用证明模板来生成证明的结构框架,而无需知道“紧致”、“封闭”和“有界”的定义。以下是证明概要:

Assume X R k . Prove X is compact iff X is closed and bounded.

假设 X⊆Rk。证明 X 是紧集当且仅当 X 是闭集且有界集。

  • Part 1. Assume X is compact.

    第一部分。假设 X 是紧致的。

    • Part 1a. Prove X is closed. (…)
      第一部分(a):证明 X 是封闭的。(……)
    • Part 1b. Prove X is bounded. (…)
      第一部分(b):证明 X 有界。(……)
  • Part 2. Assume X is closed and bounded. Prove X is compact. (…)

    第二部分:假设 X 是闭集且有界集。证明 X 是紧集。(…)

Comments: The main logical operator in the given statement is IF, so we generate the first line of the solution using the direct proof template for IF. This reduces us to proving an IFF-statement, which requires a two-part proof. We use the direct proof template two more times (along with the template for AND in part 1) to finish the proof outline. We write (…) for omitted portions of the proof, which would require us to know the definitions of the terms involved. Note that the assumption X R k is available throughout the entire proof, whereas the assumption that X is compact can be used only in part 1. Other outlines are possible. For instance, we could use a contrapositive proof in part 2 and write “ Assume X is not compact; prove X is not closed or not bounded.” This new goal could be expanded further using the OR-proof template, given later.

注释:给定语句中的主要逻辑运算符是 IF,因此我们使用 IF 的直接证明模板生成解决方案的第一行。这使我们简化为证明一个 IFF 语句,这需要两部分证明。我们再使用两次直接证明模板(以及第一部分中的 AND 模板)来完成证明大纲。对于省略的证明部分,我们用 (…) 表示,这些部分需要我们了解相关术语的定义。请注意,假设 X⊆Rk 贯穿整个证明,而假设 X 是紧集只能在第一部分中使用。还有其他可能的证明大纲。例如,我们可以在第二部分中使用逆否命题证明,并写成“假设 X 不是紧集;证明 X 不是闭集或无界集”。这个新目标可以使用后面给出的 OR 证明模板进一步扩展。

Formally Justifying the Proof Templates (Optional)

正式论证证明模板(可选)

We have given intuitive justifications for the new proof templates in this section based on truth tables. We can give a completely rigorous derivation of these templates based on certain tautologies and the Inference Rule for IF. For example, consider the template for proving the AND-statement P Q R . Suppose we can prove P by itself, Q by itself, and R by itself. Note that ( P ( Q ( R ( P Q R ) ) ) ) is an axiom, because it is a tautology (as one may check with a truth table). Applying the Inference Rule for IF to this tautology and the known statement P , we get ( Q ( R ( P Q R ) ) ) . Applying the Inference Rule for IF to this new statement and the known statement Q , we get ( R ( P Q R ) ) . Finally, applying the Inference Rule for IF to this statement and the known statement R , we get the sought-for theorem P Q R . This chain of deductions, using the Inference Rule for IF three times in a row, is the same for any AND-statement P Q R . The proof template for AND lets us condense this argument by just proving P , Q , and R separately.

本节中,我们基于真值表对新的证明模板给出了直观的解释。基于某些重言式和“如果”推理规则,我们可以对这些模板进行完全严格的推导。例如,考虑证明“与”语句 P∧Q∧R 的模板。假设我们可以分别证明 P、Q 和 R。注意,(P⇒(Q⇒(R⇒(P∧Q∧R)))) 是一个公理,因为它是一个重言式(可以通过真值表验证)。将“如果”推理规则应用于这个重言式和已知的语句 P,我们得到 (Q⇒(R⇒(P∧Q∧R)))。将“如果”推理规则应用于这个新的语句和已知的语句 Q,我们得到 (R⇒(P∧Q∧R))。最后,将“如果”推理规则应用于该命题和已知命题 R,我们得到所求定理 P∧Q∧R。对于任何“与”命题 P∧Q∧R,连续三次使用“如果”推理规则的这一系列推导都相同。“与”命题的证明模板允许我们通过分别证明 P、Q 和 R 来简化论证。

Next, consider the contrapositive proof template for proving P Q . Assume the direct proof template for IF-statements is already available. Then assuming ∼ Q and proving ∼ P proves the new theorem (∼ Q ) ⇒ (∼ P ). Using this theorem and the tautology

接下来,考虑证明 P ⇒ Q 的逆否命题证明模板。假设 IF 语句的直接证明模板已经可用。那么,假设 ∼Q 并证明 ∼P 即可证明新定理 (∼Q) ⇒ (∼P)。利用该定理和重言式

( ( Q ) ( P P ) ) ( P P Q ) ,

the Inference Rule for IF gives the new theorem P Q . The proof template for IFF can be justified similarly (see Exercise 24).

IF 的推理规则给出了新的定理 P ⇒ Q。IFF 的证明模板可以类似地证明(参见练习 24)。

Section Summary

章节概要

  1. Contrapositive Proof Template. To prove P Q via a contrapositive proof: Assume Q . Prove P .
    逆否命题证明模板。要用逆否命题证明 P ⇒ Q: 假设 ∼Q。证明 ∼P。
  2. AND Proof Template. To prove P 1 P 2 P n : Part 1. Prove P 1 . Part 2. Prove P 2 .
    AND 证明模板。证明 P1∧P2∧⋯∧Pn: 第 1 部分:证明 P 1 . 第 2 部分:证明 P 2 .
  3. Part n . Prove P n . IFF Proof Template. To prove P Q : Part 1. Prove P Q (by any method). Part 2. Prove Q P (by any method).
    第 n 部分。证明 P n 当且仅当证明模板。证明 P ⇔ Q: 第 1 部分。证明 P ⇒ Q(用任意方法)。 第 2 部分。证明 Q ⇒ P(用任意方法)。
  4. Points of Advice. (a) Memorize the new proof templates! (b) Given a complex statement built up using multiple logical operators, you need to combine multiple proof templates to generate the structure of the proof. (c) When do we use contrapositive proofs? Try this method when a direct proof fails, or when the original statement contains negative logic, or when the hypothesis of the given IF-statement seems unwieldy. (d) When proving an IFF-statement, do not forget that both implications must be proved! Also, the two parts of the proof are completely independent, so that assumptions made in part 1 cannot be used in part 2, and vice versa. But letters used in part 1 can be reused in part 2, with new meanings.
    建议要点:(a) 记住新的证明模板!(b) 给定一个由多个逻辑运算符构成的复杂语句,你需要组合多个证明模板来生成证明的结构。(c) 何时使用逆否命题证明?当直接证明失败、原始语句包含否定逻辑,或者给定条件语句的假设显得笨拙时,可以尝试使用这种方法。(d) 证明条件语句时,不要忘记两个蕴含都必须被证明!此外,证明的两个部分完全独立,因此第一部分中的假设不能用于第二部分,反之亦然。但是,第一部分中使用的字母可以在第二部分中重复使用,并赋予新的含义。

Exercises

练习

  1. Use the proof templates studied so far to write just the structural outline of a proof of each statement. Do not expand definitions of unfamiliar words. (a) If G is a cyclic group, then G is commutative. [use contrapositive proof] (b) f is a bijection iff f is injective and surjective. [use direct proof for both parts] (c) If f is continuous or monotonic, then f is integrable. [use contrapositive proof]
    使用目前为止学习过的证明模板,仅写出每个命题证明的结构概要。不要展开不熟悉的词语的定义。 (a) 如果 G 是循环群,则 G 是交换群。[使用逆否命题证明] (b) f 是双射当且仅当 f 是单射且是满射。[两部分均使用直接证明] (c) 如果 f 是连续的或单调的,则 f 是可积的。[使用逆否命题证明]
  2. Let a , b , c be fixed integers. Write a contrapositive proof of each statement. (a) If ab is even, then a is even or b is even. (b) If a does not divide bc , then a does not divide c .
    设 a、b、c 为固定整数。写出下列各命题的逆否命题证明。 (a) 如果 ab 是偶数,则 a 是偶数或 b 是偶数。 (b) 如果 a 不能整除 bc,则 a 不能整除 c。
  3. Let x and y be fixed real numbers. Prove: if x y is irrational, then x is irrational or y is irrational.
    设 x 和 y 为固定的实数。证明:如果 x − y 是无理数,那么 x 是无理数或 y 是无理数。
  4. Let m , n be fixed integers, and let x be a fixed real number. Prove each IFF-statement. (a) 2 x + 3 > 11 iff x > 4. (b) n is odd if and only if −3 n is odd. (c) n divides m iff n divides − m .
    设 m、n 为固定整数,x 为固定实数。证明下列当且仅当命题成立。 (a) 2x + 3 > 11 当且仅当 x > 4。 (b) n 为奇数当且仅当 −3n 为奇数。 (c) n 整除 m 当且仅当 n 整除 −m。
  5. Let x be a fixed real number. Prove: (a) 3 x + 7 = 5 x − 4 iff x = 11/2; (b) ( x + 3) 2 = ( x − 1) 2 iff x = −1.
    设 x 为一个固定的实数。证明: (a) 3x + 7 = 5x − 4 当且仅当 x = 11/2; (b) (x + 3) 2 = (x − 1) 2 当且仅当 x = −1。
  6. Let x be a fixed real number. Prove: (a) 9 x + 4 < 2 x − 10 iff x < −2; (b) 2 < x < 5 iff −6 < 4 − 2 x < 0; (c) ( x + 2) 2 > ( x + 5) 2 iff x < −3.5.
    设 x 为一个固定的实数。证明: (a) 9x + 4 < 2x − 10 当且仅当 x < −2; (b) 2 < x < 5 当且仅当 −6 < 4 − 2x < 0; (c) (x + 2) 2 > (x + 5) 2 当且仅当 x < −3.5。
  7. Let x be a fixed real number. Prove: if x 5 + 2 x 3 + x < 50, then x < 2.
    设 x 为一个固定的实数。证明:如果 x 5 + 2x 3 + x < 50,则 x < 2。
  8. Let x and y be fixed real numbers. Prove: if x < y then x < (2 x + y )/3 < ( x + 2 y )/3 < y .
    设 x 和 y 为固定的实数。证明:如果 x < y,则 x < (2x + y)/3 < (x + 2y)/3 < y。
  9. Let x be a fixed integer. Prove that x is odd iff x 2 is odd: (a) using ( 2.1 ) and the definitions; (b) using ( 2.1 ) and the theorem proved in Example 2.30 .
    设 x 为一个固定的整数。证明 x 为奇数当且仅当 x 2 为奇数:(a)使用(2.1)和定义;(b)使用(2.1)和例 2.30 中证明的定理。
  10. Let c be a fixed integer. Prove a Z , c = 10 a + 2 iff b Z , c = 5 b 3 and b is odd.
    设 c 为固定整数。证明:∃a∈Z,c=10a+2 当且仅当 ∃b∈Z,c=5b−3 且 b 为奇数。
  11. Let a , j , r , s be fixed integers. Prove: if j divides r + s , then: a = bj + r for some integer b iff a = cj s for some integer c .
    设 a、j、r、s 为固定的整数。证明:如果 j 整除 r + s,则:对于某个整数 b,a = bj + r 当且仅当对于某个整数 c,a = cj − s。
  12. Let a , b , d , q , r be fixed integers. Using only the definitions, prove: if b = aq + r , then: ( d divides a and d divides r ) iff ( d divides a and d divides b ).
    设 a、b、d、q、r 为固定整数。仅使用定义证明:如果 b = aq + r,则:(d 整除 a 且 d 整除 r) 当且仅当 (d 整除 a 且 d 整除 b)。
  13. (a) Informally justify the contrapositive proof template for proving P Q by analyzing the truth table for P Q . (b) Informally justify the proof template for proving P Q by analyzing the truth table for P Q .
    (a)通过分析 P ⇒ Q 的真值表,非正式地证明证明 P ⇒ Q 的逆否命题证明模板的合理性。(b)通过分析 P ⇔ Q 的真值表,非正式地证明证明 P ⇔ Q 的证明模板的合理性。
  14. Write a proof outline for each statement based on proof templates studied so far. (a) ( P Q ) ⇒ ( S T ). (b) ( x , P ( x ) ) ( Q y , R ( y ) ) . (c) x U , [ P ( x ) ( Q ( x ) R ( x ) ) ] .
    根据目前为止学习过的证明模板,为每个语句编写证明概要。 (a) (P ⇔ Q) ⇒ (S ⇔ T). (b) (∃x,P(x))⇔(Q∧∃y,R(y)). (c) ∃x∈U,[P(x)⇒(∼Q(x)∨R(x))].
  15. Find the errors (if any) in each proof outline below. (a) Statement: If ab P , then a P or b P . Proof Outline. We use a contrapositive proof. Assume a P or b P . Prove a b P . (…) (b) Statement: F is closed iff X F is open. Proof Outline. Part 1 . Assume F is closed. Prove X F is open. (…) Part 2 . Assume F is not closed. Prove X F is not open. (…) (c) Statement: A is invertible iff det ( A ) 0 . Proof Outline. Part 1 . Assume A is invertible. Prove det ( A ) 0 . (…) Part 2 . Assume det ( A ) = 0 . Prove A is not invertible. (…)
    找出以下每个证明大纲中的错误(如有)。 (a) 命题:如果 ab ∈ P,则 a ∈ P 或 b ∈ P。证明大纲:我们使用逆否命题证明。假设 a∉P 或 b∉P。证明 ab∉P。(…) (b) 命题:F 是闭集当且仅当 X − F 是开集。证明大纲:第一部分:假设 F 是闭集。证明 X − F 是开集。(…) 第二部分:假设 F 不是闭集。证明 X − F 不是开集。(…) (c) 命题:A 可逆当且仅当 det(A)≠0。证明大纲:第一部分:假设 A 可逆。证明 det(A)≠0。(…) 第二部分:假设 det(A)=0。证明 A 不可逆。(…)
  16. Identify and correct all errors in the following proof that x is odd iff 3 x + 7 is even (where x is a fixed integer). Proof. Assume x is odd, which means x = 2 k + 1. We must prove 3 x + 7 is even, which means 3 x + 7 = 2 k for some integer k . By algebra, we know 3 x + 7 = 3(2 k + 1) + 7 = 6 k + 8 = 2(3 k + 4). So we can choose k = 3 k + 4.
    找出并改正以下证明中的所有错误,证明 x 为奇数当且仅当 3x + 7 为偶数(其中 x 为固定整数)。证明:假设 x 为奇数,即 x = 2k + 1。我们需要证明 3x + 7 为偶数,即对于某个整数 k,有 3x + 7 = 2k。根据代数运算,我们知道 3x + 7 = 3(2k + 1) + 7 = 6k + 8 = 2(3k + 4)。因此,我们可以选择 k = 3k + 4。
  17. (a) For fixed positive real numbers x and y , prove: if x y then x 2 y 2 . (b) State the converse of part (a). Is the converse always true? (c) Is x , y R , x = y x 2 = y 2 true or false? Explain.
    (a) 对于给定的正实数 x 和 y,证明:如果 x ≠ y,则 x 2 ≠ y 2 。(b) 写出 (a) 部分的逆命题。逆命题总是成立吗?(c) ∀x,y∈R,x=y⇔x²=y² 是真还是假?解释原因。
  18. Prove ( 2.2 ) and ( 2.3 ) using only the ordering axioms for the set of real numbers (see axioms O1 through O6 in §8.1 ).
    仅使用实数集的排序公理(参见第 8.1 节中的公理 O1 至 O6)证明 (2.2) 和 (2.3)。
  19. Use ( 2.3 ) (and the analogous fact for >) to prove: for any fixed x , y R 0 , x y iff x 2 y 2 .
    利用 (2.3)(以及 > 的类似事实)证明:对于任何固定的 x,y∈R≥0,x ≤ y 当且仅当 x 2 ≤ y 2
  20. Let r and s be fixed positive real numbers. Prove: if r s , then r r s s .
    设 r 和 s 为固定的正实数。证明:如果 r ≤ s,则 r≤rs≤s。
  21. Suppose U = { z 1 , z 2 , …, z n } is a finite set. Create a proof template for proving x U , P ( x ) by converting this quantified statement to an AND-statement.
    假设 U = {z 1 , z 2 , …, z n 是一个有限集合。创建一个证明模板,通过将量化语句转换为 AND 语句来证明 ∀x∈U,P(x)。
  22. Create and justify two proof templates for proving P Q by replacing this OR-statement by an equivalent IF-statement.
    创建并论证两个证明模板,通过将这个 OR 语句替换为等价的 IF 语句来证明 P∨Q。
  23. Create and justify a proof template for proving P Q by replacing this XOR-statement by an equivalent IFF-statement.
    创建并证明一个证明模板,通过将这个异或语句替换为等价的因果关系语句来证明 P⊕Q。
  24. Carefully justify the proof template for IFF using the Inference Rule for IF and an appropriate tautology.
    使用 IF 推理规则和适当的重言式,仔细论证 IFF 的证明模板。

2.4 Proofs by Contradiction and Proofs of OR-Statements

2.4 反证法和或语句的证明

This section introduces a powerful general technique for proving statements called proof by contradiction . After studying some proofs using this technique, we introduce some proof templates for proving XOR-statements and OR-statements. We will see that previous proof templates for proving IF-statements (the direct proof and contrapositive proof methods) can be viewed as special cases of the OR-proof templates. This should not be too surprising, since we know an IF-statement can always be converted to an equivalent OR-statement.

本节介绍一种强大的通用证明技巧,称为反证法。在研究了一些运用此技巧的证明之后,我们将介绍一些用于证明异或语句和或语句的证明模板。我们将看到,之前用于证明条件语句的证明模板(直接证明法和逆否命题证明法)可以看作是或语句证明模板的特例。这并不令人意外,因为我们知道一个条件语句总是可以转化为一个等价的或语句。

Template for Proof by Contradiction

反证法的模板

The method of proof by contradiction is a versatile, but sometimes confusing, technique for proving many different kinds of statements. The intuition for this proof method is as follows. Suppose we are trying to prove that some statement P is true. One way to proceed is to assume that P is false, and see what goes wrong. More specifically, if the assumption that P is false leads to a contradiction, then P must be true after all. These ideas suggest the following proof template .

反证法是一种用途广泛但有时令人困惑的证明技巧,可用于证明多种不同类型的命题。这种证明方法的直觉如下:假设我们要证明某个命题 P 为真。一种方法是先假设 P 为假,然后观察结果是否矛盾。更具体地说,如果假设 P 为假导致矛盾,那么 P 必然为真。这些思路启发我们构建了以下证明模板。

2.33. Proof Template to Prove Any Statement P by Contradiction.

2.33. 用反证法证明任何命题 P 的证明模板。

Assume, to get a contradiction, P .

为了得到矛盾,假设∼P。

Use this assumption and other known facts to deduce a contradiction C .

#

利用这一假设和其他已知事实,推导出矛盾 C。

Conclude that P must be true after all.

由此得出结论,P 必然为真。

2.34. Remarks on the Contradiction Proof Template

2.34. 关于矛盾证明模板的说明

(a) As always, the logical status word “ assume ” in Step 1 is mandatory ! The next phrase in Step 1, “to get a contradiction,” while not absolutely mandatory, is highly recommended. The reader of your proof is likely to get confused if you suddenly assume a false statement ∼ P , which is the very opposite of the statement to be proved. You can allay this confusion by stating at the outset that the sole purpose of this assumption is to reach a contradiction. Similarly, upon reaching a contradiction at the end of the proof, it is good to finish with a line such as: “This contradiction shows that the original statement P is true.”

(a) 和往常一样,步骤 1 中的逻辑状态词“假设”是必需的!步骤 1 中的下一句“为了得出矛盾”虽然并非绝对必要,但强烈建议使用。如果您突然假设一个错误的命题 ∼P(这与待证明的命题完全相反),读者很可能会感到困惑。您可以通过在一开始就说明此假设的唯一目的是为了得出矛盾来消除这种困惑。同样地,在证明的最后得出矛盾时,最好以类似这样的语句结尾:“这个矛盾表明原命题 P 为真。”

(b) In Step 1, we almost always replace the assumption ∼ P with a useful denial of P , found using the denial rules.

(b)在步骤 1 中,我们几乎总是用否定规则找到的对 P 的有用否定来替换假设 ∼P。

(c) Step 2 of the proof template tells you to “deduce a contradiction,” but it does not tell you exactly what contradiction you are aiming for. This makes proofs by contradiction harder to finish compared to other types of proofs, since you do not have a specific goal to reach. You must explore the consequences of the false assumption ∼ P , combining it with other known facts to see what happens.

(c) 证明模板的第二步要求你“推导出矛盾”,但它并没有明确指出你要推导出哪个矛盾。这使得反证法比其他类型的证明更难完成,因为你没有一个明确的目标。你必须探究错误假设∼P的后果,并将其与其他已知事实结合起来,才能得出结论。

(d) Remember that a contradiction is a propositional form that evaluates to false in every row of its truth table; a typical example is P ( P ) . To be perfectly precise, Step 2 is really deducing an instance of a contradiction, which is a proposition obtained by replacing every propositional variable in the contradiction by specific statements.

(d) 请记住,矛盾式是一种命题形式,其真值表的每一行都为假;一个典型的例子是 P∧(∼P)。更准确地说,步骤 2 实际上是推导出一个矛盾式的实例,即通过将矛盾式中的每个命题变量替换为具体的语句而得到的命题。

Before looking at examples of proof by contradiction, we give a technical justification for why the proof method works. Suppose we succeed in completing Steps 1 and 2 of the contradiction proof template. Then, according to the direct proof template discussed earlier, we have really proved the statement ( P ) C . Now, recall from Example 1.31 that ( ( P ) C ) P is a tautology and hence an axiom. Since we have also just proved ( P ) C , it follows from the Inference Rule for IF that P is a theorem, as needed.

在考察反证法的例子之前,我们先给出该证明方法的原理。假设我们成功完成了反证法模板的步骤 1 和 2。那么,根据前面讨论的直接证明模板,我们实际上已经证明了命题 (∼P)⇒C。现在,回顾例 1.31,((∼P)⇒C)⇒P 是一个重言式,因此是一个公理。由于我们刚刚也证明了 (∼P)⇒C,根据 IF 的推理规则,P 是一个定理,这符合要求。

Examples of Proof by Contradiction

反证法的例子

We now give examples of the method of proof by contradiction.

现在我们给出反证法的例子。

2.35. Example. Prove that no integer is both even and odd.

2.35. 例题。证明不存在既是偶数又是奇数的整数。

Proof. We must prove: x Z , x is even x is odd . Assume, to get a contradiction, that there does exist x Z such that x is even and x is odd. This assumption means that for some integer k , x = 2 k , and for some integer m , x = 2 m + 1. Combining these assumptions, we see that 2 k = 2 m + 1, hence 2( k m ) = 1, so k m = 1/2. On one hand, we know 1/2 is not an integer. On the other hand, since k , m Z , 1/2 = k m is an integer by closure properties of Z . We have reached the contradiction “1/2 is not an integer and 1/2 is an integer.” Thus, our assumption is wrong, meaning that no integer is both even and odd.

证明。我们需要证明:∼∃x∈Z,x为偶数∧x为奇数。为了得出矛盾,假设存在x∈Z,使得x既是偶数又是奇数。这个假设意味着对于某个整数k,x = 2k;对于某个整数m,x = 2m + 1。结合这些假设,我们发现2k = 2m + 1,因此2(k − m) = 1,所以k − m = 1/2。一方面,我们知道1/2不是整数。另一方面,由于k,m∈Z,根据Z的封闭性质,1/2 = k − m是整数。我们得到了矛盾“1/2不是整数且1/2是整数”。因此,我们的假设是错误的,这意味着不存在既是偶数又是奇数的整数。

Comments: The given statement P begins with NOT, so we began the proof by using the double negation rule to deny P . We then expanded definitions until discovering a contradiction.

注释:给定的命题 P 以 NOT 开头,因此我们首先使用双重否定规则否定 P。然后我们扩展定义,直到发现矛盾。

2.36. Example. Prove: for all positive integers a and b , if a divides b then a b .

2.36. 例。证明:对于所有正整数 a 和 b,如果 a 整除 b,则 a ≤ b。

Proof. Assume, to get a contradiction, that there exist positive integers a , b such that a divides b and a > b . We know there is an integer c with b = ac . Since a and b are both positive, we see that c = b / a must also be positive. As c is an integer, c ≥ 1 follows . But now, by a known theorem (see ( 2.3 ) in §2.3), we deduce

证明。假设为了得出矛盾,存在正整数 a 和 b,使得 a 整除 b 且 a > b。我们知道存在整数 c 使得 b = ac。由于 a 和 b 均为正数,因此 c = b/a 也必须为正数。因为 c 是整数,所以 c ≥ 1。但现在,根据一个已知的定理(参见 §2.3 中的 (2.3)),我们可以推导出

b b = a 一个 c c a 一个 1 = a 一个 > b b .

We have reached the contradiction b > b , so the original statement is true. Comments: To see the contradiction more explicitly, note that b > b is the negation of b b . Since b b is a known fact, the proof has derived the contradiction “ ( b b ) ( b b ) .”

我们已经得出矛盾 b > b,因此原命题为真。注释:为了更清楚地说明矛盾,请注意 b > b 是 b ≤ b 的否定。由于 b ≤ b 是已知事实,证明得出了矛盾“(b≤b)∧∼(b≤b)。

2.37. Example. Prove: for all x , y R , if x is rational and y is irrational, then x + y is irrational.

2.37. 例。证明:对于所有 x,y∈R,如果 x 是有理数且 y 是无理数,则 x + y 是无理数。

Proof. Assume, to get a contradiction, that there exist real numbers x , y such that x is rational and y is not rational and x + y is rational. Expanding definitions, our assumption means that x = m / n and x + y = p / q for some integers m , n , p , q with n ≠ 0 ≠ q . By algebra, we deduce that

证明。为了得出矛盾,假设存在实数 x 和 y,使得 x 是有理数且 y 不是有理数,并且 x + y 是有理数。扩展定义,我们的假设意味着对于某些整数 m、n、p、q,有 x = m/n 且 x + y = p/q,其中 n ≠ 0 ≠ q。通过代数运算,我们推断出:

y = ( x x + y ) x x = ( p p / q q ) ( m / n n ) = p p n n q q m q q n n ,

where p n q m Z , q n Z , and qn ≠ 0. Comparing to the definition of “ y is rational,” this calculation proves that y is rational. We now have the contradiction “ y is rational and y is not rational.” So the original statement must be true.

其中 pn−qm∈Z,qn∈Z,且 qn≠0。与“y 是有理数”的定义比较,该计算证明了 y 是有理数。现在我们得到了“y 是有理数且 y 不是有理数”的矛盾。因此,原命题必然为真。

Next we prove a famous theorem using proof by contradiction. We need one fact in the proof that we have not yet justified from basic principles: any fraction m / n with m , n Z and n ≠ 0 can be written in lowest terms , meaning that no integer larger than 1 divides both m and n . In particular, this means that m and n can be chosen so that m and n are not both even (divisible by 2). For one proof of this fact, see Exercise 9 in § 4.6 .

接下来,我们将用反证法证明一个著名的定理。证明中需要一个尚未从基本原理推导出的事实:对于任意满足 m,n∈Z 且 n≠0 的分数 m/n,都可以化简为最简分数,这意味着不存在大于 1 的整数能同时整除 m 和 n。特别地,这意味着可以选取 m 和 n,使得它们不都是偶数(能被 2 整除)。关于这一事实的一个证明,请参见 §4.6 中的练习 9。

2.38. Theorem: 2 is irrational.

2.38. 定理:2 是无理数。

Proof. Assume, to get a contradiction , that 2 is rational. This assumption means that for some integers m , n with n ≠ 0, 2 = m / n . By the fact mentioned above, we are allowed to assume that m and n are not both even. We reach a contradiction by proving that, in fact, m and n must both be even. Squaring 2 = m / n and rearranging, we deduce that m 2 = 2 n 2 , where n 2 is an integer. This equation means that m 2 must be even. Recall a theorem already proved in the last section: “if x 2 is even, then x is even.” Using this result and the Inference Rule for IF, we see that m is even. So , there is an integer k with m = 2 k . Now, m 2 = 2 n 2 becomes (2 k ) 2 = 2 n 2 , so 4 k 2 = 2 n 2 , so 2 k 2 = n 2 . Here, k 2 is an integer, so n 2 is even. By the same theorem used earlier, we deduce that n is even. We now have the contradiction “ m and n are not both even, and m is even, and n is even.” The only way out of the contradiction is to admit that 2 is not rational.

证明。为了得出矛盾,我们假设 2 是有理数。这个假设意味着对于某些整数 m 和 n(n ≠ 0),2 = m/n。根据前面提到的事实,我们可以假设 m 和 n 不能同时是偶数。我们通过证明 m 和 n 实际上必须都是偶数来得出矛盾。将 2 = m/n 平方并重新整理,我们得出 m = 2n,其中 n 是一个整数。这个等式意味着 m 必须为偶数。回顾上一节中已经证明的一个定理:“如果 x 是偶数,那么 x 也是偶数。”利用这个结果和 IF 的推理规则,我们发现 m 是偶数。因此,存在一个整数 k,使得 m = 2k。现在,m 2 = 2n 2 变为 (2k) 2 = 2n 2 ,所以 4k 2 = 2n 2 ,因此 2k 2 = n 2 。这里,k 2 是整数,所以 n 2 是偶数。根据之前用到的定理,我们推断 n 是偶数。现在我们遇到了矛盾:“m 和 n 不能同时是偶数,而 m 是偶数,n 也是偶数。” 解决这个矛盾的唯一方法是承认 2 不是有理数。

So far, we have checked that propositional forms are tautologies by writing out truth tables. The next example shows that proof by contradiction can also be used to demonstrate that a propositional form is a tautology without making a giant truth table.

到目前为止,我们已经通过编写真值表来验证命题形式是否为重言式。下一个例子表明,反证法也可以用来证明命题形式是重言式,而无需编写庞大的真值表。

2.39. Example. Use proof by contradiction to show that

2.39. 示例。用反证法证明:

[ ( P P Q R ) ( P P S S ) ( Q S S ) ( R S S ) ] S S

(2.4)

is a tautology.

这是一个同义反复。

Proof. [Recall, by definition, that A is a tautology iff every row of the truth table for A evaluates to true. Note the universal quantifier in the definition text. So, to begin the proof by contradiction, we deny this and say:] Assume, to get a contradiction, that some row of the truth table for the propositional form ( 2.4 ) evaluates to false. In such a row, let us see what we can deduce about the truth values of various subexpressions of ( 2.4 ). The overall propositional form is a false IF-statement in this row, which (by the truth table for IF) forces

证明。[回想一下,根据定义,A 是重言式当且仅当 A 的真值表中每一行都为真。注意定义文本中的全称量词。因此,为了用反证法证明,我们否定这一点并说:] 假设,为了得到矛盾,命题形式 (2.4) 的真值表中有一行为假。在这样一行中,让我们看看我们可以推断出 (2.4) 的各个子表达式的真值。在这一行中,整个命题形式是一个假的 IF 语句,这(根据 IF 语句的真值表)迫使

( P P Q R ) ( P P S S ) ( Q S S ) ( R S S )

to be true and S to be false. Next, the truth of the AND-statement just written forces each input to the AND to be true, so we see that

为真,S 为假。接下来,刚才写的 AND 语句为真,强制 AND 的每个输入都为真,所以我们看到

P P Q R , P P S S , Q S S , R S S

are all true in this row. Now, since P S is true and S is false, it must be that P is false in this row (by the truth table for IF or the Contrapositive Inference Rule for IF). Similarly, Q must be false and R must be false in this row. But now P Q R must be false in this row, and yet also true. This contradiction shows that the original propositional form must be a tautology.

这一行中的 P ⇒ S 都为真。现在,由于 P ⇒ S 为真且 S 为假,因此这一行中 P 必定为假(根据 IF 的真值表或 IF 的逆否命题推理规则)。类似地,这一行中 Q 必定为假且 R 必定为假。但现在 P∨Q∨R 在这一行中必定为假,却又为真。这个矛盾表明,原命题形式必定是一个重言式。

Proving XOR Statements and OR Statements

证明异或语句和或语句

We close this section by introducing proof templates for proving XOR-statements and OR-statements. The template for XOR arises from the logical equivalence P Q P ( Q ) , together with the IFF template discussed earlier .

本节最后,我们将介绍用于证明异或语句和或语句的证明模板。异或语句的证明模板源于逻辑等价式 P⊕Q≡P⇔(∼Q),以及前面讨论过的 IFF 模板。

2.40. Proof Template to Prove P Q .

2.40.证明 P⊕Q 的证明模板。

Part 1. Prove P ⇒ (∼ Q ) (by any method).

第一部分。证明 P ⇒ (∼Q)(用任何方法)。

Part 2. Prove (∼ Q ) ⇒ P (by any method).

第二部分。证明(∼Q)⇒ P(用任何方法)。

Similarly, we can create templates for proving P Q by converting the OR-statement to an IF-statement and invoking previous proof templates. Specifically, note that

类似地,我们可以通过将 OR 语句转换为 IF 语句并调用之前的证明模板来创建证明 P∨Q 的模板。具体来说,请注意:

P P Q ( P P ) Q ( Q ) P P .

This leads to the following templates .

由此产生了以下模板。

2.41. First Proof Template to Prove P Q .

2.41. 证明 P∨Q 的第一个证明模板。

Assume P . Prove Q .

假设 ∼P。证明 Q。

2.42. Second Proof Template to Prove P Q .

2.42. 证明 P∨Q 的第二个证明模板。

Assume Q . Prove P .

假设 ∼Q。证明 P。

2.43. Example. Let x , y be fixed real numbers. Prove: if x ≠ 0 and y ≠ 0, then xy ≠ 0.

2.43. 例。设 x、y 为固定的实数。证明:如果 x ≠ 0 且 y ≠ 0,则 xy ≠ 0。

Proof. [The inequalities in the given statement seem hard to work with; we can convert these to equalities by doing a contrapositive proof. So we begin:] Assume xy = 0. Prove x = 0 or y = 0. Our new goal is an OR-statement, so we use the first template and assume x ≠ 0. We must prove y = 0. [We must now combine our two assumptions ( xy = 0 and x ≠ 0) to reach the new goal y = 0. We do so as follows:] It is known that the nonzero real number x has a multiplicative inverse x 1 R satisfying x −1 x = 1. Multiplying both sides of xy = 0 by x −1 gives x −1 ( xy ) = x −1 0. By algebra, this becomes 1 y = 0, giving y = 0.

证明。[给定语句中的不等式似乎难以处理;我们可以通过逆否命题证明将其转换为等式。因此,我们开始:] 假设 xy = 0。证明 x = 0 或 y = 0。我们的新目标是“或”命题,因此我们使用第一个模板并假设 x ≠ 0。我们必须证明 y = 0。[现在我们必须将两个假设(xy = 0 和 x ≠ 0)结合起来以达到新目标 y = 0。我们这样做如下:] 已知非零实数 x 存在乘法逆元 x−1∈R,满足 x −1 x = 1。将 xy = 0 两边同时乘以 x −1 ,得到 x −1 (xy) = x −1 0。通过代数运算,这变为 1y = 0,从而得出 y = 0。

We can prove an OR-statement with more than two alternatives as follows.

我们可以按如下方式证明具有两个以上选项的 OR 语句。

2.44. Proof Template to Prove P Q R S .

2.44.证明P∨Q∨R∨S的证明模板。

Assume P and ∼ Q and ∼ R . Prove S .

假设∼P、∼Q和∼R。证明S。

This proof template can be justified via truth tables, or by the Inference Rule for IF and the tautology

这个证明模板可以通过真值表来证明,也可以通过“如果成立”的推理规则和重言式来证明。

( [ ( P P ) ( Q ) ( R ) ] S S ) ( P P Q R S S ) .

When using the template, we can assume the negations of any three of the four alternatives and then prove the remaining alternative. The pattern seen in this template extends to a template for proving A 1 A 2 ∨ · · · ∨ A k where k ≥ 2. To prove such an OR-statement, assume all but one of the A j is false, and then prove the remaining A j .

使用该模板时,我们可以假设四个备择假设中任意三个的否定,然后证明剩余的备择假设。此模板中的模式可以扩展到证明 A 1 ∨A 2 ∨ · · · ∨A k (其中 k ≥ 2)的模板。要证明这样的 OR 语句,假设除一个 A j 外,其余所有 A j 都为假,然后证明剩余的 A j

Section Summary

章节概要

  1. Template for Proof by Contradiction. To prove a statement P by contradiction: Assume, to get a contradiction, P . Combine the assumption with other known facts to deduce a contradiction. Say: “This contradiction shows that P is true, after all.”
    反证法模板。要用反证法证明命题 P,首先假设 ∼P,从而得出矛盾。将该假设与其他已知事实结合起来,推导出矛盾。然后说:“这个矛盾表明 P 最终是正确的。”
  2. Advice on Contradiction Proofs. (a) Signal the start and end of a contradiction proof with phrases similar to those in the proof template, to lessen the risk of confusing the reader (or writer) of the proof. (b) Use the denial rules to find a useful form of ∼ P . (c) The proof template does not give you a specific contradiction to aim for, which makes this template hard to use compared to others. You need creativity and patience to explore the consequences of the given assumptions until a contradiction is found.
    关于矛盾证明的建议。(a) 使用类似于证明模板中的短语来标记矛盾证明的开始和结束,以降低证明的读者(或作者)感到困惑的风险。(b) 使用否定规则找到 ∼P 的有效形式。(c) 证明模板没有提供具体的矛盾目标,这使得它与其他模板相比更难使用。你需要创造力和耐心来探索给定假设的推论,直到找到矛盾为止。
  3. Proof Template for XOR. To prove P Q : Part 1. Prove P ⇒ (∼ Q ) by any method. Part 2. Prove (∼ Q ) ⇒ P by any method.
    异或 (XOR) 证明模板。证明 P⊕Q: 第 1 部分:用任意方法证明 P ⇒ (∼Q)。 第 2 部分:用任意方法证明 (∼Q) ⇒ P。
  4. Proof Template for OR. To prove P Q R S : Assume P and ∼ Q and ∼ R . Prove S . (Similarly for OR-statements with any number of alternatives.)
    OR 语句的证明模板。证明 P∨Q∨R∨S: 假设 P ≠ Q ≠ R ≠。证明 S: (类似地,对于任意数量选项的 OR 语句也适用。)

Exercises

练习

  1. Use proof templates for IF, OR, etc., to write a proof outline for each statement. (a) If f is continuous or monotonic, then f is integrable. [use contradiction proof] (b) If P is a prime ideal and ab P , then a P or b P . [use direct proof] (c) If P is a prime ideal and ab P , then a P or b P . [use contrapositive proof] (d) x Z , y Z , z Z , n Z , ( x > 0 and y > 0 and n > 2) ⇒ x n + y n z n . [use proof by contradiction] (e) n Z , p Z , p > n p is prime p + 2 is prime [use proof by contradiction]
    使用 IF、OR 等证明模板,为每个语句编写证明大纲。 (a) 如果 f 是连续的或单调的,则 f 是可积的。[使用反证法] (b) 如果 P 是素理想且 ab ∈ P,则 a ∈ P 或 b ∈ P。[使用直接证明] (c) 如果 P 是素理想且 ab ∈ P,则 a ∈ P 或 b ∈ P。[使用逆否命题证明] (d) ∀x∈Z,∀y∈Z,∀z∈Z,∀n∈Z, (x > 0 且 y > 0 且 n > 2) ⇒ x n + y n ≠ z n 。 [使用反证法] (e) ∀n∈Z,∃p∈Z,p>n∧p是素数∧p+2是素数 [使用反证法]
  2. Let x be a fixed real number. Prove each statement by contradiction. (a) If x ≠ 0, then − x ≠ 0. (b) If x ≠ 0, then 1/ x ≠ 0.
    设 x 为一个固定的实数。用反证法证明下列各命题。 (a) 若 x ≠ 0,则 −x ≠ 0。 (b) 若 x ≠ 0,则 1/x ≠ 0。
  3. Prove by contradiction: x R , y R , if x Q and y Q and x ≠ 0, then x y Q .
    用反证法证明:∀x∈R,∀y∈R,如果x∈Q且y∉Q且x≠0,则xy∉Q。
  4. Prove by contradiction: m Z , n Z , 6 m + 8 n 5 .
    用反证法证明:∀m∈Z,∀n∈Z,6m+8n≠5。
  5. Prove by contradiction: P ⇒ ( Q P ) is a tautology.
    用反证法证明:P ⇒ (Q ⇒ P) 是重言式。
  6. Suppose, in a certain proof, we assume ∼ P , go through a chain of reasoning, and eventually deduce P . What, if anything, can we conclude about the truth or falsehood of P ?
    假设在某个证明中,我们假设 ∼P,经过一系列推理,最终推导出 P。那么,关于 P 的真假,我们能得出什么结论呢?
  7. Let x be a fixed real number. (a) Prove: if x 2 + 4 x − 60 = 0, then x = −10 or x = 6. (b) Is the converse of part (a) true?
    设 x 为一个固定的实数。(a) 证明:如果 x + 4x − 60 = 0,则 x = −10 或 x = 6。(b) (a) 部分的逆命题是否成立?
  8. For fixed x R , prove: if x 2 x , then x ≤ 0 or x ≥ 1.
    对于固定的 x∈R,证明:如果 x 2 ≥ x,则 x ≤ 0 或 x ≥ 1。
  9. For fixed m Z , use ( 2.1 ) to prove: 2 divides m XOR 2 divides m + 7.
    对于固定的 m∈Z,利用(2.1)证明:2 整除 m 异或 2 整除 m + 7。
  10. For fixed x , y Z > 0 , prove: if x + y = 5, then x = 1 or x = 2 or x = 3 or x = 4.
    对于固定的 x,y∈Z>0,证明:如果 x + y = 5,则 x = 1 或 x = 2 或 x = 3 或 x = 4。
  11. (a) Find a statement equivalent to ( P Q ) R that does not use or ⇒. (b) Give four possible proof outlines of ( P Q ) R , each with different assumptions and a different goal.
    (a) 找出与 (P∧Q)⇒R 等价的语句,其中不使用 ∧ 或 ⇒。(b) 给出 (P∧Q)⇒R 的四种可能的证明思路,每种思路都有不同的假设和不同的目标。
  12. Give four possible proof outlines of A ⇒ ( B ∨(∼ C )).
    给出 A ⇒ (B∨(∼C)) 的四种可能的证明大纲。
  13. Suppose this fact is a known theorem: for all a , b Z , if 3 divides ab , then 3 divides a or 3 divides b . Use this to prove that 3 is irrational, by imitating a proof in this section.
    假设这个事实是一个已知的定理:对于所有 a,b∈Z,如果 3 能整除 ab,那么 3 也能整除 a 或 3 也能整除 b。利用这个定理,模仿本节中的一个证明,证明 3 是无理数。
  14. Suppose we imitate the proof of Theorem 2.38 in an attempt to prove that 4 is irrational. Find the exact point in the proof that no longer works.
    假设我们模仿定理 2.38 的证明方法来证明 4 是无理数。找出证明过程中不再成立的确切步骤。
  15. (a) Use the truth table for XOR to justify the XOR proof template. (b) Use the truth table for OR to justify the two proof templates for proving P Q . (c) Use any method to justify the proof template for proving P Q R S .
    (a) 利用异或运算的真值表证明异或证明模板的合理性。(b) 利用或运算的真值表证明证明 P∨Q 的两个证明模板的合理性。(c) 使用任意方法证明证明 P∨Q∨R∨S 的证明模板的合理性。
  16. Make a new proof template for proving P Q based on the equivalence ( P Q ) ( P Q ) ( P Q ) ; prove both parts of the AND-statement by contradiction.
    基于等价关系 (P⊕Q)≡(P∨Q)∧∼(P∧Q),创建一个证明 P⊕Q 的新证明模板;用反证法证明 AND 语句的两个部分。
  17. Assume p , a , b , c are fixed integers, and p is known to satisfy: x , y Z , if p divides xy , then p divides x or p divides y . Prove: if p divides abc , then p divides a or p divides b or p divides c .
    假设 p、a、b、c 是固定的整数,并且已知 p 满足:对于任意 x、y∈Z,如果 p 整除 xy,则 p 整除 x 或 p 整除 y。证明:如果 p 整除 abc,则 p 整除 a 或 p 整除 b 或 p 整除 c。
  18. Fix m Z . Find the error in the following proof that 6 m is odd. Proof. Assume, to get a contradiction, that 6 m is not odd. So 6 m is even, which means k Z , 6 m = 2 k . Choose k = 3 m , which is in Z by closure. By algebra, 2 k = 2(3 m ) = 6 m , proving that 6 m is even.
    固定 m∈Z。找出以下证明 6m 为奇数的错误。证明:假设 6m 不是奇数,从而得出矛盾。因此 6m 是偶数,这意味着存在 k∈Z,6m=2k。选择 k = 3m,根据闭包定理,3m 属于 Z。通过代数运算,可知 2k = 2(3m) = 6m,从而证明 6m 是偶数。
  19. Find the error in the following proof outline of the statement ( P Q ) ⇒ (∼ S ). Proof. To get a contradiction, prove ( P Q ) S . Part 1 . Assume P ; prove Q . Part 2 . Prove S .
    找出以下证明大纲中的错误,该证明大纲描述了命题 (P ⇒ Q) ⇒ (∼S)。证明:为了得出矛盾,证明 (P⇒Q)∧S。第一部分:假设 P 成立;证明 Q 成立。第二部分:证明 S 成立。
  20. Prove by contradiction: [ P ⇒ ( Q R )] ⇒ [( P Q ) ⇒ ( P R )] is a tautology.
    用反证法证明:[P ⇒ (Q ⇒ R)] ⇒ [(P ⇒ Q) ⇒ (P ⇒ R)] 是一个重言式。
  21. Let m be a fixed integer. Recall the theorem that every integer is even or odd, but not both. Using this, prove: if m is even, then ( k Z , m = 4 k ) ( j Z , m = 4 j + 2 ) ; but if m is odd, then ( a Z , m = 4 a + 1 ) ( b Z , m = 4 a + 3 ) .
    设 m 为一个固定的整数。回顾定理:每个整数要么是偶数,要么是奇数,但不能既是偶数又是奇数。利用此定理证明: 如果 m 是偶数,则 (∃k∈Z,m=4k)⊕(∃j∈Z,m=4j+2); 但如果 m 是奇数,则 (∃a∈Z,m=4a+1)⊕(∃b∈Z,m=4a+3)。
  22. Let b be a fixed odd positive integer. Prove 2 b is not rational.
    设 b 为一个固定的奇正整数。证明 2b 不是有理数。

2.5 Proofs by Cases and Disproofs

2.5 案例证明与反证法

We ended the last section by describing ways to prove OR-statements that are not yet known to be true. Now, we consider the related question of how to use an OR-statement that is already known to be true. We introduce a new proof method called proof by cases that is often needed in situations where we already know (or have assumed) an OR-statement and need to prove some other statement. The second half of the section considers the question of how we may disprove statements we believe to be false.

上一节我们讨论了如何证明尚未确定为真的“或”命题。现在,我们将探讨一个相关问题:如何使用已知为真的“或”命题。我们将介绍一种新的证明方法,称为“按情况证明”,这种方法常用于已知(或假设)某个“或”命题为真,而需要证明其他命题的情况。本节后半部分将讨论如何反驳我们认为为假的命题。

Proof by Cases

案例证明

We introduce proof by cases in the situation where we know (or have assumed) an OR-statement with three alternatives .

我们引入了案例证明法,用于在已知(或假设)具有三个备选方案的 OR 语句的情况下进行证明。

2.45. Proof Template for Proof by Cases. Suppose P Q R is a known OR-statement.

2.45. 情况证明的证明模板。假设 P∨Q∨R 是一个已知的 OR 语句。

To prove any statement S using proof by cases:

要用分情况证明法证明任何陈述S:

Say “Since P Q R is known , we consider 3 cases.”

“既然 P∨Q∨R 已知,我们考虑 3 种情况。”

Case 1. Assume P . Prove S .

情况 1. 假设 P。证明 S。

Case 2. Assume Q . Prove S .

情况 2. 假设 Q。证明 S。

Case 3. Assume R . Prove S .

情况 3. 假设 R。证明 S。

Conclude by saying “So, S is true in all cases.”

最后总结说:“所以,S 在所有情况下都是正确的。”

2.46. Remarks on Proof by Cases

2.46. 关于案例证明的说明

(a) The proof outlined above has three completely separate subproofs, called the three cases . In each subproof, we must prove the target statement S . Each of these three subproofs begins with a different temporary assumption ( P , Q , or R ), so the details of these subproofs are not the same. Assumptions and conclusions appearing in the proof of one case cannot be reused in another case unless they are reproved under the assumptions for that case. Temporary variables used in one case can be reused (with new meanings) in other cases.

(a) 上述证明包含三个完全独立的子证明,称为三个情况。在每个子证明中,我们都必须证明目标命题 S。这三个子证明分别以不同的临时假设(P、Q 或 R)开始,因此它们的具体细节并不相同。除非在特定情况的假设下重新证明,否则在一个情况的证明中出现的假设和结论不能在另一个情况中重复使用。在一个情况中使用的临时变量可以在其他情况中重复使用(但含义不同)。

(b) The proof template extends to the situation where we start with a known OR-statement P 1 P 2 ∨ · · · ∨ P k with k alternatives, where k ≥ 2, and we are trying to prove S . In this setting, there are k cases. The proof outline looks like this:

(b) 该证明模板可以扩展到以下情况:我们已知一个 OR 语句 P 1 ∨P 2 ∨ · · · ∨P k ,它有 k 个备选方案,其中 k ≥ 2,并且我们试图证明 S。在这种情况下,共有 k 种情况。证明流程如下:

  • Since P 1 ∨ · · · ∨ P k is known , consider k cases.
    由于 P 1 ∨ · · · ∨P k 已知,考虑 k 个案例。
  • Case 1. Assume P 1 . Prove S .
    情况 1. 假设 P 1 。证明 S。
  • Case i . Assume P i . Prove S .
    情况一:假设 P i 。证明 S。
  • Case k . Assume P k . Prove S .
    情况 k. 假设 P k 。证明 S。
  • So, S is true in all cases.
    所以,S 在所有情况下都为真。

(c) Here is a technical justification for why the proof-by-cases template is legitimate. On one hand, in order to use this template, we must already know (or have proved) P Q R . On the other hand, Case 1 of the template proves the IF-statement P S ; Case 2 proves Q S ; and Case 3 proves R S . By the AND-template, we have proved

(c) 以下是对“逐个情况证明”模板合法性的技术性论证。一方面,为了使用此模板,我们必须已经知道(或已经证明)P∨Q∨R。另一方面,该模板的案例 1 证明了 IF 语句 P ⇒ S;案例 2 证明了 Q ⇒ S;案例 3 证明了 R ⇒ S。根据“与”模板,我们已经证明了

( P P Q R ) ( P P S S ) ( Q S S ) ( R S S ) .

We also have the following axiom (a tautology proved in ( 2.4 ) of the last section):

我们还有以下公理(上一节 (2.4) 中证明的重言式):

[ ( P P Q R ) ( P P S S ) ( Q S S ) ( R S S ) ] S S .

Applying the Inference Rule for IF to the two displayed theorems, we see that S is a new theorem, as needed.

将 IF 推理规则应用于所展示的两个定理,我们发现 S 是一个新定理,符合要求。

Examples of Proof by Cases

案例证明示例

Let us look at some sample proofs where proof by cases can be used. We remark that you will generally not be told: “use proof by cases here.” Writers of proofs must recognize situations where proof by cases can be applied. As a general rule, if you encounter a situation where you know or have assumed an OR-statement (say because a proof template directed you to write one), then proof by cases gives you one potential way to utilize that OR-statement.

让我们来看一些可以使用“按情况证明”的证明示例。需要注意的是,通常不会有人直接告诉你:“这里使用按情况证明”。证明的编写者必须能够识别出哪些情况下可以使用“按情况证明”。一般来说,如果你遇到已知或假定存在“或”关系的情况(例如,因为证明模板要求你编写这样的关系),那么“按情况证明”就提供了一种利用该“或”关系的方法。

2.47. Example. Let a and b be fixed integers. Prove: if a is even or b is even, then ab is even. Proof. Assume a is even or b is even. Prove ab is even. [Noting that we have just assumed an OR-statement, we continue by using proof by cases.]

2.47. 例。设 a 和 b 为固定整数。证明:如果 a 是偶数或 b 是偶数,则 ab 也是偶数。证明:假设 a 是偶数或 b 是偶数。证明 ab 是偶数。[注意到我们刚刚假设了一个“或”关系,我们继续使用分情况证明法。]

Case 1. Assume a is even. Prove ab is even. We have assumed there is an integer k with a = 2 k . We must prove m Z , a b = 2 m . Multiplying our assumption by b gives ab = 2 kb . Choosing m to be kb , which is an integer, we get ab = 2 m as needed.

情况 1. 假设 a 为偶数。证明 ab 为偶数。我们假设存在整数 k 使得 a = 2k。我们需要证明存在 m∈Z,ab=2m。将我们的假设乘以 b,得到 ab = 2kb。选择 m 为整数 kb,我们得到 ab = 2m,符合要求。

Case 2. Assume b is even. Prove ab is even. We have assumed there is an integer k with b = 2 k . We must prove m Z , a b = 2 m . Multiplying our assumption by a gives ab = 2 ka . Choosing m to be ka , which is an integer, we get ab = 2 m as needed.

情况 2. 假设 b 为偶数。证明 ab 为偶数。我们假设存在整数 k 使得 b = 2k。我们需要证明存在 m∈Z,ab=2m。将我们的假设乘以 a,得到 ab = 2ka。选择 m 为 ka(一个整数),我们得到 ab = 2m,符合要求。

Comment: Although the two cases had nearly identical proofs, both were logically necessary.

评论:虽然这两个案例的证明几乎完全相同,但从逻辑上讲,两者都是必然的。

2.48. Example. For a fixed integer n , prove: if 30 divides n or 70 divides n , then 10 divides n . Proof. Assume 30 divides n or 70 divides n . Prove 10 divides n . [We have assumed an OR-statement, so we decide to use proof by cases.]

2.48. 例。对于给定的整数 n,证明:如果 30 能整除 n 或 70 能整除 n,则 10 能整除 n。证明。假设 30 能整除 n 或 70 能整除 n。证明 10 能整除 n。[我们假设这是一个 OR 语句,因此我们决定使用分情况证明。]

Case 1. Assume 30 divides n . Prove 10 divides n . We have assumed n = 30 k for some integer k . We must prove n = 10 m for some integer m . Note n = 30 k = 10(3 k ). So we choose m = 3 k , which is an integer such that n = 10 m .

情况 1. 假设 30 能整除 n。证明 10 能整除 n。我们假设 n = 30k,其中 k 为某个整数。我们需要证明 n = 10m,其中 m 为某个整数。注意 n = 30k = 10(3k)。因此我们选择 m = 3k,这是一个满足 n = 10m 的整数。

Case 2. Assume 70 divides n . Prove 10 divides n . We have assumed n = 70 k for some integer k . We must prove n = 10 m for some integer m . Note n = 70 k = 10(7 k ). So we choose m = 7 k , which is an integer such that n = 10 m .

情况 2. 假设 70 能整除 n。证明 10 能整除 n。我们假设 n = 70k,其中 k 为某个整数。我们需要证明 n = 10m,其中 m 为某个整数。注意 n = 70k = 10(7k)。因此我们选择 m = 7k,这是一个满足 n = 10m 的整数。

2.49. Example. Let n be a fixed integer. Prove: n 2 n is even.

2.49. 例。设 n 为一个固定的整数。证明:n 2 − n 是偶数。

Proof. We must prove k Z , n 2 n = 2 k . [Where do we go from here? Thinking of the next sentence in this proof requires a leap of creativity, even though the sentence itself seems to be stating a rather mundane fact.] We know n is even or n is not even (this is an axiom). Since we have stated a known OR-statement, we can try proof by cases.

证明。我们必须证明 ∃k∈Z, n²−n=2k。[接下来该怎么做呢?构思证明中的下一句话需要一些创造力,即使这句话本身似乎只是在陈述一个相当普通的结论。] 我们知道 n 要么是偶数,要么不是偶数(这是一个公理)。既然我们已经陈述了一个已知的“或”命题,我们就可以尝试分情况证明。

Case 1. Assume n is even. Prove k Z , n 2 n = 2 k . We have assumed n = 2 m for some integer m . We deduce from this that n 2 n = (2 m ) 2 − 2 m = 4 m 2 − 2 m = 2(2 m 2 m ). We now see that we can choose k = 2 m 2 m (which is an integer since m is) in order to have n 2 n = 2 k .

情况 1. 假设 n 为偶数。证明 ∃k∈Z,n2−n=2k。我们假设 n = 2m,其中 m 为某个整数。由此可得 n 2 − n = (2m) 2 − 2m = 4m 2 − 2m = 2(2m 2 − m)。现在我们可以看出,我们可以选择 k = 2m 2 − m(因为 m 是整数),使得 n 2 − n = 2k。

Case 2. Assume n is not even, so n is odd by a known theorem. Prove k Z , n 2 n = 2 k . We have assumed n = 2 m + 1 for some integer m . Now, algebra gives

情况 2. 假设 n 不是偶数,根据已知定理可知 n 是奇数。证明 ∃k∈Z, n²−n=2k。我们假设 n = 2m + 1,其中 m 为某个整数。现在,代数运算表明

n n 2 n n = ( 2 m + 1 ) 2 ( 2 m + 1 ) = 4 m 2 + 4 m + 1 2 m 1 = 4 m 2 + 2 m = 2 ( 2 m 2 + m ) .

Choose k = 2 m 2 + m ; this is an integer such that n 2 n = 2 k . So, n 2 n is even in all cases.

选择 k = 2m 2 + m;这是一个整数,使得 n 2 − n = 2k。因此,n 2 − n 在所有情况下都是偶数。

2.50. Example. Let x be a fixed real number. Prove: x 3 = 4 x iff x = −2 or x = 0 or x = 2.

2.50. 例。设 x 为一个固定的实数。证明:x 3 = 4x 当且仅当 x = −2 或 x = 0 或 x = 2。

Proof. This IFF-statement requires a two-part proof.

证明。此逆反馈证明语句需要两部分证明。

Part 1. Assume x 3 = 4 x . Prove x = −2 or x = 0 or x = 2. We are proving an OR-statement now, so we follow the OR-template. Assume x ≠ −2 and x ≠ 0. Prove x = 2. Since x ≠ 0, we can divide both sides of the assumption x 3 = 4 x by x to deduce x 2 = 4, so x 2 − 4 = 0, so ( x − 2)( x + 2) = 0. Since x ≠ −2, we can divide both sides of this equation by the nonzero quantity x + 2 to get x − 2 = 0, hence x = 2, as needed.

第一部分。假设 x 3 = 4x。证明 x = −2 或 x = 0 或 x = 2。我们现在要证明一个 OR 命题,所以我们遵循 OR 命题的模板。假设 x ≠ −2 且 x ≠ 0。证明 x = 2。由于 x ≠ 0,我们可以将假设 x 3 = 4x 的两边同时除以 x,从而得出 x 2 = 4,因此 x 2 − 4 = 0,所以 (x − 2)(x + 2) = 0。由于 x ≠ −2,我们可以将这个等式的两边同时除以非零量 x + 2,得到 x − 2 = 0,因此 x = 2,符合要求。

Part 2. Assume x = −2 or x = 0 or x = 2. Prove x 3 = 4 x . Here we have assumed an OR-statement, so we use proof by cases.

第二部分。假设 x = −2 或 x = 0 或 x = 2。证明 x 3 = 4x。这里我们假设了一个 OR 语句,所以我们使用分情况证明法。

  • Case 1. Assume x = −2. Prove x 3 = 4 x . By arithmetic, x 3 = −8 = 4 x in this case.
    情况 1. 假设 x = −2。证明 x 3 = 4x。根据算术,在这种情况下,x 3 = −8 = 4x。
  • Case 2. Assume x = 0. Prove x 3 = 4 x . By arithmetic, x 3 = 0 = 4 x in this case.
    情况 2. 假设 x = 0。证明 x 3 = 4x。根据算术,在这种情况下,x 3 = 0 = 4x。
  • Case 3. Assume x = 2. Prove x 3 = 4 x . By arithmetic, x 3 = 8 = 4 x in this case.
    情况 3. 假设 x = 2。证明 x 3 = 4x。根据算术,在这种情况下,x 3 = 8 = 4x。

Since x 3 = 4 x in all cases, we are done with part 2.

由于在所有情况下 x 3 = 4x,因此第 2 部分已完成。

Definition by Cases

案例定义

Sometimes cases are used to define a new mathematical concept. Proofs involving such concepts often rely on proof by cases. We give a few examples here involving the sign and absolute value of real numbers.

有时,我们会使用个案来定义新的数学概念。涉及这类概念的证明通常依赖于个案证明法。这里我们给出几个关于实数符号和绝对值的例子。

2.51. Definition: sgn ( x ) . For each real number x , define sgn ( x ) = + 1 if x > 0, sgn ( x ) = 0 if x = 0, and sgn ( x ) = 1 if x < 0.

2.51. 定义:sgn(x)。对于每个实数 x,定义 sgn(x)=+1 当 x > 0 时,sgn(x)=0 当 x = 0 时,sgn(x)=-1 当 x < 0 时。

For this definition by cases to make sense, we need to know that for each real x , exactly one of the statements “ x > 0,” “ x = 0,” or “ x < 0” is true. We take this to be a known fact from algebra. When we prove theorems about sgn ( x ) , we often start the proof with the known fact “ x > 0 or x = 0 or x < 0,” which leads into a proof by cases.

为了使这种分情况定义有意义,我们需要知道对于每个实数 x,语句“x > 0”、“x = 0”或“x < 0”中恰好有一个为真。我们将其视为代数中的一个已知事实。当我们证明关于 sgn(x) 的定理时,我们通常从已知事实“x > 0 或 x = 0 或 x < 0”开始证明,然后进行分情况证明。

2.52. Example. For fixed z R , prove sgn ( z ) = sgn ( z ) .

2.52. 例。对于固定的 z∈R,证明 sgn(−z)=−sgn(z)。

Proof. We know z > 0 or z = 0 or z < 0, so use cases.

证明。我们知道 z > 0 或 z = 0 或 z < 0,因此有多种用例。

Case 1. Assume z > 0; prove sgn ( z ) = sgn ( z ) . Because z > 0, we know − z < 0 by algebra. By definition of sgn , we know sgn ( z ) = 1 and sgn ( z ) = + 1 , so sgn ( z ) = 1 = sgn ( z ) holds in this case.

情况 1. 假设 z > 0;证明 sgn(−z)=−sgn(z)。因为 z > 0,我们根据代数可知 −z < 0。根据 sgn 的定义,我们知道 sgn(−z)=−1 且 sgn(z)=+1,所以 sgn(−z)=−1=−sgn(z) 在此情况下成立。

Case 2. Assume z = 0; prove sgn ( z ) = sgn ( z ) . Here − z = −0 = 0, so sgn ( z ) = sgn ( 0 ) = 0 = 0 = sgn ( 0 ) = sgn ( z ) follows.

情况 2. 假设 z = 0;证明 sgn(−z)=−sgn(z)。这里 −z = −0 = 0,所以 sgn(−z)=sgn(0)=0=−0=−sgn(0)=−sgn(z) 成立。

Case 3. Assume z < 0; prove sgn ( z ) = sgn ( z ) . Because z < 0, we know − z > 0 by algebra. Therefore, sgn ( z ) = + 1 = ( 1 ) = sgn ( z ) in this case. We see that sgn ( z ) = sgn ( z ) holds in all cases.

情况 3. 假设 z < 0;证明 sgn(−z)=−sgn(z)。因为 z < 0,我们根据代数可知 −z > 0。因此,在这种情况下,sgn(−z)=+1=−(−1)=−sgn(z)。由此可见,sgn(−z)=−sgn(z) 在所有情况下都成立。

2.53. Definition: Absolute Value.. For each real number r , define | r | = r if r ≥ 0, and define | r | = − r if r < 0.

2.53. 定义:绝对值.. 对于每个实数 r,定义 |r| = r,如果 r ≥ 0,定义 |r| = −r,如果 r < 0。

For example, |0.23| = 0.23, whereas | − 3| = −( − 3) = 3. Note carefully that when r is negative, the expression − r denotes a positive real number.

例如,|0.23| = 0.23,而| − 3| = −( − 3) = 3。请注意,当 r 为负数时,表达式 −r 表示一个正实数。

2.54. Example. For fixed x , y R , prove | xy | = | x | · | y |. Proof. [Here we need cases for both x and y . It turns out to be easier to create special cases when one variable is zero.] We know x = 0 or y = 0 or x , y > 0 or x > 0 > y or y > 0 > x or x , y < 0, so use cases.

2.54. 例。对于固定的 x,y∈R,证明 |xy| = |x| · |y|。证明。[这里我们需要同时考虑 x 和 y 的情况。当其中一个变量为零时,创建特殊情况会更容易。] 我们知道 x = 0 或 y = 0 或 x, y > 0 或 x > 0 > y 或 y > 0 > x 或 x, y < 0,因此使用特殊情况。

Case 1. Assume x = 0; prove | xy | = | x | · | y |. In this case, we know xy = 0 y = 0, so | xy | = |0| = 0 and | x | · | y | = 0| y | = 0.

情况 1. 假设 x = 0;证明 |xy| = |x| · |y|。在这种情况下,我们知道 xy = 0y = 0,所以 |xy| = |0| = 0 且 |x| · |y| = 0|y| = 0。

Case 2. Assume y = 0; prove | xy | = | x | · | y |. As in Case 1, xy = 0, so | xy | = 0 = | x |0 = | x | · | y |.

情况 2. 假设 y = 0;证明 |xy| = |x| · |y|。与情况 1 一样,xy = 0,所以 |xy| = 0 = |x|0 = |x| · |y|。

Case 3. Assume x > 0 and y > 0; prove | xy | = | x | · | y |. By algebra, xy > 0. Using the definition of absolute value three times, we see that | xy | = xy = | x | · | y |.

情况 3. 假设 x > 0 且 y > 0;证明 |xy| = |x| · |y|。根据代数,xy > 0。三次使用绝对值的定义,我们看到 |xy| = xy = |x| · |y|。

Case 4. Assume x > 0 and y < 0, so xy < 0. We compute | xy | = −( xy ) = x · ( − y ) = | x | · | y |.

情况 4. 假设 x > 0 且 y < 0,则 xy < 0。我们计算 |xy| = −(xy) = x · ( − y) = |x| · |y|。

Case 5. Assume x < 0 and y > 0, so xy < 0. We compute | xy | = −( xy ) = ( − x ) · y = | x | · | y |.

情况 5. 假设 x < 0 且 y > 0,则 xy < 0。我们计算 |xy| = −(xy) = ( − x) · y = |x| · |y|。

Case 6. Assume x < 0 and y < 0, so xy > 0. We compute | xy | = xy = ( − x ) · ( − y ) = | x | · | y |. So | xy | = | x | · | y | holds in all cases, as needed.

情况 6. 假设 x < 0 且 y < 0,则 xy > 0。我们计算 |xy| = xy = ( − x) · ( − y) = |x| · |y|。因此,|xy| = |x| · |y| 在所有情况下都成立,符合要求。

Disproving Statements

反驳陈述

So far, we have given many proof templates that enable us to prove true statements P . Of course, many statements P are false, and we cannot prove such statements. Instead, we would like a method for disproving a statement we believe to be false. Since P is false iff ∼ P is true, we are led to the following template .

到目前为止,我们已经给出了许多证明模板,这些模板使我们能够证明真命题 P。当然,许多命题 P 是假的,我们无法证明这些命题。相反,我们需要一种方法来反驳我们认为是假的命题。由于 P 为假当且仅当 ∼P 为真,我们得到了以下模板。

2.55. Proof Template to Disprove a False Statement P.

2.55. 反驳错误陈述的证明模板 P.

Prove P .

证明∼P。

The next question is, how do we prove a true statement of the form ∼ P ? We offer the following template as an answer.

下一个问题是,我们如何证明形如∼P的命题为真?我们提供以下模板作为答案。

2.56. Proof Template to Prove ∼ P.

2.56. 证明 ∼P 的证明模板。

Use the denial rules to replace ∼ P with an equivalent statement Q that does not begin with NOT. Prove Q .

利用否定规则,将 ∼P 替换为一个等价的语句 Q,该语句不以 NOT 开头。证明 Q。

To complete Step 2, we recursively apply other templates, expand definitions appearing in Q , and so on.

#

为了完成步骤 2,我们递归地应用其他模板,展开 Q 中出现的定义,依此类推。

2.57. Example. Disprove: x R , x 2 x .

2.57. 示例。反证:∀x∈R,x2≥x。

Solution. By the disproof template, we must prove x R , x 2 x . Using the denial rules, we find that we must prove the statement: x R , x 2 < x . [If you cannot initially find an example of this existential assertion, try drawing the graphs of y = x 2 and y = x .] We now see that we may choose x = 1/2, which is a real number such that x 2 = 1/4 < 1/2 = x .

解:根据反证模板,我们必须证明 ∼∀x∈R,x2≥x。利用否定规则,我们发现必须证明命题:∃x∈R,x2<x。[如果您一开始找不到这个存在断言的例子,请尝试绘制 y = x 2 和 y = x 的图像。] 现在我们看到,我们可以选择 x = 1/2,这是一个实数,使得 x 2 = 1/4 < 1/2 = x。

2.58. Example. Disprove: for all a , b Z , if a divides b , then a b .

2.58. 例。反证:对于所有 a,b∈Z,如果 a 整除 b,则 a ≤ b。

Solution. Using the denial rules, we must prove there exist integers a , b such that a divides b and a > b . [Before reading further, can you think of an example of this situation? The key is to realize that a and b come from the universe Z , which includes negative integers.] We can now choose a = 4 and b = −4 (say). Note that −4 = 4( − 1) where −1 is an integer, verifying that 4 divides −4. We also know 4 > −4, finishing the proof.

解:利用否定规则,我们需要证明存在整数 a 和 b,使得 a 整除 b 且 a > b。[在继续阅读之前,你能想到这种情况的例子吗?关键在于意识到 a 和 b 来自包含负整数的宇宙 Z。] 现在我们可以选择 a = 4 和 b = −4(例如)。注意 −4 = 4( − 1),其中 −1 是一个整数,这验证了 4 整除 −4。我们还知道 4 > −4,证明完毕。

In the next example, the notation “ x Q ” means “ ( x Q ) ,” i.e., x is not rational.

在下一个例子中,“x∉Q”表示“∼(x∈Q)”,即x不是有理数。

2.59. Example. Disprove: for all real x , y , if x Q and y Q , then x + y Q .

2.59. 例。反证:对于所有实数 x、y,如果 x∉Q 且 y∉Q,则 x+y∉Q。

Solution. We must prove there exist x , y R with x Q and y Q and x + y Q . Recall from Theorem 2.38 that 2 is not rational. So, let us choose x = 2 and y = 2 . The fact that y is also irrational follows from Example 2.31 , taking r = 2 . We know x + y = 2 + 2 = 0 = 0 / 1 , where 0 and 1 are integers and 1 ≠ 0. So x + y Q , as needed.

解:我们需要证明存在 x, y∈R,使得 x∉Q 且 y∉Q,并且 x+y∈Q。回顾定理 2.38,2 不是有理数。因此,我们取 x=2 和 y=−2。y 也是无理数这一事实可由例 2.31 得出,其中 r=2。我们知道 x+y=2+−2=0=0/1,其中 0 和 1 是整数且 1≠ 0。所以 x+y∈Q,证毕。

The last example shows that the set of irrational real numbers is not closed under addition. Similarly, we ask the reader to check below that the set of irrational numbers is not closed under multiplication.

最后一个例子表明,无理实数集对加法不封闭。类似地,我们请读者验证下面的例子,证明无理数集对乘法也不封闭。

2.60. Remark. Beginners sometimes confuse disproofs and proofs by contradiction. Let us contrast these two situations. To prove a true statement P by the contradiction method, we begin with the line “ Assume P to get a contradiction.” Our goal in the proof is to deduce a contradiction (not specified in advance). In contrast, to disprove a false statement P by the disproof method, we begin with the line “ Prove P .” The next step in both proofs is to find a useful denial of ∼ P by the denial rules. But the two proofs use this denial in totally different ways, since ∼ P is the assumption in the contradiction proof and the goal in the disproof. To avoid confusion, always write (and pay attention to!) these logical status words.

2.60. 备注。初学者有时会将反证法和反证法混淆。让我们对比一下这两种情况。要用反证法证明一个真命题 P,我们首先写“假设 ∼P 以得出矛盾”。证明的目标是推导出矛盾(事先未指定)。相反,要用反证法反驳一个假命题 P,我们首先写“证明 ∼P”。两种证明的下一步都是根据否定规则找到一个对 ∼P 有效的否定。但两种证明对这个否定的使用方式截然不同,因为在反证法中 ∼P 是假设,而在反证法中则是目标。为了避免混淆,务必写出(并注意!)这些逻辑状态词。

Section Summary

章节概要

  1. Proof by Cases. When you know or have assumed an OR-statement P Q R and need to prove another statement S , say this: “Since P Q R is known/assumed, we consider 3 cases. Case 1. Assume P . Prove S . Case 2. Assume Q . Prove S . Case 3. Assume R . Prove S . So, S is true in all cases.” (Similarly, a known OR-statement P 1 P 2 ∨ · · · ∨ P n leads to a proof with n cases.)
    分情况证明法。当你已知或假设一个 OR 语句 P∨Q∨R,并且需要证明另一个语句 S 时,可以这样表述: “由于 P∨Q∨R 已知/假设,我们考虑 3 种情况。 情况 1:假设 P。证明 S。 情况 2:假设 Q。证明 S。 情况 3:假设 R。证明 S。 因此,S 在所有情况下都为真。”(类似地,已知 OR 语句 P 1 ∨P 2 ∨ · · · ∨P n 则需要 n 种情况的证明。)
  2. When to Use Proof by Cases. If some proof template directs you to assume an OR-statement, you often need proof by cases to use this assumption. To prove properties of concepts defined via cases, you typically must use proof by cases based on the same cases appearing in the definition. Axioms that are instances of the tautology P ∨(∼ P ) are known OR-statements that often lead to proof by cases. For example, given a fixed integer n 0 , we know “ n 0 is even or n 0 is not even.” Given a fixed real number r 0 , we know “ r 0 ≥ 0 or r 0 < 0.” Given a fixed object x 0 and universe U , we know “ x 0 U or x 0 U .”
    何时使用分情况证明。如果某个证明模板要求你假设一个“或”语句,那么你通常需要使用分情况证明来支持这个假设。要证明通过分情况定义的概念的属性,你通常必须使用基于定义中相同分情况的分情况证明。公理是重言式 P∨(∼P) 的实例,它们是已知的“或”语句,通常会导致分情况证明。例如,给定一个固定的整数 n 0 ,我们知道“n 0 是偶数或 n 0 不是偶数”。给定一个固定的实数 r 0 ,我们知道“r 0 ≥ 0 或 r 0 < 0”。给定一个固定的对象 x 0 和宇宙 U,我们知道“x 0 ∈ U 或 x0∉U”。
  3. Disproving False Statements. To disprove a false statement P , prove the negation ∼ P . To prove a statement ∼ P that begins with NOT, use the denial rules to replace this statement with an equivalent statement not starting with NOT. Remember to flip any quantifiers that occur at the beginning of the statement being denied.
    反驳错误陈述。要反驳一个错误陈述 P,证明其否定式 ∼P。要证明以 NOT 开头的陈述 ∼P,请使用否定规则,将其替换为一个不以 NOT 开头的等价陈述。请记住,要将被否定陈述开头出现的任何量词互换位置。

Exercises

练习

  1. Give proof outlines of the following statements. (a) If f is continuous or monotonic, then f is integrable. [use direct proof] (b) It is false that every injective function has a left inverse. (c) ( P Q ) ( R ( S ) ) .
    给出下列陈述的证明概要。 (a) 如果 f 是连续的或单调的,那么 f 是可积的。[使用直接证明] (b) 并非每个单射函数都有左逆。 (c) (P∨Q)⊕(R∧(∼S)).
  2. Consider the statement ( A B ( C ) ) ( D E ) . Give proof outlines of this statement based on: (a) a direct proof and proof by cases; (b) a contrapositive proof and proof by cases; (c) a contradiction proof.
    考虑命题 (A∨B∨(∼C))⇒(D∧E)。请根据以下方法给出该命题的证明概要:(a) 直接证明和分情况证明;(b) 逆否命题证明和分情况证明;(c) 矛盾证明。
  3. Let n be a fixed integer. Prove: if 6 divides n or n + 5 is odd, then n is even.
    设 n 为一个固定的整数。证明:如果 6 能整除 n 或 n + 5 为奇数,则 n 为偶数。
  4. Let k be a fixed integer. Prove: k 2 + 3 k is even.
    设 k 为一个固定的整数。证明:k 2 + 3k 是偶数。
  5. (a) Let x and y be fixed integers. Prove: if x is even or y is even, then x 2 y 3 is even. (b) Disprove: x Z , y Z , if x is odd or y is odd, then x 2 y 3 is odd.
    (a) 设 x 和 y 为固定整数。证明:若 x 为偶数或 y 为偶数,则 xy1 为偶数。(b) 反证:对于任意 x∈Z,任意 y∈Z,若 x 为奇数或 y 为奇数,则 xy3 为奇数。
  6. Let x be a fixed real number. (a) Prove: if x < 1 or x > 3, then x 2 − 4 x + 3 > 0. Must the converse hold? (b) Prove: x 3 x > 0 iff −1 < x < 0 or x > 1.
    设 x 为一个固定的实数。 (a) 证明:如果 x < 1 或 x > 3,则 x 2 − 4x + 3 > 0。逆命题是否成立? (b) 证明:x 3 − x > 0 当且仅当 −1 < x < 0 或 x > 1。
  7. Let z R be fixed. Prove: for z 2 − 7 z + 10 to be positive, it is necessary and sufficient that z < 2 or z > 5.
    设 z∈R 为固定值。证明:为了使 z 2 − 7z + 10 为正数,充要条件是 z < 2 或 z > 5。
  8. Let x be a fixed real number. (a) Prove: if x 2 is rational or x 3 is rational, then x 6 is rational. (b) Prove: if x 5 Q , then x 2 Q or x 3 Q .
    设 x 为一个固定的实数。 (a) 证明:如果 x 2 是有理数或 x 3 是有理数,则 x 6 是有理数。 (b) 证明:如果 x5∉Q,则 x2∉Q 或 x3∉Q。
  9. (a) Disprove: for all m , n Z , if 10 divides mn , then 10 divides m or 10 divides n . (b) Disprove: for all a , b Z 0 , if a divides b , then a b .
    (a)反证:对于所有 m,n∈Z,如果 10 整除 mn,则 10 整除 m 或 10 整除 n。(b)反证:对于所有 a,b∈Z≥0,如果 a 整除 b,则 a ≤ b。
  10. (a) Disprove: for all x , y R , if x Q and y Q , then x y Q . (b) Disprove: for all x , y R > 0 , if x Q and y Q , then x + y Q .
    (a)反证:对于所有 x,y∈R,如果 x∉Q 且 y∉Q,则 xy∉Q。(b)反证:对于所有 x,y∈R>0,如果 x∉Q 且 y∉Q,则 x+y∉Q。
  11. For fixed x , y R , prove: sgn ( x y ) = sgn ( x ) sgn ( y ) .
    对于固定的 x,y∈R,证明:sgn(xy)=sgn(x)sgn(y)。
  12. For fixed x R , is sgn ( x 3 ) = sgn ( x ) true or false? Explain.
    对于固定的 x∈R,sgn(x³)=sgn(x) 是否成立?请解释。
  13. For fixed x R , prove: (a) | x | ≥ 0; (b) | − x | = | x |; (c) x = sgn ( x ) | x | .
    对于固定的 x∈R,证明:(a)|x|≥0;(b)|−x|=|x|;(c)x=sgn(x)|x|。
  14. (a) Disprove: for all x , y R , | x + y | = | x | + | y |. (b) Disprove: x R , sgn ( x 2 ) = + 1 . (c) Disprove: for all x , y R , sgn ( x + y ) = sgn ( sgn ( x ) + sgn ( y ) ) .
    (a) 反证:对于所有 x,y∈R,|x + y| = |x| + |y|。 (b) 反证:∀x∈R,sgn(x2)=+1。 (c) 反证:对于所有 x,y∈R,sgn(x+y)=sgn(sgn(x)+sgn(y))。
  15. For fixed x , y R , prove: | x + y | ≤ | x | + | y |.
    对于固定的 x,y∈R,证明:|x + y| ≤ |x| + |y|。
  16. Use the previous exercise (which holds for all x , y R ) to deduce the following facts for fixed real numbers r , s , t . (a) | r + s + t | ≤ | r | + | s | + | t |. (b) | r t | ≤ | r s | + | s t |. (c) | r s | ≤ | r | + | s |.
    利用前面的练习(对所有 x,y∈R 都成立),推导出以下关于固定实数 r, s, t 的事实。 (a) |r + s + t| ≤ |r| + |s| + |t|。 (b) |r − t| ≤ |r − s| + |s − t|。 (c) |r − s| ≤ |r| + |s|。
  17. For fixed x , y R , prove: || x | − | y || ≤ | x y |.
    对于固定的 x,y∈R,证明:||x| − |y|| ≤ |x − y|。
  18. Let x , a , r be fixed real numbers. (a) Prove: | x a | < r iff a r < x < a + r . (b) Use (a) and logic to deduce a condition equivalent to | x a | ≥ r .
    设 x、a、r 为固定的实数。(a) 证明:|x − a| < r 当且仅当 a − r < x < a + r。(b) 利用 (a) 的结论和逻辑推理,推导出与 |x − a| ≥ r 等价的条件。
  19. Find the error in this attempted disproof of “for some positive integers x , y , z , x 3 + y 3 = z 3 .” Disproof: We must prove that for some positive integers x , y , z , x 3 + y 3 z 3 . Choose x = y = 1 and z = 2. We compute x 3 + y 3 = 1 + 1 = 2, while z 3 = 2 3 = 8.
    找出以下反证“对于某些正整数 x、y、z,x 3 + y 3 = z 3 ”的错误之处。反证:我们必须证明对于某些正整数 x、y、z,x 3 + y 3 ≠ z 3 。选择 x = y = 1 和 z = 2。我们计算 x 3 + y 3 = 1 + 1 = 2,而 z 3 = 2 3 = 8。
  20. Find and correct the error in this outline of a disproof of “ P ⇒ ( Q R ).” Disproof: Assume P and Q are true and R is false. Deduce a contradiction.
    找出并改正以下反证“P ⇒ (Q ⇒ R)”的概要中的错误。反证:假设 P 和 Q 为真,R 为假。推导出矛盾。
  21. Define f ( x ) = x 2 + 1 for 0 ≤ x ≤ 2, and f ( x ) = 3 − x for 2 < x ≤ 3. For a fixed x R , prove: if 0 ≤ x ≤ 3, then 0 ≤ f ( x ) ≤ 5.
    定义函数 f(x) = x 2 + 1,其中 0 ≤ x ≤ 2;定义函数 f(x) = 3 − x,其中 2 < x ≤ 3。对于固定的 x∈R,证明:如果 0 ≤ x ≤ 3,则 0 ≤ f(x) ≤ 5。
  22. For a fixed integer m , prove m 2 + 5 m − 3 is odd.
    对于固定的整数 m,证明 m 2 + 5m − 3 是奇数。
  23. Suppose we know this theorem: m Z , j Z , m = 4 j m = 4 j + 1 m = 4 j + 2 m = 4 j + 3 . Let b be a fixed integer. Prove: 4 divides b 2 or 8 divides b 2 − 1.
    假设我们已知以下定理:∀m∈Z,∃j∈Z,m=4j∨m=4j+1∨m=4j+2∨m=4j+3。设b为一个固定的整数。证明:4能整除b 2 或8能整除b 2 − 1。
  24. For fixed n Z , prove: if n is odd, then 8 divides n 3 n .
    对于固定的 n∈Z,证明:如果 n 是奇数,则 8 整除 n 3 − n。
  25. Suppose “ P Q R ” is a known theorem. Justify the following proof template, which proves a statement S by combining the contradiction method with proof by cases. Proof of S . We know P Q R , so use cases. Case 1. Assume P and ∼ S . Deduce a contradiction C 1 . Case 2. Assume Q and ∼ S . Deduce a contradiction C 2 . Case 3. Assume R and ∼ S . Deduce a contradiction C 3 . Since all three cases led to a contradiction, S must be true.
    假设“P∨Q∨R”是一个已知定理。请证明以下证明模板的合理性,该模板结合了反证法和分情况证明法来证明命题S。 S的证明。我们已知P∨Q∨R,因此使用分情况证明法。 情况1:假设P且∼S。得出矛盾 C1。 情况2:假设Q且∼S。得出矛盾 C2。 情况3:假设R且∼S。得出矛盾 C3。 由于这三种情况都导致了矛盾,因此S必定为真。

2.6 Proving Quantified Statements

2.6 证明量化陈述

So far, we have discussed proof templates involving the following logical operators: , ⇒, , ⇔, , ∨, and ∼. It is finally time to present templates for proving a universally quantified statement. This leads us to a tricky but essential topic: how to manipulate statements containing multiple quantifiers.

到目前为止,我们已经讨论了涉及以下逻辑运算符的证明模板:∃、⇒、∧、⇔、⊕、∨ 和 ∼。现在终于到了介绍全称量词语句证明模板的时候了。这引出了一个棘手但至关重要的主题:如何处理包含多个量词的语句。

Proof by Exhaustion

穷竭证明

Our first proof template for proving a universal statement “ x U , P ( x ) ” is called proof by exhaustion . It can only be used when the universe U is a finite set.

我们第一个用于证明全称命题“∀x∈U,P(x)”的证明模板称为穷举证明法。它只能用于全集U为有限集的情况。

2.61. Proof Template for Proof by Exhaustion. Suppose U = { z 1 , z 2 , …, z n } is a finite set. To prove x U , P ( x ) in this situation:

2.61. 穷举法证明模板。假设 U = {z 1 , z 2 , …, z n 是一个有限集。要证明在此情况下对于所有 x∈U,P(x) 成立:

  • Step 1. Prove P ( z 1 ).
    步骤 1. 证明 P(z 1 )。
  • Step 2. Prove P ( z 2 ).
    步骤 2. 证明 P(z 2 )。
  • Step n . Prove P ( z n ).

    步骤 n. 证明 P(z n )。

Intuitively, we can verify the universal statement by explicitly testing the truth of P ( c ) for every possible object c in the finite universe U . We can justify this proof method more formally by recalling the following rule from § 1.4 :

直观地说,我们可以通过显式地检验有限宇宙 U 中每个可能的对象 c 的 P(c) 的真假来验证普遍命题。我们可以通过回顾 §1.4 中的以下规则来更正式地证明这种证明方法:

x x { z z 1 , z z 2 , , z z n n } , P P ( x x ) P P ( z z 1 ) P P ( z z 2 ) P P ( z z n n ) .

To prove the universal statement, we may instead prove the equivalent AND-statement. Following the proof template for an AND-statement, we obtain the proof outline written above. The name “exhaustion” is used because we are exhaustively checking all possible cases, one at a time.

为了证明全称命题,我们可以转而证明等价的“与”命题。遵循“与”命题的证明模板,我们得到上述证明概要。之所以称之为“穷举法”,是因为我们要逐一穷举地检查所有可能的情况。

2.62. Example. Every time we use a truth table to verify that a propositional form is a tautology, we are really doing a proof by exhaustion. To see why, recall that a propositional form T is a tautology iff every row in the truth table for T evaluates to true. This definition has the form x U , P ( x ) , where U is the set of rows in the truth table, and P ( x ) is the statement “the propositional form T has output TRUE in row x of the truth table.” Since there are only finitely many rows in the truth table, U is a finite set. By filling in the truth table, we are checking the truth of P ( x ) for each row x , one at a time.

2.62. 示例。每次我们使用真值表来验证一个命题形式是否为重言式时,实际上都是在进行穷举证明。为了理解这一点,我们回顾一下命题形式 T 是重言式当且仅当 T 的真值表中的每一行都为真。这个定义的形式为 ∀x∈U,P(x),其中 U 是真值表的行集合,P(x) 表示“命题形式 T 在真值表的第 x 行的输出为真”。由于真值表只有有限行,因此 U 是一个有限集合。通过填充真值表,我们实际上是在逐行检查 P(x) 的真假。

Generic Element Proofs of Universal Statements

通用命题的一般元素证明

When the universe U is not finite, proof by exhaustion does not work. We need a new strategy for proving a universal statement “ x U , P ( x ) ,” which says that P ( x 0 ) is true for all objects x 0 in the potentially infinite set U . This strategy is called the generic-element method.

当论域 U 不是有限集时,穷举法不再适用。我们需要一种新的策略来证明全称命题“∀x∈U,P(x)”,该命题表示对于潜在无限集 U 中的所有对象 x 0 ,P(x 0 ) 都成立。这种策略被称为泛元素法。

2.63. Generic-Element Proof Template to Prove x U , P ( x ) .

2.63. 通用元素证明模板,用于证明 ∀x∈U,P(x)。

Let x 0 be a fixed , but arbitrary , object in U .

设 x 0 为 U 中的一个固定但任意的对象。

Prove P ( x 0 ).

证明 P(x 0 )。

Before using this template, let us discuss its logical nuances by contrasting it with the template for proving x U , P ( x ) . When we prove the existence statement x U , P ( x ) , we do so by choosing one specific , constant object x 0 in U that makes P ( x 0 ) true. For instance, we might say: “choose x 0 = 7.” More generally, x 0 could be an expression involving other constants introduced earlier in the proof. We complete the existence proof by showing that x 0 really is in U and proving that P ( x 0 ) really is true.

在使用此模板之前,让我们通过将其与证明 ∃x∈U,P(x) 的模板进行对比,来讨论其逻辑上的细微差别。当我们证明存在性命题 ∃x∈U,P(x) 时,我们通过选择 U 中的一个特定常量对象 x 0 来使 P(x 0 ) 为真。例如,我们可以说:“选择 x 0 = 7。” 更一般地,x 0 可以是包含证明中先前引入的其他常量的表达式。我们通过证明 x 0 确实在 U 中,并证明 P(x 0 ) 确实为真,来完成存在性证明。

In the new template to prove x U , P ( x ) , we also introduce a constant object x 0 in U , but we do not choose a specific value for this object. The letter x 0 used here represents a “generic” object in U that we do not control. We know nothing about x 0 initially, except that it comes from the universe U . If we can prove the proposition P ( x 0 ) using only this information, we may safely conclude that P ( x ) does hold for every object x in U . Here is a basic example.

在证明 ∀x∈U,P(x) 的新模板中,我们也引入了 U 中的一个常量对象 x 0 ,但我们不为其选择特定值。这里使用的字母 x 0 代表 U 中一个我们无法控制的“通用”对象。我们最初对 x 0 一无所知,只知道它来自宇宙 U。如果我们仅使用这些信息就能证明命题 P(x 0 ),那么我们可以安全地得出结论:P(x) 对 U 中的每个对象 x 都成立。以下是一个基本示例。

2.64. Example. Prove: x Z , x divides 3 x 2 .

2.64. 示例。证明:∀x∈Z,x 整除 3x 2

Proof. Let x 0 be a fixed, but arbitrary integer. We must prove x 0 divides 3 x 0 2 . Expanding the definition, we must prove k Z , 3 x 0 2 = k x 0 . Choose k 0 = 3 x 0 , which is an integer since x 0 Z and 3 Z and Z is closed under multiplication. We know k 0 x 0 = ( 3 x 0 ) x 0 = 3 x 0 2 , so the proof is done.

证明。设 x 0 为任意整数。我们需要证明 x 0 整除 3x02。展开定义,我们需要证明存在 k∈Z,3x02=kx0。取 k 0 = 3x 0 ,由于 x0∈Z 且 3∈Z,Z 对乘法封闭,因此 k 为整数。我们知道 k0x0=(3x0)x0=3x02,所以证明完毕。

In this proof, we used subscripts ( x 0 and k 0 ) to distinguish the constants x 0 and k 0 from the quantified variables x and k . Observe that x 0 (coming from the universally quantified variable x ) is a generic constant, whose value is not chosen or controlled by us. But k 0 (coming from the existentially quantified variable k in the definition of “divides”) is a specific constant that we choose in a way to make the proof work. Observe that k 0 can depend on the previously introduced constant x 0 ; the expression 3 x 0 used for k 0 is a specific choice depending on the previous generic constant x 0 .

在本证明中,我们使用下标 (x 0 和 k 0 ) 来区分常量 x 0 和 k 0 与量化变量 x 和 k。请注意,x 0 (源自全称量化变量 x)是一个通用常量,其值并非由我们选择或控制。而 k 0 (源自“除法”定义中的存在量化变量 k)是一个特定的常量,我们选择它是为了使证明成立。请注意,k 0 可以依赖于先前引入的常量 x 0 ;用于表示 k 0 的表达式 3x 0 是一个基于先前通用常量 x 0 的特定选择。

We remark that many of the examples in previous sections were really proving universally quantified statements. Since we had not yet introduced the generic-element template, we introduced those examples with statements like “Let n be a fixed integer.” So, for instance, Example 2.20 on page 59 actually proves the theorem

我们注意到,前面几节中的许多例子实际上都在证明全称量词命题。由于我们当时还没有引入泛元素模板,所以我们用类似“设 n 为一个固定的整数”这样的语句来引入这些例子。例如,第 59 页的例 2.20 实际上证明了该定理。

a 一个 Z Z , b b Z Z , c c Z Z , if 如果 a 一个 divides 分割 b b and b b divides 分割 c c then 然后 a 一个 divides 分割 c c .

Statements with Multiple Quantifiers

包含多个量词的语句

We now discuss statements with multiple quantifiers, which reveal further distinctions between the proof templates for universal and existential quantifiers. The first point is that we may safely reorder two universal quantifiers or two existential quantifiers, as stated below.

现在我们讨论包含多个量词的陈述,这将进一步揭示全称量词和存在量词的证明模板之间的区别。首先,我们可以安全地重新排列两个全称量词或两个存在量词,如下所述。

2.65. Quantifier Reordering Rules. For any open sentence P ( x , y ):

2.65. 量词重排序规则。对于任意开放语句 P(x, y):

x x , y , P P ( x x , y ) is equivalent to 相当于 y , x x , P P ( x x , y ) ;
x x , y , P P ( x x , y ) is equivalent to 相当于 y , x x , P P ( x x , y ) .

Intuitively, the statements x , y , P ( x , y ) and y , x , P ( x , y ) are equivalent since both statements assert that P ( x 0 , y 0 ) is true for all possible choices of the objects x 0 and y 0 . Similarly, x , y , P ( x , y ) and y , x , P ( x , y ) are equivalent since both statements assert the existence of two objects x 0 and y 0 (possibly equal) such that P ( x 0 , y 0 ) is true. We can give more formal proofs of these rules using the proof templates we have developed; see § 2.7 for details. We can also use restricted quantifiers here; for instance,

直观地说,语句 ∀x,∀y,P(x,y) 和 ∀y,∀x,P(x,y) 是等价的,因为这两个语句都断言对于对象 x 0 和 y 0 的所有可能选择,P(x 0 , y 0 ) 都为真。类似地,∃x,∃y,P(x,y) 和 ∃y,∃x,P(x,y) 也是等价的,因为这两个语句都断言存在两个对象 x 0 和 y 0 (可能相等),使得 P(x 0 , y 0 ) 为真。我们可以使用已开发的证明模板对这些规则给出更正式的证明;详情请参见第 2.7 节。我们也可以在这里使用限定量词;例如,

x x U U , y V V , P P ( x x , y ) is equivalent to 相当于 y V V , x x U U , P P ( x x , y ) .

Recall from our discussion of definitions that we can change the letter used for a quantified variable, as long as the new letter does not already have a different meaning . For instance, the three quantified statements z , P ( z ) and w , P ( w ) and s , P ( s ) are equivalent, since each statement says that P ( c ) is true for every specific object c . We can do the same thing for statements with multiple quantifiers. For example,

回想一下我们之前对定义的讨论,我们可以更改量化变量所用的字母,只要新字母没有改变原有的含义即可。例如,三个量化语句 ∀z,P(z)、∀w,P(w) 和 ∀s,P(s) 是等价的,因为每个语句都表示 P(c) 对每个特定对象 c 都成立。对于包含多个量词的语句,我们也可以做同样的事情。例如,

x x , y , z z , P P ( x x , y , z z ) is equivalent to 相当于 r r , s s , t t , P P ( r r , s s , t t ) .

Mixed Quantifiers

混合量词

A crucial fact of quantified logic is that the relative order of universal and existential quantifiers in a multiply-quantified statement has a dramatic effect on the meaning of the statement. The next examples illustrate this fact.

量化逻辑的一个关键事实是,在多重量化语句中,全称量词和存在量词的相对顺序对语句的含义有着显著的影响。以下示例将阐明这一事实。

2.66. Example. Prove or disprove: x Z , y Z , y > x .

2.66. 示例。证明或反证:∀x∈Z,∃y∈Z,y>x。

Solution. This statement is true; it can be translated into English by saying: “for every integer, there is a larger integer.” To prove it with the templates, we say: let x 0 be a fixed, but arbitrary integer. We must prove y Z , y > x 0 . Choose y 0 = x 0 + 1. We know y 0 is an integer (by closure), and x 0 + 1 > x 0 , completing the proof. In this proof, note carefully that the specific constant y 0 chosen in the existence proof was allowed to depend on the previously introduced generic constant x 0 . Contrast this situation to the next example.

解:这个命题是正确的;它可以翻译成英文:“对于每一个整数,都存在一个更大的整数。” 为了用模板证明它,我们令 x 0 为一个固定的任意整数。我们需要证明存在 y∈Z,y>x0。选择 y 0 = x 0 + 1。我们知道 y 0 是一个整数(由闭包定理可知),并且 x 0 + 1 > x 0 ,证明完毕。在这个证明中,请仔细注意,存在性证明中选择的特定常数 y 0 可以依赖于之前引入的通用常数 x 0 。请将此情况与下一个例子进行对比。

2.67. Example. Prove or disprove: y Z , x Z , y > x . (The only difference from the previous example is the order of the initial quantifiers.)

2.67. 示例。证明或反驳:∃y∈Z,∀x∈Z,y>x。(与前一个例子的唯一区别在于初始量词的顺序。)

Solution. If we tentatively try to prove this statement by using the templates, we must begin by choosing a specific, constant integer y 0 , and continue by proving x Z , y 0 > x . What y 0 shall we pick? If we try y 0 = 7, we must prove the statement x Z , 7 > x , which is evidently false. If we try y 0 = 100, we must prove the statement x Z , 100 > x , which is also false. We might try y 0 = ∞, but ∞ is not an integer. Why can’t we choose y 0 = x 0 + 1, as we did in the previous proof? The reason is that in this proof, the fixed constant integer x 0 has not been introduced yet, so we cannot use it when choosing y 0 . That is the key difference between the two statements. In fact, the current statement is false; in English, it says that “there is an integer larger than every integer.”

解:如果我们尝试用模板来证明这个命题,首先必须选择一个特定的常数整数 y 0 ,然后证明 ∀x∈Z,y0>x。我们应该选择什么 y 0 呢?如果我们尝试 y 0 = 7,就必须证明命题 ∀x∈Z,7>x,这显然是错误的。如果我们尝试 y 0 = 100,就必须证明命题 ∀x∈Z,100>x,这同样是错误的。我们或许可以尝试 y 0 = ∞,但 ∞ 不是整数。为什么我们不能像之前的证明那样选择 y 0 = x 0 + 1 呢?原因在于,在这个证明中,固定的常数整数 x 0 尚未引入,因此我们在选择 y 0 时无法使用它。这是两个陈述之间的关键区别。事实上,当前的陈述是错误的;用通俗的话来说,它表示“存在一个大于所有整数的整数”。

Now, since we think the statement is false, we ought to be able to disprove it. We do so by proving a useful denial of the statement. Applying the denial rules, we must prove y Z , x Z , y x . This denial resembles the original example, since it starts with followed by . To prove the denial, we fix an arbitrary integer y 0 , and prove x Z , y 0 x . We now choose x 0 to be y 0 , which works since y 0 Z and y 0 y 0 . We could have also chosen x 0 to be y 0 + 1 or y 0 + 2, for instance. Note that x 0 is allowed to depend on the previously introduced constant y 0 .

既然我们认为这个陈述是错误的,我们就应该能够反驳它。我们通过证明一个有用的否定来做到这一点。应用否定规则,我们必须证明 ∀y∈Z,∃x∈Z,y≤x。这个否定与原例类似,因为它以 ∀ 开头,后面跟着 ∃。为了证明这个否定,我们固定一个任意整数 y 0 ,并证明 ∃x∈Z,y0≤x。现在我们选择 x 0 为 y 0 ,这符合条件,因为 y0∈Z 且 y 0 ≤ y 0 。我们还可以选择 x 0 为 y 0 + 1 或 y 0 + 2,例如。请注意,x 0 可以依赖于先前引入的常数 y 0

In general, the statement y , x , P ( x , y ) is harder to prove than the statement x , y , P ( x , y ) , because the y we choose is allowed to depend on x in the latter statement, but y cannot depend on x in the former statement. Nevertheless, statements of the first type are sometimes true, as we see next.

一般来说,命题 ∃y,∀x,P(x,y) 比命题 ∀x,∃y,P(x,y) 更难证明,因为在后一个命题中,我们选择的 y 可以依赖于 x,而在前一个命题中,y 不能依赖于 x。然而,正如我们接下来看到的,第一类命题有时也是成立的。

2.68. Example. Prove or disprove: y Z , x Z , y < x 2 .

2.68. 例。证明或反证:∃y∈Z,∀x∈Z,y<x2。

Proof. The existential quantifier comes first, so we must begin by choosing y . We choose y 0 = −1, which is a specific integer. We must prove x Z , 1 < x 2 . Let x 0 be a fixed, but arbitrary integer. We must prove 1 < x 0 2 . By algebra, x 0 2 0 > 1 . [We could give a more detailed proof that x 0 2 0 by looking at the two cases where x 0 ≥ 0 or x 0 < 0, and using properties of inequalities to deduce x 0 2 0 in each case.]

证明。存在量词在前,所以我们必须先选择 y。我们选择 y 0 = −1,这是一个特定的整数。我们必须证明对于所有 x∈Z,−1<x2。设 x 0 是一个固定的任意整数。我们必须证明 −1<x02。根据代数性质,x02≥0>−1。[我们可以通过考察 x 0 ≥ 0 或 x 0 < 0 这两种情况,并利用不等式的性质推导出每种情况下的 x02≥0,来给出更详细的 x02≥0 的证明。]

2.69. Example. Prove or disprove: x Z , y Z , y < x 2 .

2.69. 例。证明或反证:∀x∈Z,∃y∈Z,y<x2。

Proof. [This statement is true; we proved an even stronger statement in the previous example. To see why, let us prove the current statement.] Let x 0 be a fixed, but arbitrary integer. We must prove y Z , y < x 0 2 . Choose y 0 to be −1 [note that y 0 is allowed to depend on x 0 , but our choice is not required to use x 0 .] We know 1 Z and 1 < 0 x 0 2 , as before.      □

证明。[此命题为真;我们在前一个例子中证明了一个更强的命题。为了说明原因,让我们证明当前的命题。] 设 x 0 为一个固定的任意整数。我们必须证明存在 y∈Z,y<x02。选择 y 0 为 −1。[注意,y 0 可以依赖于 x 0 ,但我们的选择并非必须使用 x 0 。] 我们知道 −1∈Z 且 −1<0≤x02,与之前相同。□

The last two examples illustrate the following general fact.

最后两个例子说明了以下一般事实。

2.70. Theorem. For any open sentence P ( x , y ) and any universes U and V ,

2.70. 定理。对于任意开语句 P(x, y) 和任意全集 U 和 V,

y U U , x x V V , P P ( x x , y ) x x V V , y U U , P P ( x x , y ) .

We prove this theorem in the next section.

我们将在下一节证明这个定理。

Proving Statements with Multiple Quantifiers

证明含有多个量词的陈述

It is very common in mathematics to have statements in which three or more quantifiers (some existential, some universal) appear at the beginning of a complex statement. The next example illustrates how the rules above generalize to such a situation. The main fact to remember is that when proving an existential statement by choosing a value for the variable, that value is only allowed to depend on all previously introduced constants, which correspond to quantified variables appearing to the left of the given variable.

在数学中,经常会出现这样的情况:一个复杂的语句开头出现三个或更多量词(有些是存在量词,有些是全称量词)。下一个例子说明了上述规则如何推广到这种情况。需要记住的关键一点是,当通过为变量选择一个值来证明一个存在命题时,该值只能依赖于之前引入的所有常量,这些常量对应于出现在给定变量左侧的量化变量。

2.71. Example. Outline the proof of this statement:

2.71. 示例。简述以下命题的证明:

u , δ d , w 西 , x x , y , z z , P P ( u , δ d , w 西 , x x , y , z z ) .
#

The second quantified variable is the Greek letter δ (delta). Solution. We follow the proof templates, processing the quantifiers from left to right, obtaining the following lines in this order:

第二个量化变量是希腊字母δ(delta)。解法:我们按照证明模板,从左到右处理量词,得到以下几行:

Suppose someone tries to shorten the preceding text by writing the following variation of the proof outline:

假设有人试图通过编写以下证明大纲的变体来缩短前面的文本:

This variation is not correct, since the order in which variables are introduced indicates that w 0 and y 0 might potentially depend on all the previously mentioned constants u 0 , δ 0 , x 0 , and z 0 . In fact, the alternate proof outline would prove the statement

这种变体并不正确,因为变量的引入顺序表明 w 0 和 y 0 可能依赖于所有前面提到的常量 u 0 、δ 0 、x 0 和 z 0 。事实上,另一种证明思路可以证明该命题。

u , δ d , x x , z z , w 西 , y , P P ( u , δ d , w 西 , x x , y , z z ) ,

which is weaker than the original statement in general.

总体而言,这比原声明的力度要弱。

2.72. Example. Prove: x R , y R , if x < y , then z R , x < z < y .

2.72. 例。证明:对于任意 x∈R,任意 y∈R,如果 x < y,则存在 z∈R,x < z < y。

Proof. Let x 0 and y 0 be fixed, but arbitrary real numbers. Assume x 0 < y 0 . Prove z R , x 0 < z < y 0 . Intuitively, the average of x 0 and y 0 should lie between these two numbers. So, we choose z 0 = ( x 0 + y 0 )/2, which is a real number since R is closed under addition and division by 2. We must prove x 0 < z 0 < y 0 , which is really the AND-statement x 0 < z 0 and z 0 < y 0 . To do this, divide the assumed inequality x 0 < y 0 by 2 to get x 0 /2 < y 0 /2. On one hand, adding x 0 /2 to both sides produces x 0 /2 + x 0 /2 < x 0 /2 + y 0 /2, so x 0 < z 0 . On the other hand, adding y 0 /2 to both sides produces x 0 /2 + y 0 /2 < y 0 /2 + y 0 /2, so z 0 < y 0 . This completes the proof.

证明。设 x 0 和 y 0 为任意实数。假设 x 0 < y 0 。证明:存在 z∈R,x0<z<y0。直观地讲,x 0 和 y 0 的平均值应该介于这两个数之间。因此,我们选择 z 0 = (x 0 + y 0 )/2,由于 R 对加法和除以 2 封闭,所以 z 0 是一个实数。我们必须证明 x 0 < z 0 < y 0 ,这实际上是与语句 x 0 < z 0 且 z 0 < y 0 相等。为此,将假设的不等式 x 0 < y 0 除以 2,得到 x 0 /2 < y 0 /2。一方面,等式两边同时加上 x 0 /2,得到 x 0 /2 + x 0 /2 < x 0 /2 + y 0 /2,所以 x 0 < z 0 。另一方面,等式两边同时加上 y 0 /2,得到 x 0 /2 + y 0 /2 < y 0 /2 + y 0 /2,所以 z 0 < y 0 。证明完毕。

Section Summary

章节概要

  1. Proof by Exhaustion. To prove x U , P ( x ) , where U is the finite set { z 1 , …, z n }: Prove P ( z 1 ). Prove P ( z 2 ). … Prove P ( z n ).
    穷举法。证明对于所有 x∈U,P(x),其中 U 是有限集 {z 1 , …, z n :证明 P(z 1 )。证明 P(z 2 )。…证明 P(z n )。
  2. Generic-Element Template for Universal Statements. To prove x U , P ( x ) : Step 1 . Let x 0 be a fixed , but arbitrary , object in U . Step 2 . Prove P ( x 0 ). Here x 0 is a new letter denoting a generic constant. We cannot choose a specific value for x 0 . We have no specific initial knowledge about x 0 , except we know x 0 U . Step 1 is often abbreviated by saying: “fix x 0 U .”
    通用语句的通用元素模板。证明 ∀x∈U,P(x): 步骤 1. 设 x 0 为 U 中一个固定的任意对象。 步骤 2. 证明 P(x 0 )。 这里 x 0 是一个新字母,表示一个通用常量。我们不能为 x 0 选择特定值。我们对 x 0 没有具体的初始信息,只知道 x 0 ∈ U。步骤 1 通常简写为:“固定 x 0 ∈ U。”
  3. Constructive Proof of x U , P ( x ) . Choose a specific object x 0 ; prove x 0 is in U ; prove P ( x 0 ). In contrast to the generic-element method, we must choose a specific object x 0 here. This object is allowed to depend on constants previously introduced in the proof, but it cannot involve constants introduced later.
    证明存在 x∈U,P(x)。选择一个具体对象 x 0 ;证明 x 0 属于 U;证明 P(x 0 )。与通用元素方法不同,这里我们必须选择一个具体对象 x 0 。该对象可以依赖于证明中先前引入的常量,但不能包含之后引入的常量。
  4. Quantifier Reordering Rules. The following rules hold for all open sentences R ( x , y ):
    1. x , y , R ( x , y ) y , x , R ( x , y ) . 2. x , y , R ( x , y ) y , x , R ( x , y ) . 3. x , y , R ( x , y ) y , x , R ( x , y ) .
    The converse of rule 3 is not universally valid: in general, changing the order of the quantifiers and affects the meaning of the statement.
    量词重排序规则。以下规则适用于所有开放语句 R(x, y): :1.∀x,∀y,R(x,y)⇔∀y,∀x,R(x,y)。2.∃x,∃y,R(x,y)⇔∃y,∃x,R(x,y)。3.∃x,∀y,R(x,y)⇒∀y,∃x,R(x,y)。规则 3 的逆命题并非普遍成立:一般来说,改变量词 ∃ 和 ∀ 的顺序会影响语句的含义。
  5. Sample Proof Outlines of Statements with Mixed Quantifiers. Observe how differences in the types and ordering of quantifiers leads to contrasting proof structures in the following examples. (a) To prove x , y , z , P ( x , y , z ) : Fix x 0 . Fix y 0 . Choose z 0 = (…) (an expression that can involve x 0 and y 0 ). Prove P ( x 0 , y 0 , z 0 ). (b) To prove x , z , y , P ( x , y , z ) : Fix x 0 . Choose z 0 = (…) (an expression that can involve x 0 only). Fix y 0 . Prove P ( x 0 , y 0 , z 0 ). (c) To prove z , x , y , P ( x , y , z ) : Choose z 0 = (…) (a specific object not depending on x or y ). Fix x 0 . Fix y 0 . Prove P ( x 0 , y 0 , z 0 ). (d) To disprove x , z , y , P ( x , y , z ) : Prove x , z , y , P ( x , y , z ) . Choose x 0 = (…) (a specific object not depending on anything else). Fix z 0 . Choose y 0 = (…) (an expression that can involve x 0 and z 0 ). Prove ∼ P ( x 0 , y 0 , z 0 ).
    混合量词语句的证明大纲示例。观察以下示例中量词的类型和顺序如何导致不同的证明结构。 (a) 证明 ∀x,∀y,∃z,P(x,y,z):固定 x 0 。固定 y 0 。选择 z 0 = (…)(一个可以包含 x 0 和 y 0 的表达式)。证明 P(x 0 , y 0 , z 0 ). (b) 证明 ∀x,∃z,∀y,P(x,y,z): 固定 x 0 。选择 z 0 = (…)(一个仅包含 x 0 的表达式)。固定 y 0 。证明 P(x 0 , y 0 , z 0 ). (c) 为了证明 ∃z,∀x,∀y,P(x,y,z):选择 z 0 = (…)(一个不依赖于 x 或 y 的特定对象)。固定 x 0 。固定 y 0 。证明 P(x 0 , y 0 , z 0 ). (d) 为了反证 ∀x,∃z,∀y,P(x,y,z):证明 ∃x,∀z,∃y,∼P(x,y,z)。选择 x 0 = (…)(一个不依赖于任何其他对象的特定对象)。固定 z 0 。选择 y 0 = (…)(一个可以包含 x 0 和 z 0 的表达式)。证明 ∼P(x 0 , y 0 , z 0 )。

Exercises

练习

  1. Prove each universal statement. (a) For all x , y Z , if x divides y , then x 2 divides y 2 . (b) For all x ∈ {1, 3, 5, 7}, 8 divides x 2 − 1. (c) For all x R , if x 3 Q , then x Q .
    证明下列全称命题。 (a) 对于所有 x,y∈Z,如果 x 整除 y,则 x 2 整除 y 2 (b) 对于所有 x ∈ {1, 3, 5, 7},8 整除 x 2 − 1。 (c) 对于所有 x∈R,如果 x3∉Q,则 x∉Q。
  2. Give proof outlines for each statement similar to those in summary item 5 on page 86. (a) r , s , t , u , Q ( r , s , t , u ) . (b) s , t , r , u , Q ( r , s , t , u ) . (c) s , t , u , r , Q ( r , s , t , u ) . (d) u , t , r , s , Q ( r , s , t , u ) . (e) Find all pairs of statements in (a) through (d) such that the first statement always implies the second statement.
    为每个语句给出类似于第 86 页摘要第 5 项中的证明概要。 (a) ∀r,∀s,∃t,∃u,Q(r,s,t,u). (b) ∀s,∃t,∀r,∃u,Q(r,s,t,u). (c) ∀s,∃t,∃u,∀r,Q(r,s,t,u). (d) ∃u,∃t,∀r,∀s,Q(r,s,t,u). (e) 找出 (a) 到 (d) 中所有语句对,使得第一个语句总是蕴含第二个语句。
  3. Outline disproofs of statements (a) through (d) in the previous exercise.
    简述上一练习中陈述 (a) 至 (d) 的反证。
  4. True or false? Prove each true statement, and disprove each false statement. (a) y Z , x Z , x y = 0 . (b) x R , y R , x y = 1 . (c) y R , x R , x y = 1 . (d) x R > 0 , y R > 0 , y < x . (e) y R > 0 , x R > 0 , y x . (f) x Z , y Z 0 , x 2 = y 2 .
    判断正误。证明每个陈述为真,并反驳每个陈述为假。 (a) ∃y∈Z,∀x∈Z,xy=0. (b) ∀x∈R,∃y∈R,xy=1. (c) ∃y∈R,∀x∈R,xy=1. (d) ∀x∈R>0,∃y∈R>0,y<x. (e) ∃y∈R>0,∀x∈R>0,y≤x. (f) ∀x∈Z,∃y∈Z≥0,x²=y².
  5. Prove or disprove each statement. (a) m Z , n Z , mn is even. (b) m Z , n Z , mn is odd. (c) n Z , m Z , mn is even. (d) n Z , m Z , mn is odd. (e) n Z , m Z , m + n is odd.
    证明或反驳下列陈述。 (a) ∃m∈Z,∀n∈Z, mn 为偶数。 (b) ∃m∈Z,∀n∈Z, mn 为奇数。 (c) ∀n∈Z,∃m∈Z, mn 为偶数。 (d) ∀n∈Z,∃m∈Z, mn 为奇数。 (e) ∀n∈Z,∃m∈Z, m + n 为奇数。
  6. Prove or disprove each statement. (a) a Z , b Z , a divides b . (b) a Z , b Z , b divides a . (c) a Z , b Z , c Z , a divides b + c . (d) a Z , c Z , b Z , a divides b + c . (e) a Z , b Z , c Z , b + c divides a . (f) a Z , c Z , b Z , b + c divides a .
    证明或反驳下列陈述。 (a) ∃a∈Z,∀b∈Z, a 整除 b。 (b) ∃a∈Z,∀b∈Z, b 整除 a。 (c) ∀a∈Z,∀b∈Z,∃c∈Z, a 整除 b + c。 (d) ∀a∈Z,∃c∈Z,∀b∈Z, a 整除 b + c。 (e) ∀a∈Z,∀b∈Z,∃c∈Z, b + c 整除 a。 (f) ∀a∈Z,∃c∈Z,∀b∈Z, b + c 整除 a。
  7. Prove or disprove each statement. (a) x R , y R , xy is rational. (b) x R , y R , x + y is rational. (c) q Q , n Z > 0 , n q Z . (d) n Z > 0 , q Q , n q Z . (e) q Q , n Z , n q Z .
    证明或反驳下列陈述。 (a) ∃x∈R,∀y∈R, xy 是有理数。 (b) ∀x∈R,∃y∈R, x + y 是有理数。 (c) ∀q∈Q,∃n∈Z>0,nq∈Z。 (d) ∃n∈Z>0,∀q∈Q,nq∈Z。 (e) ∃q∈Q,∀n∈Z,nq∈Z。
  8. Prove or disprove each statement. (a) x R , y R , y x | x | = | y | . (b) z R , w R , | z | | w | . (c) x R > 0 , y R , | x + y | | x | + | y | . (d) y R , x R > 0 , | x + y | | x | + | y | .
    证明或反驳下列陈述。 (a) ∀x∈R,∃y∈R,y≠x∧|x|=|y|. (b) ∃z∈R,∀w∈R,|z|≤|w|. (c) ∀x∈R>0,∃y∈R,|x+y|≠|x|+|y|. (d) ∃y∈R,∀x∈R>0,|x+y|≠|x|+|y|.
  9. Prove: y R , x R , y = 2 x 7 . (b) Disprove: x R , y R , y = 2 x 7 . (c) Disprove: y Z , x Z , y = 2 x 7 .
    证明:∀y∈R,∃x∈R,y=2x−7。(b)反证:∃x∈R,∀y∈R,y=2x−7。(c)反证:∀y∈Z,∃x∈Z,y=2x−7。
  10. (a) Prove: y R 3 , x R 5 , x y 5 y 3 x + 13 = 0 . (b) Prove: x R 5 , y R 3 , x y 5 y 3 x + 13 = 0 . (c) Prove: x R , b R , y R , x y 5 y 3 x + 13 = b .
    (a) 证明:∀y∈R≠3,∃x∈R≠5,xy−5y−3x+13=0. (b) 证明:∀x∈R≠5,∃y∈R≠3,xy−5y−3x+13=0. (c) 证明:∃x∈R,∃b∈R,∀y∈R,xy−5y−3x+13=b.
  11. Prove or disprove the following statements, which are variations of Example 2.72 . (a) x R , z R , y R , x < y x < z < y . (b) x R , z R , y R , x < y x < z < y . (c) x R , z R , y R , x < z x < z l t y . (d) z R , x R < 0 , y R > 0 , x < z < y . (e) x Q , y Q , z Q , x z y y z x .
    证明或反驳下列陈述,它们是例 2.72 的变体。(a) ∀x∈R,∃z∈R,∀y∈R,x<y⇒x<z<y. (b) ∀x∈R,∀z∈R,∃y∈R,x<y⇒x<z<y. (c) ∀x∈R,∀z∈R,∃y∈R,x<z⇒x<zlty. (d) ∃z∈R,∀x∈R<0,∀y∈R>0,x<z<y. (e) ∀x∈Q,∀y∈Q,∃z∈Q,x≤z≤y∨y≤z≤x.
  12. Prove or disprove each statement. (a) x R , y R , z R , x + y + z = 0 . (b) x R , y R , z R , x + y + z = 0 . (c) x R , y R , z R , x + y + z = 0 . (d) y R , z R , x R , x + y z = 3 . (e) x R , y R , z R , x + y z = 3 .
    证明或反驳下列陈述。 (a) ∀x∈R,∀y∈R,∃z∈R,x+y+z=0. (b) ∀x∈R,∃y∈R,∀z∈R,x+y+z=0. (c) ∃x∈R,∀y∈R,∀z∈R,x+y+z=0. (d) ∃y∈R,∀z∈R,∃x∈R,x+yz=3. (e) ∃x∈R,∃y∈R,∀z∈R,x+yz=3.
  13. When choosing objects in this problem, indicate explicitly which other objects the chosen object may depend upon. (a) Outline a proof of this statement: ϵ R > 0 , x R , δ R > 0 , y R , | x y | < δ | f ( x ) f ( y ) | < ϵ . (b) Outline a proof of this statement: ϵ R > 0 , δ R > 0 , x R , y R , | x y | < δ | f ( x ) f ( y ) | < ϵ . (c) Outline a disproof of the statement in (b). [ Remark: (a) is the definition of continuity of the function f , whereas (b) is the definition of uniform continuity of the function f . These concepts are studied in advanced calculus.]
    在本题中选择对象时,请明确指出所选对象可能依赖于哪些其他对象。 (a) 简述以下命题的证明:∀ϵ∈R>0,∀x∈R,∃δ∈R>0,∀y∈R,|x−y|<δ⇒|f(x)−f(y)|<ϵ。 (b) 简述以下命题的证明:∀ϵ∈R>0,∃δ∈R>0,∀x∈R,∀y∈R,|x−y|<δ⇒|f(x)−f(y)|<ϵ。 (c) 简述 (b) 中命题的反证。 [注:(a) 是函数 f 的连续性定义,而 (b) 是函数 f 的一致连续性定义。这些概念在高等微积分中会学习。
  14. Let ( a 1 , a 2 , …, a m , …) be a fixed sequence of real numbers. Let L R be fixed. (a) Outline a proof of: ϵ R > 0 , N Z > 0 , m Z > 0 , ( m N | a m L | < ϵ ) . (b) Outline a disproof of the statement in (a).
    设 (a, a, …, a, …) 为一个固定的实数序列。设 L∈R 为固定值。(a) 简述以下命题的证明:∀ϵ∈R>0,∃N∈Z>0,∀m∈Z>0,(m≥N⇒|am−L|<ϵ)。(b) 简述 (a) 中命题的反证。
  15. (a) Outline a proof of: e G , x G , ( e x = x y G , y x = e ) . (b) Outline a disproof of the statement in (a).
    (a)简述证明:∃e∈G,∀x∈G,(e⋆x=x∧∃y∈G,y⋆x=e)。(b)简述反证(a)中的陈述。
  16. (a) Prove or disprove the statement in part (a) of the previous problem, taking G = Z , and replacing by + (integer addition). (b) Repeat (a) taking G = R and replacing by · (real multiplication). (c) Repeat (a) taking G = Q > 0 and replacing by · (multiplication of rational numbers).
    (a) 证明或反驳上一题 (a) 部分的陈述,令 G=Z,并将 ⋆ 替换为 +(整数加法)。 (b) 重复 (a),令 G=R,并将 ⋆ 替换为 ·(实数乘法)。 (c) 重复 (a),令 G=Q>0,并将 ⋆ 替换为 ·(有理数乘法)。
  17. (a) Is x , y , P ( x , y ) equivalent to x , y , P ( y , x ) for all P ( x , y )? Explain. (b) Is x , y , P ( x , y ) equivalent to x , y , P ( y , x ) for all P ( x , y )? Explain. (c) Is x , y , P ( x , y ) equivalent to x , y , P ( y , x ) for all P ( x , y )? Explain.
    (a) 对于所有 P(x, y),∃x,∃y,P(x,y) 是否等价于 ∃x,∃y,P(y,x)?请解释。 (b) 对于所有 P(x, y),∀x,∀y,P(x,y) 是否等价于 ∀x,∀y,P(y,x)?请解释。 (c) 对于所有 P(x, y),∀x,∃y,P(x,y) 是否等价于 ∀x,∃y,P(y,x)?请解释。
  18. Create your own proof template to prove a uniqueness statement of the form ! x U , P ( x ) . [ Hint: Use the elimination rule for the uniqueness symbol, and then combine several previously discussed proof templates to outline a proof of the new statement].
    创建你自己的证明模板,以证明形如 ∃!x∈U,P(x) 的唯一性命题。[提示:使用唯一性符号的消去规则,然后结合几个前面讨论过的证明模板来概述新命题的证明]。
  19. Use the template in the previous problem to prove each uniqueness statement. (a) ! x R , 3 x + 5 = 0 . (b) y R , a R 0 , b R , ! x R , a x + b = y . (c) ! x Z > 0 , x 2 5 x 14 = 0 .
    使用上一题中的模板证明每个唯一性陈述。 (a) ∃!x∈R,3x+5=0. (b) ∀y∈R,∀a∈R≠0,∀b∈R,∃!x∈R,ax+b=y. (c) ∃!x∈Z>0,x2−5x−14=0.
  20. Let P ( x , y ) be a fixed open sentence. (a) Outline a proof of: x R , ! y R , P ( x , y ) . (b) Outline a disproof of the statement in (a).
    设 P(x, y) 为一个固定的开语句。 (a) 简述以下证明:∀x∈R,∃!y∈R,P(x,y)。 (b) 简述 (a) 中陈述的反证。
  21. For each choice of P ( x , y ), prove or disprove the statement in part (a) of the previous problem. (a) P ( x , y ) is “ y = x 2 .” (b) P ( x , y ) is “ x = y 2 .” (c) P ( x , y ) is “ x = y 3 .” (d) P ( x , y ) is “ xy = 1.” (e) P ( x , y ) is “ x 2 + y 2 = 4 y 0 .”
    对于 P(x, y) 的每种取值,证明或反驳上一题 (a) 部分的陈述。(a) P(x, y) 是“y = x 2 ”。(b) P(x, y) 是“x = y 2 ”。(c) P(x, y) 是“x = y 3 ”。(d) P(x, y) 是“xy = 1”。(e) P(x, y) 是“x²+y²=4∧y≥0”。
  22. Let S be a fixed set of real numbers. (a) Outline a proof of: y R , [ ( x S , x y ) ( z R , z < y u S , z < u ) ] . (b) Outline a disproof of the statement in (a).
    设 S 为实数集。 (a) 简述以下命题的证明:∃y∈R,[(∀x∈S,x≤y)∧(∀z∈R,z<y⇒∃u∈S,z<u)]. (b) 简述 (a) 中命题的反证。
  23. For each choice of the set S , prove or disprove the statement in part (a) of the previous problem. (a) S = R . (b) S is the set of x R with 0 ≤ x ≤ 1. (c) S is the set of x R with x < 3. (d) S is the set of real numbers of the form 2 − 1/ n for some n Z > 0 . (e) S is the empty set. (f) S = Z .
    对于集合 S 的每种选择,证明或反驳上一题 (a) 部分的陈述。(a) S=R。(b) S 是 x∈R 且 0 ≤ x ≤ 1 的集合。(c) S 是 x∈R 且 x < 3 的集合。(d) S 是形如 2 − 1/n 的实数集合,其中 n∈Z>0。(e) S 是空集。(f) S=Z。
  24. Prove ϵ R > 0 , δ R > 0 , x R , y R , | x y | < δ | ( 3 x + 5 ) ( 3 y + 5 ) | < ϵ by completing the outline in part (b) of Exercise 13.
    通过完成练习 13 (b) 部分的提纲,证明 ∀ϵ∈R>0,∃δ∈R>0,∀x∈R,∀y∈R,|x−y|<δ⇒|(3x+5)−(3y+5)|<ϵ。
  25. Prove ϵ R > 0 , x R , δ R > 0 , y R , | x y | < δ | x 2 y 2 | < ϵ by completing the outline in part (a) of Exercise 13.
    通过完成练习 13 (a) 部分的提纲,证明 ∀ϵ∈R>0,∀x∈R,∃δ∈R>0,∀y∈R,|x−y|<δ⇒|x2−y2|<ϵ。
  26. Disprove ϵ R > 0 , δ R > 0 , x R , y R , | x y | < δ | x 2 y 2 | < ϵ by completing the outline in part (c) of Exercise 13.
    通过完成练习 13 的 (c) 部分中的提纲,反驳 ∀ϵ∈R>0,∃δ∈R>0,∀x∈R,∀y∈R,|x−y|<δ⇒|x2−y2|<ϵ。
  27. Prove: ϵ R > 0 , x R > 0 , δ R > 0 , y R > 0 , | x y | < δ | ( 1 / x ) ( 1 / y ) | < ϵ .
    证明:∀ϵεR>0,∀xεR>0,∃δεR>0,∀yεR>0,|x−y|<δ⇒|(1/x)−(1/y)|<ϵ。
  28. Disprove:
    ϵ R > 0 , δ R > 0 , x R > 0 , y R > 0 , | x y | < δ | ( 1 / x ) ( 1 / y ) | < ϵ .

    反证:∀ϵ∈R>0,∃δεR>0,∀x∈R>0,∀y∈R>0,|x−y|<δ⇒|(1/x)−(1/y)|<ϵ。
  29. (a) Prove: ! y R , ! x R , x 2 = y . (b) Prove or disprove: ! x R , ! y R , x 2 = y .
    (a)证明:∃!y∈R,∃!x∈R,x2=y。(b)证明或反证:∃!x∈R,∃!y∈R,x2=y。

2.7 More Quantifier Properties and Proofs (Optional)

2.7 量词的更多性质和证明(可选)

We begin this section with some logical rules showing how the quantifiers and interact with propositional operators such as and ∨. After some informal discussion, we use the proof templates to prove these rules and some other properties stated in the last section. These proofs highlight the difference between how multiple quantifiers are treated in known statements vs. statements to be proved . The material presented here is a bit more abstract and challenging compared to earlier sections, so I have designated this section as optional. But it is a great opportunity to understand, at a deeper level, the nuances of how quantifiers and proof templates work.

本节首先介绍一些逻辑规则,说明量词 ∀ 和 ∃ 如何与命题运算符(例如 ∧ 和 ∨)交互。经过一些非正式的讨论后,我们将使用证明模板来证明这些规则以及上一节中提到的其他一些性质。这些证明突出了已知语句和待证明语句中处理多个量词的方式的差异。与前面的章节相比,本节内容更加抽象和具有挑战性,因此我将其列为可选内容。但这仍然是一个深入理解量词和证明模板运作方式细微差别的绝佳机会。

How Quantifiers Interact with Propositional Operators

量词如何与命题运算符交互

We have remarked that the universal quantifier can be viewed as a generalization of the logical AND operator , whereas the existential quantifier generalizes the OR operator ∨. This observation may give some algebraic intuition for the following result.

我们已经指出,全称量词 ∀ 可以看作是逻辑与运算符 ∧ 的推广,而存在量词 ∃ 则是或运算符 ∨ 的推广。这一观察或许能为以下结果提供一些代数上的直观理解。

2.73. Theorem. For any open sentences P ( x ) and Q ( x ) and universe U :

2.73. 定理。对于任意开语句 P(x) 和 Q(x) 以及论域 U:

  1. (a) x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) .
    (a)∀x∈U,[P(x)∧Q(x)]⇔(∀y∈U,P(y))∧(∀z∈U,Q(z))。
  2. (b) x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) .
    (b)∃x∈U,[P(x)∨Q(x)]⇔(∃y∈U,P(y))∨(∃z∈U,Q(z))。

We give an intuitive explanation for this theorem here, deferring a careful proof in the technical notes to this section. The universal statement on the left side of statement (a) asserts that every object c in U makes both P ( c ) and Q ( c ) true. The right side of statement (a) is an AND-statement asserting that, first, every object c in U makes P ( c ) true; and second, every object c in U makes Q ( c ) true. Intuitively, the two sides of statement (a) mean the same thing, so they should be logically equivalent. Statement (b) has a similar intuitive justification: the left side says that at least one object c in U makes P ( c ) true or Q ( c ) true; the right side says that either some object c in U makes P ( c ) true, or some object c in U makes Q ( c ) true.

我们在此对该定理给出直观的解释,而将详细的证明放在技术说明中。语句 (a) 左侧的全称命题断言,U 中的每个对象 c 都使 P(c) 和 Q(c) 为真。语句 (a) 右侧是一个“与”语句,它断言,首先,U 中的每个对象 c 都使 P(c) 为真;其次,U 中的每个对象 c 都使 Q(c) 为真。直观上,语句 (a) 的两部分含义相同,因此它们在逻辑上应该是等价的。语句 (b) 也有类似的直观解释:左侧表明,U 中至少有一个对象 c 使 P(c) 为真或 Q(c) 为真;右侧表明,U 中要么存在某个对象 c 使 P(c) 为真,要么存在某个对象 c 使 Q(c) 为真。

Next we present two more subtle rules, involving the interaction of and ∨ on one hand, and the interaction of and on the other hand.

接下来,我们将介绍两条更微妙的规则,一条涉及∀和∨的相互作用,另一条涉及∃和∧的相互作用。

2.74. Theorem. For any open sentences P ( x ) and Q ( x ) and universe U :

2.74. 定理。对于任意开语句 P(x) 和 Q(x) 以及论域 U:

  1. (a) [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] [ x U , ( P ( x ) Q ( x ) ) ] .
    (a) [(∀y∈U,P(y))∨(∀z∈U,Q(z))]⇒[∀x∈U,(P(x)∨Q(x))].
  2. (b) [ x U , ( P ( x ) Q ( x ) ) ] [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] .

    (b) [∃x∈U,(P(x)∧Q(x))]⇒[(∃y∈U,P(y))∧(∃z∈U,Q(z))].

Warning: The converse of (a) is NOT always true; the converse of (b) is NOT always true.

警告:(a) 的逆命题并不总是成立;(b) 的逆命题并不总是成立。

Let us begin with specific examples illustrating the failure of the converse statements. For statement (a), let U be the set of all humans, let P ( y ) mean “ y is female,” and let Q ( z ) mean “ z is male.” The statement y U , P ( y ) says that every human is female, which is false. The statement z U , Q ( z ) says that every human is male, which is also false. So the OR-statement appearing as the hypothesis of statement (a) is false in this example. On the other hand, for any fixed, particular human x 0 , the statement that “ x 0 is female or x 0 is male” is true, so that the universal statement appearing as the conclusion of statement (a) is true in this example. So, statement (a) is indeed true ( F T is true), but the converse of statement (a) is false in this example.

让我们先来看一些具体的例子来说明逆命题的失败。对于命题 (a),设 U 为所有人类的集合,P(y) 表示“y 是女性”,Q(z) 表示“z 是男性”。命题 ∀y∈U,P(y) 表示所有人类都是女性,这是错误的。命题 ∀z∈U,Q(z) 表示所有人类都是男性,这也是错误的。因此,作为命题 (a) 假设的 OR 命题在这个例子中是错误的。另一方面,对于任意给定的特定人类 x 0 ,“x 0 是女性或 x 0 是男性”这一命题为真,因此作为命题 (a) 结论的全称命题在这个例子中为真。所以,命题 (a) 的确为真(F ⇒ T 为真),但命题 (a) 的逆命题在这个例子中为假。

For statement (b), we give a more mathematical example. Let U = Z , let P ( x ) mean “ x > 0,” and let Q ( x ) mean “ x < 0.” Does there exist at least one fixed integer x 0 making the statement “ x 0 > 0 and x 0 < 0” true? The answer is no, so the hypothesis of statement (b) is false in this example. On the other hand, consider the AND-statement in the conclusion of statement (b). Does there exist an integer y 0 such that y 0 > 0? Yes, for example, y 0 = 1. So the first half of the AND-statement is true. Does there exist an integer z 0 such that z 0 < 0? Yes, for example, z 0 = −1. So the second half of the AND-statement is true. Thus, statement (b) has a true conclusion. We see now that statement (b) is true but its converse is false, in this example.

对于陈述 (b),我们给出一个更数学化的例子。设 U=Z,P(x) 表示“x > 0”,Q(x) 表示“x < 0”。是否存在至少一个固定的整数 x 0 使得陈述“x 0 > 0 且 x 0 < 0”为真?答案是否定的,因此陈述 (b) 的假设在这个例子中是错误的。另一方面,考虑陈述 (b) 结论中的 AND 语句。是否存在整数 y 0 使得 y 0 > 0?是的,例如,y 0 = 1。因此 AND 语句的前半部分为真。是否存在整数 z 0 使得 z 0 < 0?例如,z 0 = −1。因此,AND语句的后半部分为真。所以,语句(b)的结论为真。现在我们看到,在这个例子中,语句(b)为真,但其逆命题为假。

We now give informal justifications of of Theorem 2.74 ; see below for a formal proof. Intuitively, the hypothesis of (a) demands that one of two very strong conditions hold: either every object in the universe satisfies P , or every object in the universe satisfies Q . The conclusion of statement (a) is the much weaker assertion that for each individual object, considered one at a time, that object satisfies either P or Q . (Which one of the two is true can vary from object to object.)

现在我们对定理 2.74 进行非正式的论证;正式证明见下文。直观地说,(a) 的假设要求两个非常强的条件之一成立:要么宇宙中的每个对象都满足 P,要么宇宙中的每个对象都满足 Q。(a) 的结论是一个弱得多的断言,即对于每个单独的对象,逐一考虑,该对象满足 P 或 Q。(这两个条件中哪一个成立可能因对象而异。)

Here is intuition for statement (b). The hypothesis of statement (b) is the rather strong statement that we can find one particular object in U that simultaneously satisfies both P and Q . The conclusion of statement (b) says that some object in U satisfies P , and some (possibly different) object in U satisfies Q .

以下是对陈述 (b) 的直观解释。陈述 (b) 的假设是一个相当强的陈述,即我们可以在 U 中找到一个特定的对象,它同时满足 P 和 Q。陈述 (b) 的结论是,U 中存在某个对象满足 P,并且 U 中存在某个(可能不同的)对象满足 Q。

Proofs of the Quantifier Interaction Properties

We now use the proof templates to generate formal proofs of Theorem 2.73 and 2.74 . Recall that letters with subscripts (like x 0 ) denote constants, whereas the corresponding letter without a subscript is a quantified variable. Throughout these proofs, we let P ( x ) and Q ( x ) be fixed, but arbitrary, open sentences, and let U be a fixed universe. [Note that we have just used the generic-element proof template three times to deal with the universal quantifiers that begin the theorem statements].

2.75. Proof of x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) .

[This is an IFF-statement, so we need a two-part proof.]

Part 1 . Assume x U , [ P ( x ) Q ( x ) ] . Prove ( y U , P ( y ) ) ( z U , Q ( z ) ) .

[The new goal is an AND-statement, so we prove each part separately.]

First, prove y U , P ( y ) . Fix an arbitrary object y 0 U . Prove P ( y 0 ). To do so, take x = y 0 in the assumed universal statement (we are using the Inference Rule for ALL here). The assumption tells us that P ( y 0 ) Q ( y 0 ) is true. By definition of AND, P ( y 0 ) is true, as needed.

Second, prove z U , Q ( z ) . Fix an arbitrary object z 0 U . Prove Q ( z 0 ). To do so, take x = z 0 in the assumed universal statement (using the Inference Rule for ALL). The assumption tells us that P ( z 0 ) Q ( z 0 ) is true. By definition of AND, Q ( z 0 ) is true, as needed. Part 1 is now complete.

Part 2 . Assume ( y U , P ( y ) ) ( z U , Q ( z ) ) . Prove x U , [ P ( x ) Q ( x ) ] .

[The goal of part 2 is a universal statement, so we use the generic-element proof template.]

Let x 0 be a fixed, but arbitrary element of U . We must prove P ( x 0 ) Q ( x 0 ) . To prove this AND-statement, we first prove P ( x 0 ), and then prove Q ( x 0 ). Note that the AND-statement we have assumed tells us that y U , P ( y ) is true, and z U , Q ( z ) is true. Taking y = x 0 in the first of these statements, we see that P ( x 0 ) is true by the Inference Rule for ALL. Taking z = x 0 in the second of these statements, we see that Q ( x 0 ) is true by the Inference Rule for ALL. Part 2 is now complete.

2.76. Proof of x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) . We could prove this statement from first principles, but it is quicker to use the statement we have just proved, namely Theorem 2.73 (a). Since this statement was proved for all open sentences P ( x ) and Q ( x ), we can replace P ( x ) by ∼ P ( x ) and Q ( x ) by ∼ Q ( x ) in part (a). We see that we already know this fact:

x U , [ ( P ( x ) ) ( Q ( x ) ) ] [ y U , P ( y ) ) ( z U , Q ( z ) ] .
Recall the logical equivalence A B ≡ (∼ A ) ⇔ (∼ B ). We can use this equivalence to transform the fact written above by negating the statements on the left and right sides of the ⇔ symbol. Using the denial rules to further simplify each side, we eventually get
x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) ,

which is precisely the result we needed to prove.

2.77. Proof of [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] [ x U , ( P ( x ) Q ( x ) ) ] .

[We give a direct proof of the IF-statement.]

Assume ( y U , P ( y ) ) ( z U , Q ( z ) ) . Prove x U , ( P ( x ) Q ( x ) ) .

[We have just assumed an OR-statement, so use proof by cases.]

Case 1 . Assume y U , P ( y ) . Prove x U , ( P ( x ) Q ( x ) ) . Let x 0 be a fixed, but arbitrary element of U . Prove P ( x 0 )∨ Q ( x 0 ). Taking y = x 0 in the assumption, the Inference Rule for ALL tells us that P ( x 0 ) is true. Now, by definition of OR, “ P ( x 0 )∨ Q ( x 0 )” must also be true.

Case 2 . Assume z U , Q ( z ) . Prove x U , ( P ( x ) Q ( x ) ) . Let x 0 be a fixed, but arbitrary element of U . Prove P ( x 0 )∨ Q ( x 0 ). Taking z = x 0 in the assumption, the Inference Rule for ALL tells us that Q ( x 0 ) is true. Now, by definition of OR, “ P ( x 0 )∨ Q ( x 0 )” must also be true.

So the needed conclusion holds in all cases.

We could deduce Theorem 2.74 (b) by taking the contrapositive of what we just proved and simplifying (see the exercises), but proving part (b) from scratch is also instructive.

2.78. Proof of [ x U , ( P ( x ) Q ( x ) ) ] [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] .

Assume x U , P ( x ) Q ( x ) . Our assumption means there is at least one constant object x 0 in U such that P ( x 0 ) Q ( x 0 ) is true [we are using existential instantiation (EI) to give a name x 0 to the object whose existence is assumed; see page 59]. Prove ( y U , P ( y ) ) ( z U , Q ( z ) ) . [To prove an AND-statement, we prove each separate part]. We must first prove y U , P ( y ) . Choose y 0 = x 0 , which is in U . By definition of AND, P ( x 0 ) is true. We must next prove z U , Q ( z ) . Choose z 0 = x 0 , which is in U . By definition of AND, Q ( x 0 ) is true.

The reader is invited to attempt to prove the converse of each statement in Theorem 2.74 for general open sentences P ( x ) and Q ( x ), to see exactly where the proof templates get stuck.

量词交互性质的证明 我们现在使用证明模板来生成定理 2.73 和 2.74 的形式化证明。回想一下,带下标的字母(例如 x 0 )表示常量,而对应的不带下标的字母表示量化变量。在这些证明中,我们令 P(x) 和 Q(x) 为固定的但任意的开语句,并令 U 为固定的论域。[注意,我们刚刚三次使用了通用元素证明模板来处理定理陈述开头的全称量词]。2.75. 证明 ∀x∈U,[P(x)∧Q(x)]⇔(∀y∈U,P(y))∧(∀z∈U,Q(z))。 [这是一个逆正则表达式,所以我们需要两部分证明。] 第一部分:假设 ∀x∈U,[P(x)∧Q(x)]。证明 (∀y∈U,P(y))∧(∀z∈U,Q(z))。[新的目标是一个与表达式,所以我们分别证明每一部分。] 首先,证明 ∀y∈U,P(y)。固定任意对象 y 0 ∈ U。证明 P(y 0 )。为此,在假设的全称表达式中取 x = y 0 (这里我们使用全称表达式的推理规则)。假设告诉我们 P(y0)∧Q(y0) 为真。根据与表达式的定义,P(y 0 ) 为真,满足要求。第二,证明 ∀z∈U,Q(z)。固定任意对象 z 0 ∈ U。证明 Q(z 0 )。为此,在假定的全称命题中取 x = z 0 (使用全称推理规则)。该假定表明 P(z0)∧Q(z0) 为真。根据 AND 的定义,Q(z 0 ) 为真,满足要求。第一部分完成。第二部分。假设 (∀y∈U,P(y))∧(∀z∈U,Q(z))。证明 ∀x∈U,[P(x)∧Q(x)]。 【第二部分的目标是证明一个全称命题,因此我们使用通用元素证明模板。】设 x 0 是 U 的一个固定但任意的元素。我们必须证明 P(x0)∧Q(x0)。为了证明这个 AND 命题,我们首先证明 P(x 0 ),然后证明 Q(x 0 )。注意,我们假设的 AND 命题告诉我们,对于所有 y∈U,P(y) 为真,并且对于所有 z∈U,Q(z) 为真。在第一个命题中,取 y = x 0 ,根据全称推理规则,我们看到 P(x 0 ) 为真。在第二个陈述中,令 z = x 0 ,根据“对所有”推理规则,Q(x 0 ) 为真。第二部分到此完成。2.76. 证明 ∃x∈U,[P(x)∨Q(x)]⇔(∃y∈U,P(y))∨(∃z∈U,Q(z))。我们可以从基本原理证明这个陈述,但使用我们刚刚证明的陈述,即定理 2.73(a) 会更快。由于该陈述已针对所有开语句 P(x) 和 Q(x) 得到证明,因此我们可以将 (a) 部分中的 P(x) 替换为 ∼P(x),将 Q(x) 替换为 ∼Q(x)。我们看到,我们已经知道这个事实:∀x∈U,[(∼P(x))∧(∼Q(x))]⇔[∀y∈U,∼P(y))∧(∀z∈U,∼Q(z)]。回想一下逻辑等价关系 A ⇔ B ≡ (∼A) ⇔ (∼B)。我们可以利用这个等价关系,通过否定 ⇔ 符号左右两侧的语句来转换上述事实。利用否定规则进一步简化等式两边,最终得到 ∃x∈U,[P(x)∨Q(x)]⇔(∃y∈U,P(y))∨(∃z∈U,Q(z)),这正是我们需要证明的结果。2.77. 证明[(∀y∈U,P(y))∨(∀z∈U,Q(z))]⇒[∀x∈U,(P(x)∨Q(x))]。[我们直接证明 IF 语句。] 假设 (∀y∈U,P(y))∨(∀z∈U,Q(z))。证明 ∀x∈U,(P(x)∨Q(x))。[我们刚刚假设了一个 OR 语句,所以使用分情况证明。] 情况 1. 假设 ∀y∈U,P(y)。证明 ∀x∈U,(P(x)∨Q(x))。设 x 0 是 U 的一个固定但任意的元素。证明 P(x 0 )∨Q(x 0 )。取 y =在假设中,对于所有元素 x 0 ,推理规则告诉我们 P(x 0 ) 为真。现在,根据 OR 的定义,“P(x 0 )∨Q(x 0 )” 也必须为真。情况 2. 假设 ∀z∈U,Q(z)。证明 ∀x∈U,(P(x)∨Q(x))。设 x 0 是 U 中一个固定的任意元素。证明 P(x 0 )∨Q(x 0 )。在假设中取 z = x 0 ,推理规则告诉我们 Q(x 0 ) 为真。现在,根据 OR 的定义, “P(x 0 )∨Q(x 0 )”也必须为真。因此,所需的结论在所有情况下都成立。我们可以通过对刚刚证明的定理取逆否命题并简化来推导出定理 2.74(b)(参见练习),但从头开始证明 (b) 部分也很有启发意义。2.78. [∃x∈U,(P(x)∧Q(x))]⇒[(∃y∈U,P(y))∧(∃z∈U,Q(z))] 的证明。假设 ∃x∈U,P(x)∧Q(x)。我们的假设意味着 U 中至少存在一个常量对象 x 0 ,使得 P(x0)∧Q(x0) 为真[我们使用存在实例化 (EI) 来命名]。将 x 0 赋值给假定存在的对象;参见第 59 页]。证明 (∃y∈U,P(y))∧(∃z∈U,Q(z))。[为了证明一个 AND 语句,我们需要分别证明每个部分]。我们首先必须证明 ∃y∈U,P(y)。选择 y 0 = x 0 ,它属于 U。根据 AND 的定义,P(x 0 ) 为真。接下来,我们必须证明 ∃z∈U,Q(z)。选择 z 0 = x 0 ,它属于 U。根据 AND 的定义,Q(x 0 ) 为真。读者可以尝试证明定理 2.74 中每个语句的逆命题,对于一般的开语句 P(x) 和 Q(x),以了解其逆命题。正是证明模板卡住的地方。

Proofs of Quantifier Reordering Rules

量词重排序规则的证明

This section provides proofs of the rules for reordering quantifiers, which were stated in the previous section. Throughout this section, we let x and y be arbitrary distinct variables, and we let P ( x , y ) be a fixed, but arbitrary, open sentence.

本节提供上一节所述量词重排序规则的证明。在本节中,我们设 x 和 y 为任意不同的变量,P(x, y) 为一个固定的但任意的开语句。

2.79. Proof of x , y , P ( x , y ) y , x , P ( x , y ) .

2.79. 证明 ∀x,∀y,P(x,y)⇔∀y,∀x,P(x,y)。

[This is an IFF-statement, so we do a two-part proof.]

[这是一个逆正则表达式,所以我们需要进行两步证明。]

Part 1 . Assume x , y , P ( x , y ) . Prove y , x , P ( x , y ) .

第一部分。假设对于所有 x,对于所有 y,P(x,y)。证明对于所有 y,对于所有 x,P(x,y)。

[The new goal is a universal statement, so use the generic-element template].

[新目标是通用的,因此请使用通用元素模板]。

Let y 0 be a fixed, but arbitrary object. Let x 0 be a fixed, but arbitrary object. Prove P ( x 0 , y 0 ) is true. [To continue, note that the assumption is a known universal statement, so we can use the Inference Rule for ALL, abbreviated IRA]. In the assumption, take x to be the object x 0 to deduce (by IRA) y , P ( x 0 , y ) . Then take y to be the object y 0 to deduce (by IRA) that P ( x 0 , y 0 ) is true. This completes part 1. We have now proved

设 y 0 为一个固定的任意对象。设 x 0 为一个固定的任意对象。证明 P(x 0 , y 0 ) 为真。[继续,注意假设是一个已知的全称命题,因此我们可以使用全集推理规则,简称 IRA]。在假设中,取 x 为对象 x 0 ,根据 IRA 可推导出 ∀y,P(x0,y)。然后取 y 为对象 y 0 ,根据 IRA 可推导出 P(x 0 , y 0 ) 为真。第一部分完成。我们现在已经证明了

x x , y , P P ( x x , y ) y , x x , P P ( x x , y ) .

Part 2 . [This part is almost identical to part 1, so we let you fill in the details. Alternatively, the IF-statement to be proved in part 2 follows from the IF-statement already proved in part 1 by interchanging x and y , after replacing P ( x , y ) by P ( y , x ).]

第二部分。[这部分几乎与第一部分完全相同,因此我们留给您自行补充细节。或者,第二部分中要证明的 IF 语句可以通过交换第一部分中已证明的 IF 语句中的 x 和 y,并将 P(x, y) 替换为 P(y, x) 来得出。]

In the exercises below, we ask you to prove

在以下练习中,我们要求你证明

x x , y , P P ( x x , y ) y , x x , P P ( x x , y )

(2.5)

in two ways.

有两种方式。

2.80. Proof of [ x , y , P ( x , y ) ] [ y , x , P ( x , y ) ] .

2.80. [∃x,∀y,P(x,y)]⇒[∀y,∃x,P(x,y)]的证明。

Assume x , y , P ( x , y ) . Prove y , x , P ( x , y ) .

假设对于任意 y,都有 P(x,y)。证明对于任意 y,都有 P(x,y)。

By Existential Instantiation (EI), the assumption tells us there is some fixed object x 0 that makes y , P ( x 0 , y ) true. To begin proving the new goal, let y 0 be a fixed, but arbitrary object. We must show x , P ( x , y 0 ) . We choose x here to be x 0 ; so now we must show P ( x 0 , y 0 ) is true. This follows by applying the Inference Rule for ALL to y , P ( x 0 , y ) , letting y be the object y 0 .

根据存在实例化(EI),假设告诉我们存在某个固定对象 x 0 ,使得 ∀y,P(x0,y) 为真。为了开始证明新的目标,设 y 0 为一个固定但任意的对象。我们必须证明存在 x,P(x,y0)。这里我们选择 x 为 x 0 ;因此,现在我们必须证明 P(x 0 , y 0 ) 为真。这可以通过将“全部”推理规则应用于 ∀y,P(x0,y) 来证明,其中 y 为对象 y 0

To help understand what happened in that last proof, let’s try to prove the converse and see what goes wrong.

为了帮助理解上一个证明中发生了什么,让我们尝试证明它的逆命题,看看哪里出了问题。

2.81. Attempted Proof of [ y , x , P ( x , y ) ] [ x , y , P ( x , y ) ] .

2.81. 尝试证明 [∀y,∃x,P(x,y)]⇒[∃x,∀y,P(x,y)].

Assume y , x , P ( x , y ) . Prove x , y , P ( x , y ) . [Here, at the very beginning of the proof, we are required to choose a specific object x 0 making y , P ( x 0 , y ) true. What are we going to choose? The assumption cannot help us unless we can figure out a helpful constant to substitute for y using the Inference Rule for ALL (IRA). Suppose we blindly try y = y 0 , where y 0 is some particular object. By IRA and then EI, the assumption does give us an object x 0 making P ( x 0 , y 0 ) true for this particular y 0 . But now, to prove y , P ( x 0 , y ) , we need to fix a new arbitrary object y 1 , which need not equal the fixed object y 0 selected earlier in the proof. So there is no obvious way to complete the proof. We have already seen examples where the original statement fails, so we will not be able to find a proof.]

假设对于所有 y,存在 x,P(x,y)。证明对于所有 y,存在 x,P(x,y)。[这里,在证明的开头,我们需要选择一个特定的对象 x 0 ,使得对于所有 y,P(x0,y) 为真。我们应该选择什么呢?除非我们能利用全集推理规则 (IRA) 找到一个有用的常数来代替 y,否则这个假设对我们没有任何帮助。假设我们盲目地尝试 y = y 0 ,其中 y 0 是某个特定的对象。根据 IRA 和 EI,这个假设确实给出了一个对象 x 0 ,使得对于这个特定的 y 0 ,P(x 0 , y 0 ) 为真。但是,为了证明 ∀y,P(x0,y),我们需要固定一个新的任意对象 y 1 ,它不必等于证明中先前选定的固定对象 y 0 。因此,没有明显的办法来完成证明。我们已经看到原命题不成立的例子,所以我们无法找到证明。

The moral of this section is that the relative ordering of quantifiers and other logical symbols often has a big impact on the meaning of the statement. So great care is needed when working with a statement containing multiple quantifiers.

本节的要点在于,量词和其他逻辑符号的相对顺序通常会对语句的含义产生重大影响。因此,在处理包含多个量词的语句时,需要格外谨慎。

Section Summary

章节概要

  1. Quantifier Rules. The following rules hold for all open sentences P ( x ) and Q ( x ):
    x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) . x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) . [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] [ x U , ( P ( x ) Q ( x ) ) ] . [ x U , ( P ( x ) Q ( x ) ) ] [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] .
    In the last two rules, there exist open sentences for which the converse statements are false.
    量词规则。以下规则适用于所有开语句 P(x) 和 Q(x):∀x∈U,[P(x)∧Q(x)]⇔(∀y∈U,P(y))∧(∀z∈U,Q(z)).∃x∈U,[P(x)∨Q(x)]⇔(∃y∈U,P(y))∨(∃z∈U,Q(z)).[(∀y∈U,P(y))∨(∀z∈U,Q(z))]⇒[∀x∈U,(P(x)∨Q(x))].[∃x∈U,(P(x)∧Q(x))]⇒[(∃y∈U,P(y))∧(∃z∈U,Q(z))].最后两条规则中存在开放语句,其逆命题为假。
  2. Proofs Involving Multiple Quantifiers. When proving complex statements involving multiple quantifiers, process the quantifiers in the order they are encountered using the proof templates for and . To use known statements involving quantifiers, carefully apply Existential Instantiation (EI) and the Inference Rule for ALL (IRA). Be sure to distinguish quantified variables from constants and note whether constants are arbitrary or chosen by the proof writer.
    涉及多个量词的证明。证明涉及多个量词的复杂命题时,应按照量词出现的顺序,使用 ∃ 和 ∀ 的证明模板进行处理。要使用已知的包含量词的命题,请仔细应用存在实例化 (EI) 和 ALL 推理规则 (IRA)。务必区分量化变量和常量,并注明常量是任意的还是由证明编写者选择的。

Exercises

练习

  1. Is each statement true or false? Explain each answer. (a) y R , y 0 . (b) z R , z < 0 . (c) ( y R , y 0 ) ( z R , z < 0 ) . (d) x 0 ≥ 0 ⇔ x 0 < 0, where x 0 is a fixed real number. (e) x R , ( x 0 x < 0 ) . (f) For all open sentences P ( x ) and Q ( x ), x R , ( P ( x ) Q ( x ) ) iff ( y R , P ( y ) ) ( z R , Q ( z ) ) .
    以下每个陈述是正确的还是错误的?解释每个答案。 (a) ∀y∈R,y≥0. (b) ∀z∈R,z<0. (c) (∀y∈R,y≥0)⇔(∀z∈R,z<0). (d) x 0 ≥ 0 ⇔ x 0 < 0,其中 x 0 是一个固定的实数。 (e) ∀x∈R,(x≥0⇔x<0). (f) 对于所有开语句 P(x) 和 Q(x),∀x∈R,(P(x)⇔Q(x)) 当且仅当(∀y∈R,P(y))⇔(∀z∈R,Q(z)).
  2. Deduce part (b) of Theorem 2.74 from part (a) by taking the contrapositive and using the denial rules.
    通过取逆否命题并使用否定规则,从定理 2.74 的 (a) 部分推导出定理 2.74 的 (b) 部分。
  3. Use proof templates to prove part (b) of Theorem 2.73 without using part (a).
    使用证明模板证明定理 2.73 的 (b) 部分,而不使用 (a) 部分。
  4. (a) Prove ( 2.5 ) as a consequence of the analogous result for reordering universal quantifiers. (b) Prove ( 2.5 ) directly, using proof templates for existential statements and existential instantiation.
    (a) 利用全称量词重排序的类似结果证明 (2.5)。(b) 使用存在性命题和存在性实例化的证明模板直接证明 (2.5)。
  5. Prove: for all nonempty universes U and all open sentences P ( x ),
    [ x U , P ( x ) ] [ x U , P ( x ) ] .

    证明:对于所有非空全集 U 和所有开语句 P(x),[∀x∈U,P(x)]⇒[∃x∈U,P(x)].
  6. (a) Prove: for all universes U and all open sentences P ( x ) and Q ( x ),
    [ x U , ( P ( x ) Q ( x ) ) ] [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] .
    (b) Disprove: for all universes U and all open sentences P ( x ) and Q ( x ),
    [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] [ x U , ( P ( x ) Q ( x ) ) ] .

    (a) 证明:对于所有宇宙 U 和所有开语句 P(x) 和 Q(x), [∀x∈U,(P(x)⇒Q(x))]⇒[(∀y∈U,P(y))⇒(∀z∈U,Q(z))]. (b) 反证:对于所有宇宙 U 和所有开语句 P(x) 和 Q(x), [(∀y∈U,P(y))⇒(∀z∈U,Q(z))]⇒[∀x∈U,(P(x)⇒Q(x))].
  7. Use part (a) of the previous problem and other known theorems to prove:
    [ x U , ( P ( x ) Q ( x ) ) ] [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] .

    利用前一个问题的 (a) 部分和其他已知定理证明:[∀x∈U,(P(x)⇔Q(x))]⇒[(∀y∈U,P(y))⇔(∀z∈U,Q(z))].
  8. Try to prove the converse of each statement in Theorem 2.74 using the proof templates. Explain what goes wrong.
    尝试使用证明模板证明定理 2.74 中每个命题的逆命题。解释哪里出了问题。
  9. Consider the two statements: (i) x U , ( P ( x ) Q ( x ) ) (ii) ( y U , P ( y ) ) ( z U , Q ( z ) ) (a) Does (i) imply (ii) in all situations? Prove your answer. (b) Does (ii) imply (i) in all situations? Prove your answer.
    考虑以下两个命题: (i) ∃x∈U,(P(x)⊕Q(x)) (ii) (∃y∈U,P(y))⊕(∃z∈U,Q(z)) (a) (i) 是否在所有情况下都蕴含 (ii)?证明你的答案。 (b) (ii) 是否在所有情况下都蕴含 (i)?证明你的答案。
  10. Consider the two statements: (i) x U , ( P ( x ) Q ( x ) ) (ii) ( y U , P ( y ) ) ( z U , Q ( z ) ) (a) Does (i) imply (ii) in all situations? Prove your answer. (b) Does (ii) imply (i) in all situations? Prove your answer.
    考虑以下两个命题: (i) ∀x∈U,(P(x)⊕Q(x)) (ii)(∀y∈U,P(y))⊕(∀z∈U,Q(z)) (a) (i) 是否在所有情况下都蕴含 (ii)?证明你的答案。 (b) (ii) 是否在所有情况下都蕴含 (i)?证明你的答案。
  11. Consider the two statements: (i) x U , ( P ( x ) Q ( x ) ) (ii) ( y U , P ( y ) ) ( z U , Q ( z ) ) (a) Does (i) imply (ii) in all situations? Prove your answer. (b) Does (ii) imply (i) in all situations? Prove your answer.
    考虑以下两个命题: (i) ∃x∈U,(P(x)⇔Q(x)) (ii) (∃y∈U,P(y))⇔(∃z∈U,Q(z)) (a) (i) 是否在所有情况下都蕴含 (ii)?证明你的答案。 (b) (ii) 是否在所有情况下都蕴含 (i)?证明你的答案。
  12. Consider the two statements: (i) x U , ( P ( x ) Q ( x ) ) (ii) ( y U , P ( y ) ) ( z U , Q ( z ) ) (a) Does (i) imply (ii) in all situations? Prove your answer. (b) Does (ii) imply (i) in all situations? Prove your answer.
    考虑以下两个命题: (i) ∃x∈U,(P(x)⇒Q(x)) (ii) (∃y∈U,P(y))⇒(∃z∈U,Q(z)) (a) (i) 是否在所有情况下都蕴含 (ii)?证明你的答案。 (b) (ii) 是否在所有情况下都蕴含 (i)?证明你的答案。
  13. (a) Prove: for all propositions P and all open sentences Q ( x ), ( P x , Q ( x ) ) x , ( P Q ( x ) ) . (b) Disprove: for all propositions Q and all open sentences P ( x ), ( x , P ( x ) ) Q x , ( P ( x ) Q ) . (c) Can you find a way to modify the right half of (b) to produce a true statement? If so, prove your answer.
    (a) 证明:对于所有命题 P 和所有开放语句 Q(x),(P⇒∀x,Q(x))⇔∀x,(P⇒Q(x)). (b) 反证:对于所有命题 Q 和所有开放语句 P(x),(∀x,P(x))⇒Q⇔∀x,(P(x)⇒Q). (c) 你能找到一种方法来修改 (b) 的右半部分,使其成立吗?如果能,请证明你的答案。
  14. Fix an arbitrary proposition P and an arbitrary open sentence Q ( x ). Prove: (a) [ x , ( P Q ( x ) ) ] P [ y , Q ( y ) ] . (b) [ x , ( P Q ( x ) ) ] P [ y , Q ( y ) ] .
    设任意命题 P 和任意开语句 Q(x)。证明: (a) [∀x,(P∧Q(x))]⇔P∧[∀y,Q(y)]. (b) [∀x,(P∨Q(x))]⇔P∨[∀y,Q(y)].
  15. Fix an arbitrary proposition P and an arbitrary open sentence Q ( x ). Prove: (a) [ x , ( P Q ( x ) ) ] P [ y , Q ( y ) ] . (b) [ x , ( P Q ( x ) ) ] P [ y , Q ( y ) ] . Do this directly, or by using the previous exercise.
    设任意命题 P 和任意开语句 Q(x)。证明: (a) [∃x,(P∧Q(x))]⇔P∧[∃y,Q(y)]. (b) [∃x,(P∨Q(x))]⇔P∨[∃y,Q(y)]. 可以直接证明,也可以使用前面的练习。
  16. Let U be a fixed, nonempty universe. Consider the two statements: (i) [ x U , ( P Q ( x ) ) ] (ii) P [ y U , Q ( y ) ] . (a) Does (i) imply (ii) in all situations? Prove your answer. (b) Does (ii) imply (i) in all situations? Prove your answer.
    设 U 为一个固定的非空全集。考虑以下两个命题: (i) [∃x∈U,(P⊕Q(x))] (ii) P⊕[∃y∈U,Q(y)]. (a) (i) 是否在所有情况下都蕴含 (ii)?证明你的答案。 (b) (ii) 是否在所有情况下都蕴含 (i)?证明你的答案。
  17. Let U be a fixed, nonempty universe. Consider the two statements [ x U , ( P Q ( x ) ) ] (ii) P [ y U , Q ( y ) ] . (a) Does (i) imply (ii) in all situations? Prove your answer. (b) Does (ii) imply (i) in all situations? Prove your answer.
    设 U 为一个固定的非空全集。考虑以下两个命题: [∀x∈U,(P⇔Q(x))] (ii)P⇔[∀y∈U,Q(y)]. (a) (i) 是否在所有情况下都蕴含 (ii)?证明你的答案。 (b) (ii) 是否在所有情况下都蕴含 (i)?证明你的答案。
  18. Prove: for all nonempty universes U and all open sentences P ( x ), y U , ( P ( y ) x U , P ( x ) ) .
    证明:对于所有非空全集 U 和所有开语句 P(x),∃y∈U,(P(y)⇒∀x∈U,P(x))。
  19. Prove: for all nonempty universes U and all open sentences P ( x ), y U , ( P ( y ) x U , P ( x ) ) .
    证明:对于所有非空全集 U 和所有开语句 P(x),∀y∈U,(P(y)⇒∃x∈U,P(x))。
  20. Does ! x , y , P ( x , y ) always imply y , ! x , P ( x , y ) ? Explain.
    ∃!x,∀y,P(x,y) 是否总是蕴含 ∀y,∃!x,P(x,y)?请解释。
  21. Does ! x , ( P ( x ) Q ( x ) ) always imply [ ! y , P ( y ) ] [ ! z , Q ( z ) ] ? Explain.
    ∃!x,(P(x)∨Q(x)) 是否总是蕴含 [∃!y,P(y)]∨[∃!z,Q(z)]?请解释。
  22. Does [ ! y , P ( y ) ] [ ! z , Q ( z ) ] always imply ! x , ( P ( x ) Q ( x ) ) ? Explain.
    [∃!y,P(y)]∨[∃!z,Q(z)] 是否总是蕴含 ∃!x,(P(x)∨Q(x))?请解释。
  23. Does ! x , ( P ( x ) Q ( x ) ) always imply [ ! y , P ( y ) ] [ ! z , Q ( z ) ] ? Explain.
    ∃!x,(P(x)∧Q(x)) 是否总是蕴含 [∃!y,P(y)]∧[∃!z,Q(z)]?请解释。
  24. Does ! x , ! y , P ( x , y ) always imply ! y , ! x , P ( x , y ) ? Explain.
    ∃!x,∃!y,P(x,y)是否总是蕴含∃!y,∃!x,P(x,y)?请解释。
  25. Which of the answers to the previous five problems change if every ! is replaced by in the conclusions?
    如果将前五个问题的结论中的所有 ∃! 替换为 ∃,哪些答案会改变?
#

Review of Logic and Proofs

逻辑与证明回顾

Logical Symbols

逻辑符号

Complicated propositions are built up from simpler ones using the following symbols.

复杂的命题是由简单的命题用以下符号构建而成的。

Formal Expression

形式表达式

English Translation

英文翻译

P

∼P

P is not true.

P 不成立。

P Q

P∧Q

P and Q .

P 和 Q。

P Q

P∨Q

P or Q .

P 或 Q。

P Q

P ⇒ Q

If P , then Q .

如果 P,则 Q。

P Q

P ⇔ Q

P if and only if Q .

当且仅当 Q 时,P 才成立。

P Q

P⊕Q

P or Q , but not both.

P 或 Q,但不能两者兼有。

x U , P ( x )

∀x∈U,P(x)

For every x 0 in U , P ( x 0 ) is true.

对于 U 中的每个 x 0 ,P(x 0 ) 为真。

x U , P ( x )

∃x∈U,P(x)

There exists at least one x 0 in U for which P ( x 0 ) is true.

U 中至少存在一个 x 0 ,使得 P(x 0 ) 为真。

! x U , P ( x )

∃!x∈U,P(x)

There exists exactly one x 0 in U for which P ( x 0 ) is true.

U 中恰好存在一个 x 0 ,使得 P(x 0 ) 为真。

The meaning of ∼, , ∨, ⇒, ⇔, and is given by the following truth tables.

∼、∧、∨、⇒、⇔ 和 ⊕ 的含义由以下真值表给出。

NOT 不是 AND OR 或者 IF 如果 IFF 敌我识别 XOR 异或 P P Q P P P P Q P P Q P P Q P P Q P P Q T T T T F F T T T T T T T T F F T T F F F F F F T T F F F F T T F F T T T T F F T T T T F F T T F F F F T T F F F F T T T T F F

Possible translations of P Q include:

P ⇒ Q 的可能翻译包括:

if P , then Q .

如果 P,则 Q。

P implies Q .

P蕴含Q。

Q if P .

Q 如果 P。

P only if Q .

P 仅当 Q 时才成立。

P is sufficient for Q .

P 足以满足 Q。

Q is necessary for P .

Q 是 P 的必要条件。

Q whenever P .

Q 当 P 时。

(not P ) or Q .

(不是 P)或 Q。

In mathematics, OR always means inclusive-OR (∨).

在数学中,OR 总是表示包含性 OR (∨)。

Denial Rules

拒绝规则

To form a useful denial of a given statement, repeatedly apply the following rules.

要对某个陈述形成有效的反驳,请反复应用以下规则。

Statement

陈述

Denial of Statement

否认声明

Symbolic Version of Rule

规则的符号版本

A and B

A 和 B

(denial of A) or (denial of B)

(否认A)或(否认B)

( A B ) ( A ) ( B )

∼(A∧B)≡(∼A)∨(∼B)

A or B

A 或 B

(denial of A) and (denial of B)

(否认A)和(否认B)

( A B ) ( A ) ( B )

∼(A∨B)≡(∼A)∧(∼B)

if A, then B

如果 A,则 B

A and (denial of B)

A 和(否认 B)

( A B ) A ( B )

∼(A⇒B)≡A∧(∼B)

not A

不是A

A

一个

∼( ∼ A ) ≡ A

∼( ∼ A) ≡ A

For all x , P ( x )

对于所有 x,P(x)

There is x , (denial of P ( x ))

存在 x,(否定 P(x))

x U , P ( x ) iff x U , P ( x )

∼∀x∈U,P(x) 当且仅当 ∃x∈U,∼P(x)

There is x , P ( x )

存在 x,P(x)

For all x , (denial of P ( x ))

对于所有 x,(否定 P(x))

x U , P ( x ) iff x U , P ( x )

∼∃x∈U,P(x) 当且仅当 ∀x∈U,∼P(x)

A iff B

A 当且仅当 B

A or B, not both

A 或 B,两者不可兼得

( A B ) ( A B )

∼(A⇔B)≡(A⊕B)

A or B, not both

A 或 B,两者不可兼得

A iff B

A 当且仅当 B

( A B ) ( A B )

∼(A⊕B)≡(A⇔B)

Note: the useful denial of an IF-statement is an AND-statement, not an IF-statement .

注意:对 IF 语句的有效否定是 AND 语句,而不是 IF 语句。

Example. A useful denial of “ ϵ > 0 , δ > 0 , x R , ( 2 δ < x < 2 + δ ) ( 4 ϵ < x 2 < 4 + ϵ ) ” is “ ϵ > 0 , δ > 0 , x R , ( 2 δ < x < 2 + δ ) and ( 4 ϵ x 2 or x 2 4 + ϵ ) .

例子。 “∀ϵ>0,∃δ>0,∀xεR,(2−δ<x<2+δ)⇒(4−ϵ<x2<4+ϵ)”的有用否定是“∃ϵ>0,∀δ>0,∀xεR,(2−δ<x<2+δ) and (4−ϵ≥x2 or x2≥4+ϵ)”。

Propositional Logic

命题逻辑

  1. Filling in Truth Tables. Make a column for each subexpression of the propositional form. Fill in each column using the defining truth tables for the logical operators (given above). Be sure to refer to the correct previous columns. Recall NOT flips true and false; AND outputs true precisely when both inputs are true; OR outputs false precisely when both inputs are false; IF (i.e., A B ) outputs false precisely when A is true and B is false; IFF outputs true precisely when the inputs agree; XOR outputs true precisely when the inputs disagree.
    填写真值表。为命题形式的每个子表达式创建一列。使用上面给出的逻辑运算符真值表填写每一列。务必参考前面正确的列。回想一下,非 (NOT) 会交换真假;与 (AND) 当且仅当两个输入都为真时输出真;或 (OR) 当且仅当两个输入都为假时输出假;如果 (IF)(即 A ⇒ B)当且仅当 A 为真且 B 为假时输出假;如果 (IFF) 当且仅当两个输入都为真时输出真;异或 (XOR) 当且仅当两个输入都为假时输出真。
  2. Logical Equivalence. Two propositional forms P and Q are logically equivalent , written P Q , iff all rows of the truth tables for P and Q match. Some frequently used logical equivalences appear in the Theorem on Logical Equivalences, the Theorem on IF, and the Theorem on IFF.
    逻辑等价性。两个命题形式 P 和 Q 逻辑等价,记作 P ≡ Q,当且仅当 P 和 Q 的真值表所有行都相等。一些常用的逻辑等价性出现在逻辑等价定理、如果成立定理和相反当且仅当定理中。
  3. Tautologies and Contradictions. A propositional form is a tautology iff all rows of its truth table have output true. A propositional form is a contradiction iff all rows of its truth table have output false. Every proposition is true or false; but most propositional forms are neither tautologies nor contradictions.
    重言式与矛盾式。 一个命题形式是重言式当且仅当其真值表的每一行输出都为真。一个命题形式是矛盾式当且仅当其真值表的每一行输出都为假。每个命题要么为真,要么为假;但大多数命题形式既不是重言式也不是矛盾式。
  4. Terminology for IF-statements. (a) A B is called a conditional with hypothesis A and conclusion B . (b) The converse of A B is B A . (c) The contrapositive of A B is ( ∼ B ) ⇒ ( ∼ A ). (d) A denial of A B is A ( B ) . (e) The inverse of A B is ( ∼ A ) ⇒ ( ∼ B ). (f) “If A then B ” is logically equivalent to its contrapositive. (g) “If A then B ” is not logically equivalent to its converse, its inverse, or its negation (in general).
    IF语句的术语。 (a) A ⇒ B 称为以 A 为假设、B 为结论的条件语句。 (b) A ⇒ B 的逆命题是 B ⇒ A。 (c) A ⇒ B 的逆否命题是 (∼ B) ⇒ (∼ A)。 (d) A ⇒ B 的否定命题是 A∧(∼B)。 (e) A ⇒ B 的逆命题是 (∼ A) ⇒ (∼ B)。 (f) “如果 A 则 B” 与其逆否命题逻辑等价。 (g) “如果 A 则 B” 与其逆命题、逆命题或否定命题(一般而言)逻辑不等价。
  5. Eliminating ⇒, ⇔, and . Use these logical equivalences to rewrite IF, IFF, and XOR in terms of other logical symbols: (a) A B ≡ (∼ A )∨ B . (b) A B ( A B ) ( B A ) ( A B ) ( ( A ) ( B ) ) . (c) A B ( A ( B ) ) ( ( A ) B ) ( A B ) ( ( A B ) ) .
    去掉 ⇒、⇔ 和 ⊕。利用这些逻辑等价关系,用其他逻辑符号重写 IF、IFF 和 XOR: (a) A ⇒ B ≡ (∼ A)∨B. (b) A⇔B≡(A⇒B)∧(B⇒A)≡(A∧B)∨((∼A)∧(∼B)). (c) A⊕B≡(A∧(∼B))∨((∼A)∧B)≡(A∨B)∧(∼(A∧B)).
  6. Inference Rules for IF. If P Q is a known theorem and P is a known theorem, you can deduce Q as a new theorem. If P Q is a known theorem and ∼ Q is a known theorem, you can deduce ∼ P as a new theorem.
    IF 推理规则。如果 P ⇒ Q 是一个已知定理,且 P 本身也是一个已知定理,则可以推导出 Q 是一个新定理。如果 P ⇒ Q 是一个已知定理,且 ∼Q 也是一个已知定理,则可以推导出 ∼P 是一个新定理。

Quantifiers

量词

Let U be a fixed universe (nonempty set of objects).

设 U 为一个固定的宇宙(非空对象集合)。

  1. Restricted and Unrestricted Quantifiers. For all open sentences P ( x ),

    受限量词和非受限量词。对于所有开放语句 P(x),

    x U , P ( x ) iff x , ( x U P ( x ) ) ;

    ∀x∈U,P(x) 当且仅当 ∀x,(x∈U⇒P(x));

    x U , P ( x ) iff x , ( x U P ( x ) ) .

    ∃x∈U,P(x) 当且仅当 ∃x,(x∈U∧P(x))。

  2. Quantifiers Rules and Pitfalls. For all open sentences P ( x ), Q ( x ), and R ( x , y ):
    (a) x , P ( x ) v , P ( v ) where v is a new letter. (b) x , P ( x ) v , P ( v ) where v is a new letter. (c) x , y , R ( x , y ) y , x , R ( x , y ) . (d) x , y , R ( x , y ) y , x , R ( x , y ) . (e) x , y , R ( x , y ) y , x , R ( x , y ) . (f) x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) . (g) x U , [ P ( x ) Q ( x ) ] ( y U , P ( y ) ) ( z U , Q ( z ) ) . (h) [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] [ x U , ( P ( x ) Q ( x ) ) ] . (i) [ x U , ( P ( x ) Q ( x ) ) ] [ ( y U , P ( y ) ) ( z U , Q ( z ) ) ] .
    The right side of (e) says for all y , there exists an x (which can be chosen differently for different y ) such that R ( x , y ) is true. The left side of (e) is the stronger statement that there is one choice of x (the same choice for all y ) such that R ( x , y ) is true. To see that (h) cannot be strengthened to an IFF-statement, let U be the set of humans, let P ( x ) mean “ x is male,” and let Q ( x ) mean “ x is female.” To see that (i) cannot be strengthened to an IFF-statement, let U = R , let P ( x ) mean “ x < 0,” and let Q ( x ) mean “ x > 0.”
    量词规则和陷阱。对于所有开放语句 P(x)、Q(x) 和 R(x, y):(a)∀x,P(x)⇔∀v,P(v),其中 v 是一个新的字母。(b)∃x,P(x)⇔∃v,P(v),其中 v 是一个新的字母。字母.(c)∀x,∀y,R(x,y)⇔∀y,∀x,R(x,y).(d)∃x,∃y,R(x,y)⇔∃y,∃x,R(x,y).(e)∃x,∀y,R(x,y)⇒∀y,∃x,R(x,y).(f)∀x∈U,[P(x)∧Q(x)]⇔(∀y∈U,P(y))∧(∀z∈U,Q(z)).( g)∃x∈U,[P(x)∨Q(x)]⇔(∃y∈U,P(y))∨(∃z∈U,Q(z)).(h)[(∀y∈U,P(y))∨(∀z∈U,Q(z))]⇒[∀x∈U,(P(x)∨Q(x))].(i)[∃x∈U,(P(x)∧Q(x))]⇒[(∃y∈U,P(y))∧(∃z∈U,Q(z))].(e) 的右侧表示对于所有 y,存在一个 x(对于不同的 y 可以有不同的 x 值),使得 R(x, y) 为真。 (e) 的左侧是一个更强的命题,即存在一个 x 的选择(对于所有 y 都相同),使得 R(x, y) 为真。为了证明 (h) 不能加强为一个 IFF 命题,令 U 为人类集合,令 P(x) 表示“x 是男性”,令 Q(x) 表示“x 是女性”。为了证明 (i) 不能加强为一个 IFF 命题,令 U=R,令 P(x) 表示“x < 0”,令 Q(x) 表示“x > 0”。
  3. Contrasting Translations. Let B ( x ) mean x is black; let C ( x ) mean x is a crow. “No crows are black” translates to x , C ( x ) ⇒∼ B ( x ) . “Not all crows are black” translates to x , C ( x ) B ( x ) . We can transform this translation to x , C ( x ) B ( x ) . “All crows are not black” translates to x , C ( x ) ⇒∼ B ( x ) . “Black crows do not exist” translates to x , C ( x ) B ( x ) . We can transform this translation to x , C ( x ) ⇒∼ B ( x ) .
    对比翻译。设 B(x) 表示 x 为黑色;令 C(x) 表示 x 是乌鸦。 “没有乌鸦是黑色的”可翻译为 ∀x,C(x)⇒∼B(x)。 “并非所有乌鸦都是黑色的”可翻译为 ∼∀x,C(x)⇒B(x)。 我们可以将此翻译转换为 ∃x,C(x)∧∼B(x)。 “并非所有乌鸦都是黑色的”可翻译为 ∀x,C(x)⇒∼B(x)。 “不存在黑色的乌鸦”可翻译为 ∼∃x,C(x)∧B(x)。 我们可以将此翻译转换为 ∀x,C(x)⇒∼B(x)。
  4. Inference Rule for ALL. If “ x U , P ( x ) ” is a known theorem and “ c U ” is a known theorem (where c is any expression), then you can deduce the new theorem P ( c ).
    对所有情况的推理规则。如果“∀x∈U,P(x)”是一个已知定理,并且“c ∈ U”是一个已知定理(其中c是任意表达式),那么你可以推导出新定理P(c)。
  5. Multiple Quantifier Example. To prove x , y , z , P ( x , y , z ) : fix x 0 ; choose an appropriate y 0 (which may depend on x 0 ); fix z 0 ; prove P ( x 0 , y 0 , z 0 ). To disprove x , y , z , P ( x , y , z ) , prove x , y , z , P ( x , y , z ) as follows. Choose an appropriate specific object x 0 ; fix y 0 ; choose an appropriate z 0 (which may depend on x 0 and y 0 ); prove ∼ P ( x 0 , y 0 , z 0 ).
    多重量词示例。要证明 ∀x,∃y,∀z,P(x,y,z):固定 x 0 ;选择一个合适的 y 0 (它可以依赖于 x 0 );固定 z 0 ;证明 P(x 0 , y 0 , z 0 )。要反证 ∀x,∃y,∀z,P(x,y,z),证明 ∃x,∀y,∃z,∼P(x,y,z) 的方法如下:选择一个合适的具体对象 x 0 ;固定 y 0 ;选择合适的 z 0 (可能取决于 x 0 和 y 0 );证明 ∼P(x 0 , y 0 , z 0 )。
  6. Rule to Eliminate the Uniqueness Symbol.
    消除唯一性符号的规则。
    ! x x U U , P P ( x x ) ( x x U U , P P ( x x ) ) x x U U , y U U , [ ( P P ( x x ) P P ( y ) ) x x = y ] .

Proof Templates

校对模板

General Comments:

一般性意见:

(a) Memorize the proof templates! Practice the proof templates!

(a)记住证明模板!练习证明模板!

(b) To decide which template to apply at each stage, look for the outermost logical operator in the current statement to be proved.

(b)为了决定在每个阶段应用哪个模板,查找要证明的当前语句中最外层的逻辑运算符。

(c) In each template, the numbered statements (1), (2), ... should be explicitly written down at the appropriate point in the proof.

(c)在每个模板中,编号的语句(1)、(2)、……应该在证明的适当位置明确写下来。

(d) After writing these statements, continue the proof by either using more proof templates or expanding definitions in what you have just written.

(d)写完这些陈述后,继续进行证明,可以使用更多的证明模板,或者扩展你刚刚写的定义。

(e) Remember to introduce each sentence, formula, and new variable with logical status words (“assume,” “we must prove,” “we know,” “choose,” etc.).

(e)记住要用逻辑状态词(例如“假设”、“我们必须证明”、“我们知道”、“选择”等)来引入每个句子、公式和新变量。

(f) Do not reuse a variable to mean two things.

(f)不要重复使用一个变量来表示两个意思。

(g) Do not deduce consequences from an equation (or statement) that is yet to be proved. Instead, work forward from known and assumed statements to gather information leading toward the statement to be proved.

(g) 不要从尚未证明的等式(或陈述)中推断结论。相反,应从已知和假设的陈述出发,逐步收集信息,最终得出待证明的陈述。

  1. Direct Proof of “if P then Q .” (1) Assume P . (2) Prove Q .
    “如果 P 则 Q”的直接证明。(1) 假设 P。(2) 证明 Q。
  2. Contrapositive Proof of “if P then Q .” (1) Assume ∼ Q . (2) Prove ∼ P . [Use the denial rules to find useful denials of P and Q here.]
    “如果 P 则 Q”的逆否命题证明。(1) 假设 ∼Q。(2) 证明 ∼P。[使用否定规则在此处找到 P 和 Q 的有用否定。]
  3. Contradiction Proof of “if P then Q .” (1) Assume, to get a contradiction, “ P and ∼ Q .” (2) Deduce a contradiction.
    “如果 P 则 Q”的矛盾证明。(1) 为了得到矛盾,假设“P 且 ∼Q”。(2) 推导出矛盾。
  4. Contradiction Proof of Any Statement P . (1) Assume, to get a contradiction, ∼ P [write a useful denial of P here]. (2) Use this assumption, definitions, and known results to produce a contradiction (such as S ( S )). (3) Conclude that P is true.
    矛盾证明任意命题 P。(1) 假设 ∼P 以得出矛盾 [在此处写出 P 的一个有用的否定]。(2) 利用此假设、定义和已知结果得出矛盾(例如 S∧(∼S))。(3) 得出结论,P 为真。
  5. Two-Part Proof of “ P iff Q .” (1) Prove “if P then Q ” by any method. (2) Prove “if Q then P ” by any method.
    “P 当且仅当 Q”的两步证明。(1) 用任意方法证明“如果 P 则 Q”。(2) 用任意方法证明“如果 Q 则 P”。
  6. Constructive Proof of x U , P ( x ) . (1) Describe a specific object x 0 that will make P ( x ) true (we often say: “choose x 0 = ...”). (2) Prove x 0 U . (3) Prove P ( x 0 ). Note x 0 is allowed to depend on constants introduced earlier in the proof.
    证明存在 x∈U,P(x)。(1) 描述一个特定的对象 x 0 ,使得 P(x) 为真(我们通常说:“选择 x 0 = ...”)。(2) 证明 x 0 ∈ U。(3) 证明 P(x 0 )。注意 x 0 可以依赖于证明中前面引入的常数。
  7. Generic-Element Proof of x U , P ( x ) . (1) Fix an arbitrary object x 0 U . (2) Prove P ( x 0 ). (It is not correct to choose a specific object in U here; step (2) must work for generic elements of U .)
    ∀x∈U,P(x) 的一般元素证明。(1) 固定任意对象 x 0 ∈ U。(2) 证明 P(x 0 )。(此处选择 U 中的特定对象是不正确的;步骤 (2) 必须适用于 U 的一般元素。)
  8. Proof by Cases. To prove Q when you already know P 1 P 2 ∨ · · · ∨ P k : (0) Say: “We know P 1 or P 2 or ... or P k , so use cases.”

    (1) Case 1. Assume P 1 . Prove Q .

    (2) Case 2. Assume P 2 . Prove Q .

    ...

    ( k ) Case k . Assume P k . Prove Q . (Thus Q must be proved k times, where there is a different assumption each time.)


    案例证明法。为了证明 Q,已知 P 1 ∨P 2 ∨ · · · ∨P k :(0) 说明:“我们知道 P 1 或 P 2 或 ... 或 P k ,所以使用案例证明法。” (1) 案例 1:假设 P 1 。证明 Q。(2) 案例 2:假设 P 2 。证明 Q。... (k) 案例 k:假设 P k 。证明 Q。(因此,Q 需要证明 k 次,每次的假设都不同。)
  9. Proving P Q . (1) Prove P . (2) Prove Q .
    证明 P∧Q。(1)证明 P。(2)证明 Q。
  10. Proving P Q . (1) Assume ∼ P . (2) Prove Q .
    证明 P∨Q。(1)假设 ∼P。(2)证明 Q。
  11. Proving P Q . Prove P ⇔ ( ∼ Q ).
    证明 P⊕Q。证明 P ⇔ ( ∼ Q)。
  12. Proving ∼ P . (1) Use the denial rules to find a useful denial of P not starting with NOT. (2) Prove the denial found in (1).
    证明 ∼P。(1)利用否定规则找到一个不以 NOT 开头的 P 的有用否定。(2)证明(1)中找到的否定。
  13. Disproving P when P is False. Prove ∼ P (see previous item).
    当 P 为假时,如何反驳 P?证明 ∼P(参见前一项)。
  14. Proof by Exhaustion. If U = { z 1 , …, z n } is a finite set, you can prove x U , P ( x ) by proving “ P ( z 1 ) and P ( z 2 ) and ... and P ( z n ).”
    穷举法证明。如果 U = {z 1 , …, z n 是一个有限集,你可以通过证明“P(z 1 ) 且 P(z 2 ) 且 ... 且 P(z n )”来证明 ∀x∈U,P(x)。

Review Problems

复习题

  1. Complete the following definitions: (a) for a fixed integer c , c divides c 2 + 7 iff... (b) A real number r is rational iff... (c) A propositional form A is a contradiction iff...
    完成下列定义:(a)对于固定的整数 c,c 整除 c 2 + 7 当且仅当……(b)实数 r 是有理数当且仅当……(c)命题形式 A 是矛盾式当且仅当……
  2. Give a truth table for the propositional form ( ( P ) Q ) ( P Q ) . Show all columns. Is this a tautology?
    为命题形式 ((∼P)∧Q)⇔(P∨Q) 写出真值表。列出所有列。这是一个重言式吗?
  3. Write seven different English phrases that could be used to translate P Q .
    写出七个可以用来翻译 P ⇒ Q 的不同英语短语。
  4. Let Q be the statement “If X is compact, then: X is separable or X is not closed.” (a) Write the converse of Q . (b) Write the contrapositive of Q . (c) Outline a direct proof of Q . (d) Outline a proof of Q by contradiction.
    设 Q 为命题“如果 X 是紧集,则:X 是可分集或 X 不是闭集”。(a)写出 Q 的逆命题。(b)写出 Q 的逆否命题。(c)简述 Q 的直接证明。(d)简述 Q 的反证法证明。
  5. True or false? (a) 7 is odd or 7 is odd. (b) 1 + 1 = 2 if 0 = 1. (c) 0 = 0 whenever 0 ≠ 0. (d) 5 Z implies 0.37 Z . (e) 0.5 is rational only if 8 is odd. (f) 2 R is a necessary condition for 2 Z . (g) 2 is odd iff 7 is even. (h) ! x R , ( x 3 = x ) . (i) A useful denial of “4 < x y ” is “4 ≥ x > y .” (j) Every propositional form is either a tautology or a contradiction. (k) The propositional form P P is a contradiction. (l) ( P Q ) R is logically equivalent to P ( Q R ) . (m) Given that P Q is a known theorem and Q is a known theorem, we may always deduce P as a new theorem. (n) One way to prove P Q is to assume ∼ P and prove Q .
    判断题? (a) 7 是奇数或 7 是奇数。 (b) 若 0 = 1,则 1 + 1 = 2。 (c) 若 0 ≠ 0,则 0 = 0。 (d) 若 5∈Z,则 0.37∈Z。 (e) 若 8 是奇数,则 0.5 是有理数。 (f) 若 2∈R,则 2∈Z 的必要条件。 (g) 若 7 是偶数,则 2 是奇数。 (h) 若 x∈R,则存在 x³=x。 (i) 否定“4 < x ≤ y”的一个有用反例是“4 ≥ x > y。” (j) 每个命题形式要么是重言式,要么是矛盾式。 (k) 命题形式 P⊕P 是矛盾式。 (l) (P∧Q)∨R 逻辑等价于 P∧(Q∨R)。 (m) 已知 P ⇒ Q 是一个已知定理,Q 也是一个已知定理,我们总能推导出 P 是一个新定理。 (n) 证明 P⊕Q 的一种方法是假设 ∼P 并证明 Q。
  6. (a) Make truth tables for ( P Q ) ⇒ P and P ⇔ ( Q P ). (b) Are the two forms in (a) logically equivalent? Explain briefly. (c) Is either form in (a) a tautology? Explain briefly.
    (a) 为 (P ⇔ Q) ⇒ P 和 P ⇔ (Q ⇒ P) 分别写出真值表。 (b) (a) 中的两种形式逻辑等价吗?简要解释。 (c) (a) 中的两种形式中哪一种是重言式?简要解释。
  7. Let Q be the statement “If ( X is compact and Hausdorff) or X is metrizable, then X is normal.” (a) Write the converse of Q . (b) Write the contrapositive of Q . (c) Outline a direct proof of Q .
    设 Q 为命题“如果 (X 是紧集且 Hausdorff 集) 或 X 是可度量集,则 X 是正规集。” (a) 写出 Q 的逆命题。 (b) 写出 Q 的逆否命题。 (c) 简述 Q 的直接证明。
  8. Consider the statement: ( A B ) ( C D ) . (a) Outline a direct proof of this statement. (b) Outline a contrapositive proof of this statement. (c) Outline a proof by contradiction of this statement.
    考虑命题:(A∧B)⇒(C⊕D)。 (a) 简述该命题的直接证明。 (b) 简述该命题的逆否命题证明。 (c) 简述该命题的反证法证明。
  9. Let W ( x ) mean “ x is white,” R ( x ) mean “ x is a rose,” and P ( x ) mean “ x is pretty.” (a) Using this notation, logical connectives, and unrestricted quantifiers, write this statement in symbolic form: “Although all white roses are pretty, some pretty roses are not white.” (b) Give a useful denial of the previous statement. Do not use symbols in your answer.
    设 W(x) 表示“x 是白色的”,R(x) 表示“x 是一朵玫瑰”,P(x) 表示“x 很漂亮”。(a)使用此符号、逻辑连接词和无限制量词,将以下陈述用符号形式写出:“虽然所有的白玫瑰都很漂亮,但有些漂亮的玫瑰不是白色的。”(b)对上述陈述给出一个有意义的否定。答案中不要使用符号。
  10. Prove: for all x , y R , if x Q and y Q and y ≠ 0, then x / y Q .
    证明:对于所有 x,y∈R,如果 x∈Q 且 y∈Q 且 y≠0,则 x/y∈Q。
  11. (a) Prove: For all integers x , if x 2 + 4 is even, then x is even. (b) Is the converse of (a) true?
    (a) 证明:对于所有整数 x,如果 x 2 + 4 是偶数,则 x 是偶数。(b) (a) 的逆命题是否成立?
  12. Prove: for all a , b , c , r , s Z , if a divides b and a divides c , then a divides rb + sc .
    证明:对于所有 a,b,c,r,s∈Z,如果 a 整除 b 且 a 整除 c,则 a 整除 rb + sc。
  13. Outline a direct proof of this statement, assuming P , a , b are fixed objects: “if P is a prime ideal, then ab P iff a P or b P .”
    假设 P、a、b 是固定的对象,请概述以下陈述的直接证明: “如果 P 是一个素理想,那么 ab ∈ P 当且仅当 a ∈ P 或 b ∈ P。”
  14. (a) Prove: x R , y R , y > x + 2 . (b) Disprove: y R , x R , y > x + 2 .
    (a)证明:∀x∈R,∃y∈R,y>x+2。(b)反证:∃y∈R,∀x∈R,y>x+2。
  15. (a) Prove: x R > 0 , y R > 0 , x < y < 2 x . (b) Disprove: y R > 0 , x R > 0 , x < y < 2 x .
    (a)证明:∀x∈R>0,∃y∈R>0,x<y<2x。(b)反证:∃y∈R>0,∀x∈R>0,x<y<2x。
  16. Prove: for all r R , r is rational iff 3 r + 5 is rational.
    证明:对于所有 r∈R,r 是有理数当且仅当 3r + 5 是有理数。
  17. Disprove: for all a , b , c , r , s Z , if a divides rb + sc , then a divides b or a divides c or a divides r or a divides s .
    反证:对于所有 a、b、c、r、s∈Z,如果 a 整除 rb + sc,则 a 整除 b 或 a 整除 c 或 a 整除 r 或 a 整除 s。
  18. (a) Prove: for all a , b , c Z , if a divides b or a divides c , then a divides bc 2 . (b) Prove or disprove the converse of (a).
    (a)证明:对于所有 a、b、c∈Z,如果 a 整除 b 或 a 整除 c,则 a 整除 bc。(b)证明或反证(a)的逆命题。
  19. Show that P ( Q R ) is logically equivalent to ( P Q ) ( P R ) using known logical equivalences, not truth tables.
    使用已知的逻辑等价关系,而不是真值表,证明 P⇒(Q∧R) 与 (P⇒Q)∧(P⇒R) 在逻辑上等价。
  20. Give a proof by contradiction of this statement: For all b , c Z , if bc 2 is even, then b is even or c is even.
    用反证法证明以下陈述: 对于所有 b,c∈Z,如果 bc 2 是偶数,则 b 是偶数或 c 是偶数。
#

3

Sets

3.1  Set Operations and Subset Proofs

3.1 集合运算和子集证明

At this point, we have introduced nearly all of the fundamental rules for generating proofs of logical statements. (The main rule yet to be discussed is mathematical induction, which we cover in Chapter 4 .) Armed with our knowledge of logic and proofs, we can now proceed to study whatever mathematical theories we are interested in: set theory, number theory, graph theory, abstract algebra, analysis, probability theory, complexity theory, and so on. Surprisingly, the theory of sets turns out to be a foundational subject on which every other branch of mathematics can be built. Many aspects of set theory relate very closely to the propositional logic and quantifier logic that we have just studied. So, we continue our journey up the tower of mathematics by examining elementary set theory.

至此,我们已经介绍了几乎所有逻辑命题证明的基本规则。(尚未讨论的主要规则是数学归纳法,我们将在第四章中介绍。)掌握了逻辑和证明的知识后,我们现在可以继续学习我们感兴趣的任何数学理论:集合论、数论、图论、抽象代数、分析学、概率论、复杂性理论等等。令人惊讶的是,集合论竟然是所有其他数学分支的基础学科。集合论的许多方面都与我们刚刚学习的命题逻辑和量词逻辑密切相关。因此,我们将继续攀登数学之塔,探索初等集合论。

Informal Introduction to Set Theory

集合论的非正式介绍

Most readers of this text have already seen some aspects of set theory at an intuitive level. We begin by introducing the basic ideas intuitively, to motivate the formal definitions given below. Informally, a set is any collection of objects. For example, { a , b , c , d } is a set of four letters; ℤ is the set of all integers; [0, 3] is the set of all real numbers x satisfying 0 ≤ x ≤ 3, and so on. If x is an object and S is a set, we write x S to mean that x is a member of S ; x S means that x is not a member of S . For example, 3 ∈ ℤ, b ∈ { a , b , c , d }, 5 [ 0 , 3 ] , and so on. (We already used this notation in our discussion of restricted quantifiers.) The symbol x S can be read aloud as “ x is a member of S ” or “ x belongs to S ” or “ x is in S .” Beginners sometimes read the symbol x S as “ x exists in S ,” but this phrase should not be used, as it suggests an existential quantifier that is not present.

本文的大多数读者可能已经对集合论的一些方面有了直观的了解。我们首先以直观的方式介绍基本概念,为下文给出的正式定义提供参考。通俗地说,集合是任何对象的集合。例如,{a, b, c, d} 是四个字母的集合;ℤ 是所有整数的集合;[0, 3] 是所有满足 0 ≤ x ≤ 3 的实数 x 的集合,以此类推。如果 x 是一个对象,S 是一个集合,我们记 x ∈ S 表示 x 是 S 的成员;x∉S 表示 x 不是 S 的成员。例如,3 ∈ ℤ,b ∈ {a, b, c, d},5∉[0,3],等等。 (我们在讨论受限量词时已经使用过这种符号。)符号 x ∈ S 可以读作“x 是 S 的成员”、“x 属于 S”或“x 在 S 中”。初学者有时会将符号 x ∈ S 读作“x 存在于 S 中”,但这种说法是错误的,因为它暗示了一个并不存在的存在量词。

The members of a set can be any objects whatsoever, including other sets. For example, {0, 1, 2, 3, ℤ, [0, 3]} is a set with six members: the four integers 0, 1, 2, and 3; the set ℤ of all integers; and the closed interval [0, 3], which is a set of real numbers. The following diagram, called a Venn diagram , shows two geometric sets A and B ; the members of A are the points on this page lying within the left circle.

集合的成员可以是任何对象,包括其他集合。例如,{0, 1, 2, 3, ℤ, [0, 3]} 是一个包含六个成员的集合:四个整数 0、1、2 和 3;所有整数构成的集合 ℤ;以及闭区间 [0, 3],它是一个实数集。下图(称为维恩图)展示了两个几何集合 A 和 B;集合 A 的成员是本页左侧圆圈内的点。

ufig3_1.jpg

We can combine sets to form new sets using various set operations. Let A and B be any two sets. The union of A and B , denoted A B , is formed by lumping together all the members of A and B into one big set. Symbolically, A B = { x : x A or x B } . For example, { 1 , 4 , 5 , 7 } { 2 , 3 , 5 , 7 , 11 } = { 1 , 2 , 3 , 4 , 5 , 7 , 11 } . For A and B in the Venn diagram above, A B is the shaded region below consisting of all points lying within at least one of the two circles:

我们可以使用各种集合运算将集合组合成新的集合。设 A 和 B 为任意两个集合。A 和 B 的并集,记作 A ∪ B,是将 A 和 B 中的所有元素合并成一个大集合。符号表示为 A ∪ B = {x: x∈A 或 x∈B}。例如,{1,4,5,7} ∪ {2,3,5,7,11} = {1,2,3,4,5,7,11}。对于上图中的 A 和 B,A ∪ B 是下图阴影区域,该区域包含所有位于至少一个圆内的点:

ufig3_2.jpg

The intersection of A and B , denoted A B , consists of all objects that are members of both A and B . Symbolically, A B = { x : x A and x B } . Continuing our previous example, { 1 , 4 , 5 , 7 } { 2 , 3 , 5 , 7 , 11 } = { 5 , 7 } . The Venn diagram for A B looks like this:

集合 A 和 B 的交集,记作 A ∩ B,包含所有同时属于 A 和 B 的对象。符号表示为 A∩B={x:x∈A 且 x∈B}。继续之前的例子,{1,4,5,7}∩{2,3,5,7,11}={5,7}。A ∩ B 的维恩图如下所示:

ufig3_3.jpg

The set difference A B is the set of all members of A that are not members of B . Symbolically, A B = { x : x A and x B } . For example, {1, 4, 5, 7} − {2, 3, 5, 7, 11} = {1, 4} and {2, 3, 5, 7, 11} − {1, 4, 5, 7} = {2, 3, 11}. Here is the Venn diagram for A B :

集合差集 A − B 是集合 A 中所有不属于集合 B 的元素构成的集合。符号表示为 A − B = {x: x∈A 且 x∉B}。例如,{1, 4, 5, 7} − {2, 3, 5, 7, 11} = {1, 4},{2, 3, 5, 7, 11} − {1, 4, 5, 7} = {2, 3, 11}。下图是 A − B 的维恩图:

ufig3_4.jpg

We say that A is a subset of B , written A B , to mean that every member of A is also a member of B . The notation A B means that A is not a subset of B . For example, { 1 , 4 , 5 , 7 } { 2 , 3 , 5 , 7 , 11 } since 1 is a member of {1, 4, 5, 7} that is not a member of {2, 3, 5, 7, 11}. On the other hand, {1, 4, 5, 7}⊆{1, 2, 3, 4, 5, 6, 7}, {1, 4, 5, 7} ⊆ ℤ, and ℤ ⊆ ℚ ⊆ ℝ. In the Venn diagrams displayed above, A B and B A , since each of the two circles contains some points not in the other circle. In the following Venn diagram, C D but D C :

我们说 A 是 B 的子集,记作 A ⊆ B,表示 A 中的每个元素也都是 B 中的元素。记作 A ⊈ B 表示 A 不是 B 的子集。例如,{1, 4, 5, 7} ⊈ {2, 3, 5, 7, 11},因为 1 是 {1, 4, 5, 7} 中的元素,而 1 不是 {2, 3, 5, 7, 11} 中的元素。另一方面,{1, 4, 5, 7} ⊆ {1, 2, 3, 4, 5, 6, 7},{1, 4, 5, 7} ⊆ ℤ,且 ℤ ⊆ ℚ ⊆ ℝ。在上面的维恩图中,A ⊈ B 且 B ⊈ A,因为两个圆中都包含一些不在另一个圆中的点。在以下维恩图中,C ⊆ D 但 D⊈C:

ufig3_5.jpg

It is possible for a set to have no members whatsoever. Such a set is called an empty set , and it is denoted by the symbol ∅ or {} (a set of curly braces with nothing inside). Be warned that these two notations for the empty set must not be combined: {∅} is a set with one member (namely, the empty set), so {∅} is not the same set as the empty set. We give a more detailed discussion of this point later, after introducing the appropriate definitions.

一个集合可以不包含任何元素。这样的集合称为空集,可以用符号 ∅ 或 {}(一个空的大括号)表示。需要注意的是,这两种表示空集的符号不能混用:{∅} 是一个只有一个元素的集合(即空集),因此 {∅} 与空集并非同一个集合。我们将在介绍相关定义之后,对此进行更详细的讨论。

Formal Definitions of Subsets and Set Operations

子集和集合运算的形式化定义

We now begin afresh with a formal development of set theory, which is motivated by the informal discussion in the previous section. In our theory, the following words and symbols are undefined terms (see § 2.1 ): set, membership in a set, x S . Every other concept in set theory will be defined from these undefined terms and the logical operators discussed earlier. We begin with the formal definition of subsets.

现在,我们重新开始正式阐述集合论,这源于上一节的非正式讨论。在我们的理论中,以下词语和符号是未定义术语(参见§2.1):集合、属于集合、x ∈ S。集合论中的所有其他概念都将由这些未定义术语和前面讨论过的逻辑运算符定义。我们首先给出子集的正式定义。

3.1. Definition: Subsets. For any sets A and B , A B iff x , x A x B .

3.1. 定义:子集。对于任意集合 A 和 B,A⊆B 当且仅当 ∀x,x∈A⇒x∈B。

The sentence in this definition is a new definitional axiom that lets us replace the newly defined symbol A B (“ A is a subset of B ”) by the definition text appearing after the word “iff,” and vice versa. Since P Q is equivalent to ( ∼ P ) ⇔ ( ∼ Q ), we can obtain a definition of A B by applying the denial rules. We get:

此定义中的句子是一个新的定义公理,它允许我们将新定义的符号 A ⊆ B(“A 是 B 的子集”)替换为“当且仅当”之后的定义文本,反之亦然。由于 P ⇔ Q 等价于 (∼ P) ⇔ (∼ Q),我们可以通过应用否定规则得到 A⊈B 的定义。我们得到:

For any sets A and B , A B iff x , x A x B .

对于任意集合 A 和 B,A⊈B 当且仅当存在 x,x∈A∧x∉B。

Similarly, we can deny both sides of all definitions given below to find out what the denial of a newly defined term means.

同样地,我们可以否定下面给出的所有定义的正反两面,以找出否定一个新定义的术语意味着什么。

The following variants of the subset notation are sometimes needed. For all sets A and B , we say A is a proper subset of B , written A B , iff A B and A B and A B . Do not confuse this notation with A B , which means that A is not a subset of B . We can also reverse the order of A and B . We say B is a superset of A , written B A , iff A B . Similarly, B is a proper superset of A , written B A , iff A B .

有时需要用到以下子集符号的变体。对于任意集合 A 和 B,我们称 A 是 B 的真子集,记作 A⊊B,当且仅当 A⊆B 且 A≠B。不要将此符号与 A⊈B 混淆,A⊈B 表示 A 不是 B 的子集。我们也可以颠倒 A 和 B 的顺序。我们称 B 是 A 的超集,记作 B⊇A,当且仅当 A⊆B。类似地,B 是 A 的真超集,记作 B⊋A,当且仅当 A⊊B。

Next we define union, intersection, set difference, and the empty set. These definitions tacitly require the following idea: to define a new set, we must specify exactly which objects are members of that set . (We develop this idea more formally in the next section, when we define set equality.) So, for example, to define the new set A B , we need an axiom telling us precisely which objects belong to this set. Here is the definitional axiom we need.

接下来,我们定义并集、交集、差集和空集。这些定义隐含着一个思想:要定义一个新集合,我们必须明确指出哪些对象属于该集合。(我们将在下一节定义集合相等时更正式地阐述这个思想。)例如,要定义新集合 A ∪ B,我们需要一个公理来明确指出哪些对象属于该集合。以下就是我们需要的定义公理。

3.2. Definition: Union of Two Sets. For all sets A and B and all objects x : x A B iff x A or x B .

3.2. 定义:两个集合的并集。对于所有集合 A 和 B 以及所有对象 x:x∈A∪B 当且仅当 x∈A 或 x∈B。

3.3. Definition: Intersection of Two Sets. For all sets A and B and all objects x : x A B iff x A and x B .

3.3. 定义:两个集合的交集。对于所有集合 A 和 B 以及所有对象 x:x∈A∩B 当且仅当 x∈A 且 x∈B。

Restating these definitions more symbolically, x A B means ( x A )∨( x B ), whereas x A B means ( x A ) ( x B ) . Observe that ∪ (the union symbol) and ∨ (the OR symbol) both open upwards, whereas ∩ (the intersection symbol) and ∧ (the AND symbol) both open downwards. This observation may help avoid confusing the two symbols ∪ and ∩. We should also remember that ∪ and ∨ are not the same symbol; ∪ is used to combine sets, whereas ∨ is used to combine logical statements. Similar comments apply to ∩ and ∧.

用更符号化的方式重述这些定义:x ∈ A ∪ B 表示 (x ∈ A)∨(x ∈ B),而 x∈A∩B 表示 (x∈A)∧(x∈B)。请注意,∪(并集符号)和 ∨(或符号)的开口都向上,而 ∩(交集符号)和 ∧(与符号)的开口都向下。这一观察有助于避免混淆 ∪ 和 ∩ 这两个符号。我们还应该记住,∪ 和 ∨ 不是同一个符号;∪ 用于组合集合,而 ∨ 用于组合逻辑语句。类似的说明也适用于 ∩ 和 ∧。

3.4. Definition: Set Difference. For all sets A and B and all objects x : x A B iff x A and x B .

3.4. 定义:集合差集。对于所有集合 A 和 B 以及所有对象 x:x∈A−B 当且仅当 x∈A 且 x∉B。

3.5. Definition: the Empty Set ∅. For all objects x , x .

3.5. 定义:空集∅。对于所有对象x,x∉∅。

Subset Proofs

子集证明

Let A and B be given sets. How can we prove that A B ? By definition, we must prove x , x A x B . We can obtain a proof outline of this statement by combining the generic-element proof template with the direct proof template for IF-statements. The resulting sequence of steps occurs so frequently in proofs that it is worth stating (and memorizing) as its own separate proof template:

设 A 和 B 为给定的集合。如何证明 A ⊆ B?根据定义,我们必须证明 ∀x,x∈A⇒x∈B。我们可以结合通用元素证明模板和 IF 语句的直接证明模板,得到该命题的证明概要。由此得到的步骤序列在证明中非常常见,因此值得将其作为一个独立的证明模板列出(并牢记):

3.6. Subset Proof Template to Prove A B . ( A and B are sets.)

3.6. 子集证明模板,用于证明 A ⊆ B。(A 和 B 是集合。)

Fix an arbitrary object x 0 . Assume x 0 A . Prove x 0 B .

固定任意对象 x 0 。假设 x 0 ∈ A。证明 x 0 ∈ B。

(Continue the proof by expanding the definitions of “ x 0 A ” and “ x 0 B ” and using more proof templates.)

(继续证明,扩展“x 0 ∈ A”和“x 0 ∈ B”的定义,并使用更多证明模板。)

Note that A and B can be any expressions standing for sets, not necessarily individual variables, as seen in the next example.

请注意,A 和 B 可以是任何表示集合的表达式,而不一定是单个变量,如下例所示。

3.7. Example. Let U and V be sets. Outline a proof that U × V P ( U V ) .

3.7. 示例。设 U 和 V 为集合。简述证明 U×V⊆P(U∪V)。

Solution. Although the given statement uses potentially unfamiliar symbols (to be discussed later), we note that the statement is asserting that one set is a subset of another. So we can generate the proof outline immediately:

解答。虽然给定的语句使用了可能不太熟悉的符号(稍后讨论),但我们注意到该语句断言一个集合是另一个集合的子集。因此,我们可以立即生成证明概要:

Fix an arbitrary object x 0 . Assume x 0 U × V . Prove x 0 P ( U V ) .

固定任意对象 x 0 。假设 x 0 ∈ U × V。证明 x0∈P(U∪V)。

The next theorem lists a host of properties of the subset relation.

下一个定理列举了子集关系的一系列性质。

3.8. Theorem on Subsets. For all sets A , B , C :

3.8. 子集定理。对于任意集合 A、B、C:

(a) Reflexivity: A A .

(a)自反性:A ⊆ A。

(b) Transitivity: If A B and B C , then A C .

(b)传递性:如果 A ⊆ B 且 B ⊆ C,则 A ⊆ C。

(c) Lower Bound: A B A and A B B .

(c)下界:A ∩ B ⊆ A 且 A ∩ B ⊆ B。

(d) Greatest Lower Bound: C A B iff ( C A and C B ).

(d)最大下界:C⊆A∩B 当且仅当(C ⊆ A 且 C ⊆ B)。

(e) Upper Bound: A A B and B A B .

(e)上界:A⊆A∪B 且 B⊆A∪B。

(f) Least Upper Bound: A B C iff ( A C and B C ).

(f)最小上界:A∪B⊆C 当且仅当(A ⊆ C 且 B ⊆ C)。

(g) Minimality of Empty Set: ∅ ⊆ A .

(g)空集的最小性:∅ ⊆ A。

(h) Difference Property: A B A .

(h)差分性质:A − B ⊆ A。

(i) Monotonicity: If A B , then: A C B C and A C B C and A C B C .

(i)单调性:如果 A ⊆ B,则:A∩C⊆B∩C 且 A∪C⊆B∪C 且 A − C ⊆ B − C。

(j) Inclusion Reversal: If A B , then C B C A .

(j)包含反转:如果 A ⊆ B,则 C − B ⊆ C − A。

We now use the subset template, combined with other proof templates and definitions, to prove some parts of this theorem. We ask the reader to prove the other parts in later exercises. Throughout all these proofs, we let A , B , and C be fixed, but arbitrary, sets.

现在,我们结合其他证明模板和定义,使用子集模板来证明该定理的部分内容。其余部分则留给读者在后续练习中证明。在所有这些证明中,我们假设 A、B 和 C 是固定的但任意的集合。

3.9. Proof of Reflexivity. We must prove A A . Fix an arbitrary object x 0 . Assume x 0 A . Prove x 0 A . Since we just assumed that x 0 A , the proof is done.

3.9. 自反性的证明。我们必须证明 A ⊆ A。固定一个任意对象 x 0 。假设 x 0 ∈ A。证明 x 0 ∈ A。由于我们刚刚假设 x 0 ∈ A,证明完毕。

3.10. Proof of Upper Bound. We must prove “ A A B and B A B .” To prove the AND-statement, we prove each part separately.

3.10. 上界的证明。我们必须证明“A⊆A∪B 且 B⊆A∪B”。为了证明“与”命题,我们需要分别证明每个部分。

Part 1. Prove A A B . Fix an arbitrary object x 0 . Assume x 0 Ax 0 A B . We must prove “ x 0 A or x 0 B .” Since x 0 A was assumed to be true, the truth table for OR shows that the required OR-statement is also true.

第一部分:证明 A⊆A∪B。固定一个任意对象 x 0 。假设 x 0 ∈ A,x 0 ∈ A ∪ B。我们需要证明“x 0 ∈ A 或 x 0 ∈ B”。由于假设 x 0 ∈ A 为真,根据 OR 的真值表可知,所需的 OR 语句也为真。

Part 2. Prove B A B . Fix an arbitrary object y 0 . Assume y 0 B . Prove y 0 A B . We must prove “ y 0 A or y 0 B .” Since y 0 B was assumed to be true, the truth table for OR shows that the required OR-statement is also true.

第二部分:证明 B⊆A∪B。固定一个任意对象 y 0 。假设 y 0 ∈ B。证明 y 0 ∈ A ∪ B。我们需要证明“y 0 ∈ A 或 y 0 ∈ B”。由于假设 y 0 ∈ B 为真,根据 OR 的真值表可知,所需的 OR 语句也为真。

3.11. Proof of Minimality of Empty Set. We must prove ∅ ⊆ A . [If we mechanically recite the subset proof template, we generate these lines: “Fix x 0 . Assume x 0 . Prove x 0 A .” It may be unclear how to proceed from this point. So we try the proof again, starting from the definition of subsets.] We must prove x , x x A . Fix an arbitrary object x 0 . We must prove the IF-statement “ x 0 x 0 A .” Now, the hypothesis “ x 0 ” is false by definition of the empty set. So the entire IF-statement is true, by the truth table for IF. Thus we have proved ∅ ⊆ A .

3.11. 空集最小性的证明。我们必须证明 ∅ ⊆ A。[如果我们机械地背诵子集证明模板,就会得到以下几行:“固定 x 0 。假设 x0∈∅。证明 x 0 ∈ A。” 从这一点开始,如何继续可能不太清楚。所以我们再次尝试证明,从子集的定义开始。] 我们必须证明 ∀x,x∈∅⇒x∈A。固定任意对象 x 0 。我们必须证明 IF 语句“x0∈∅⇒x0∈A”。现在,根据空集的定义,假设“x0∈∅”为假。因此,根据 IF 语句的真值表,整个 IF 语句为真。因此,我们证明了 ∅ ⊆ A。

You may find the proof just given to be somewhat cryptic. If so, you may prefer the following contrapositive proof of the IF-statement. Assume x 0 A ; prove x 0 . The goal is true by definition of the empty set [and we do not need to use the assumption here]. Similarly, you can prove ∅ ⊆ A by a contradiction proof (see Exercise 10). This example illustrates the following general observation. It is often easier to prove statements involving the empty set using negative logic, such as a contrapositive proof or a proof by contradiction.

你可能会觉得刚才给出的证明有些晦涩难懂。如果是这样,你或许会更喜欢下面这个 IF 语句的逆否命题证明。假设 x0∉A;证明 x0∉∅。根据空集的定义,这个目标成立[这里我们不需要用到假设]。类似地,你可以用反证法证明 ∅ ⊆ A(参见练习 10)。这个例子说明了以下一般性观察:对于涉及空集的命题,通常更容易用否定逻辑来证明,例如使用逆否命题证明或反证法。

We continue proving parts of the Theorem on Subsets in the next section.

下一节我们将继续证明子集定理的部分内容。

Section Summary

章节概要

1. Subsets. A B means x , x A x B .

1.子集。A⊆B 表示 ∀x,x∈A⇒x∈B。

To prove A B : fix an arbitrary object x 0 ; assume x 0 A ; prove x 0 B .

要证明 A ⊆ B:固定任意对象 x 0 ;假设 x 0 ∈ A;证明 x 0 ∈ B。

2. Set Operations. Memorize these definitions: for all sets A , B and all objects x ,

2.集合运算。记住以下定义:对于所有集合 A、B 和所有对象 x,

Union: x A B iff x A or x B .

并集:x∈A∪B 当且仅当 x∈A 或 x∈B。

Intersection: x A B iff x A and x B .

交集:x∈A∩B 当且仅当 x∈A 且 x∈B。

Set Difference: x A B iff x A and x B .

集合差集:x∈A−B 当且仅当 x∈A 且 x∉B。

Empty Set: x , x .

空集:∀x,x∉∅。

3. Theorem on Subsets. Memorize the facts listed in the Theorem on Subsets for later use. In particular, for all sets A , B , C : ∅ and A B and A are always subsets of A ; A B C implies A C ; A B is a subset of both A and B ; C A B iff C A and C B ; A and B are both subsets of A B ; A B C iff A C and B C ; and A B implies A C B C , A C B C , A C B C , and C B C A .

3.子集定理。记住子集定理中列出的事实,以便后续使用。特别地,对于任意集合A、B、C:∅ 且 A − B 和 A 始终是 A 的子集;A ⊆ B ⊆ C 蕴含 A ⊆ C;A ∩ B 是 A 和 B 的子集;C⊆A∩B 当且仅当 C ⊆ A 且 C ⊆ B;A 和 B 都是 A ∪ B 的子集;A∪B⊆C 当且仅当 A ⊆ C 且 B ⊆ C;并且 A ⊆ B 蕴含 A∩C⊆B∩C、A∪C⊆B∪C、A − C ⊆ B − C 和 C − B ⊆ C − A。

Exercises

练习

1. Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 4, 5}, and C = {2, 4, 6, 8, 10}. Informally compute the following sets: (a) A B , A B , A C , ( A B ) C , A ( B C ) . (b) B A , A B , B C , C B , A C , C A , B B . (c) X A , X B , A X , X ( A B ) , ( X A ) ( X B ) , ( X A ) − B , X − ( A B ).

1.设 X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},A = {1, 3, 5, 7, 9},B = {1, 2, 3, 4, 5},C = {2, 4, 6, 8, 10}。非正式地计算以下集合:(a) A ∪ B,A ∩ B,A ∩ C,(A∪B)∩C,A∪(B∩C)。(b) B − A,A − B,B − C,C − B,A − C,C − A,B − B。(c) X − A,X − B,A − X,X−(A∩B),(X−A)∪(X−B),(X − A) − B,X − (A − B)。

2. In the previous exercise, consider the following ten sets: ∅, A B , B C , A , B , C , A B , B C , X A , X . Make a 10 × 10 chart showing, for each pair S , T of sets from this list, whether S T .

2.在前面的练习中,考虑以下十个集合:∅、A ∩ B、B ∩ C、A、B、C、A ∪ B、B ∪ C、X − A、X。制作一个 10 × 10 的表格,显示对于此列表中的每一对集合 S、T,S 是否 ⊆ T。

3. Let S = {0, 3, 5, 9, 0.6, π }, T = { 4 , 5 , 7 , 3 / 5 , 2 } , and U = { 2 , 4 , 9 , 0.36 } . Informally compute the following sets: (a) S T , S U , T U , S T U . (b) S Z , T Q , U Z , S ™ ℚ, U ™ ℚ, ( U S ) − T , U − ( S T ).

3.设 S = {0, 3, 5, 9, 0.6, π},T = {4, 5, 7, 3/5, 2},U = {2, 4, 9, 0.36}。非正式地计算以下集合:(a) S ∩ T, S ∩ U, T ∩ U, S∩T∩U。(b) S∩Z, T∩Q, U∩Z, S ⊆ ℚ, U ⊆ ℚ, (U − S) − T, U − (S − T)。

4. In the Venn diagram shown below, express each of the seven numbered regions in terms of the sets A , B , and C . For example, region 1 can be written ( A B ) − C or A ( B C ) .

4.在下面所示的维恩图中,用集合 A、B 和 C 表示七个编号区域中的每一个。例如,区域 1 可以写成 (A − B) − C 或 A−(B∪C)。

ufig3_6.jpg

5. Make copies of the Venn diagram shown above with the following regions shaded.

5.复制上图所示的维恩图,并将以下区域涂上阴影。

(a) A − ( B C ) (b) ( B A ) C (c) A ( B C ) (d) ( A C ) B (e) ( A B ) C (f) A ( B C ) .

(a)A − (B − C) (b) (B−A)∩C (c) A∪(B−C) (d) (A∪C)−B (e) (A∪B)∩C (f) A∪(B∩C).

6. Draw a Venn diagram showing four sets A , B , C , D with the following properties: A C = ; A B B C ; D B A . Then shade in the set ( C D ) ( C D ) .

6.画一个维恩图,表示四个集合 A、B、C、D,具有以下性质:A∩C=∅;A∩B≠∅≠B∩C;D ⊆ B − A。然后涂黑集合 (C∪D)−(C∩D)。

7. Draw a Venn diagram with five sets where the intersection of any two of the sets is not empty, but the intersection of any three of the sets is empty.

7.画一个包含五个集合的维恩图,其中任意两个集合的交集不为空,但任意三个集合的交集为空。

8. Create a proof template for each situation below.

8.针对以下每种情况创建证明模板。

(a) Prove A B by expanding the definition and using a contrapositive proof.

(a)通过展开定义并使用逆否命题证明来证明 A ⊆ B。

(b) Prove A B using proof by contradiction.

(b)用反证法证明 A ⊆ B。

(c) Disprove A B .

(c)反证 A ⊆ B。

9. Prove parts (c) and (h) of the Theorem on Subsets: for all sets A , B , A B A and A B B and A B A .

9.证明子集定理的(c)和(h)部分:对于所有集合 A、B,A ∩ B ⊆ A 和 A ∩ B ⊆ B 和 A − B ⊆ A。

10. Fix an arbitrary set A . Prove ∅ ⊆ A using proof by contradiction.

10. 固定任意集合 A。用反证法证明 ∅ ⊆ A。

11. Disprove each statement.

11.反驳下列陈述。

(a) For all sets A and B , A B B .

(a)对于所有集合 A 和 B,A∪B⊆B。

(b) For all sets A and B , A A B .

(b)对于所有集合 A 和 B,A ⊆ A − B。

(c) For all sets A and B , if A B then C A C B .

(c)对于所有集合 A 和 B,如果 A ⊆ B,则 C − A ⊆ C − B。

12. Suppose we replace “For all” by “There exists” in each part of the previous exercise. Prove that the resulting statements are all true.

12. 假设我们将上一题中每一部分的“对于所有”替换为“存在”。证明所得的语句全部为真。

13. Draw Venn diagrams illustrating (as best you can) the following items from the Theorem on Subsets: (a) lower bound; (b) difference property; (c) least upper bound; (d) monotonicity.

13.绘制维恩图,尽可能清晰地说明子集定理中的下列各项:(a)下界;(b)差集性质;(c)最小上界;(d)单调性。

14. Outline the proof of each statement; do not expand unfamiliar definitions.

14. 列出每个陈述的证明;不要扩展不熟悉的定义。

(a) For all sets S , Int( S )⊆Clo( S ).

(a)对于所有集合 S,Int(S)⊆Clo(S)。

(b) For all sets F and S , if F is closed and F S , then F Clo ( S ) .

(b)对于所有集合 F 和 S,如果 F 是闭集且 F⊇S,则 F⊇Clo(S)。

(c) For all sets S and all x , if x ∈ Int( S ), then r R > 0 , B ( x ; r ) S .

(c)对于所有集合 S 和所有 x,如果 x ∈ Int(S),则存在 r∈R>0,B(x;r)⊆S。

15. Prove this part of the Theorem on Subsets: for all sets A , B , if A B , then A C B C and A C B C .

15.证明子集定理的这一部分:对于所有集合 A、B,如果 A ⊆ B,则 A∩C⊆B∩C 且 A − C ⊆ B − C。

16. Give a specific example of sets A , B , C , D , E where A C , B C , C D , and C E are all true, but A B , B A , D E , and E D are all false.

16.举例说明集合 A、B、C、D、E,其中 A ⊆ C、B ⊆ C、C ⊆ D 和 C ⊆ E 都为真,但 A ⊆ B、B ⊆ A、D ⊆ E 和 E ⊆ D 都为假。

17. (a) Prove: for all sets A , B , C , D , if A B and B C and C D and D A then C B . (b) Suppose we omit the hypothesis A B in (a). Prove or disprove the resulting statement. (c) Suppose we omit the hypothesis B C in (a). Prove or disprove the resulting statement.

17.(a) 证明:对于任意集合 A、B、C、D,若 A ⊆ B 且 B ⊆ C 且 C ⊆ D 且 D ⊆ A,则 C ⊆ B。(b) 假设我们省略 (a) 中的假设 A ⊆ B。证明或反驳所得命题。(c) 假设我们省略 (a) 中的假设 B ⊆ C。证明或反驳所得命题。

18. Ten sets created from sets A , B , and C are listed below. Make a 10 × 10 chart showing, for each pair of sets S , T from this list, whether S T must be true for all choices of A , B , and C .

18. 下面列出了由集合 A、B 和 C 构成的十个集合。请制作一个 10 × 10 的表格,对于列表中的每一对集合 S 和 T,说明对于 A、B 和 C 的所有选择,S ⊆ T 是否必然成立。

( A 一个 B B ) C C , ( A 一个 B B ) C C , ( A 一个 B B ) C C , ( A 一个 B B ) C C , ( A 一个 B B ) C C , ( A 一个 B B ) C C ,
( A 一个 B B ) C C , ( A 一个 B B ) C C , ( A 一个 B B ) C C , A 一个 ( B B C C ) .

19. For each pair of sets S , T listed in the previous exercise, decide whether S T must be the empty set for all choices of A , B , and C .

19.对于上一练习中列出的每一对集合 S、T,判断对于 A、B 和 C 的所有选择,S ∩ T 是否必须为空集。

3.2 Subset Proofs and Set Equality Proofs

3.2 子集证明和集合相等性证明

This section continues our discussion of subset proofs, as well as the related idea of using a known statement A B to prove other facts. We then define set equality and look at various ways to prove that two sets are equal.

本节继续讨论子集证明,以及如何利用已知命题 A ⊆ B 来证明其他事实。接下来,我们将定义集合相等,并探讨证明两个集合相等的各种方法。

Subset Proofs, Continued

子集证明(续)

In the last section, we described how to prove statements of the form A B . Recall the main template used for such proofs goes like this:

上一节中,我们介绍了如何证明形如 A ⊆ B 的命题。回顾一下,此类证明的主要模板如下:

Fix x 0 . Assume x 0 A . Prove x 0 B .

固定 x 0 。假设 x 0 ∈ A。证明 x 0 ∈ B。

Typically, the proof continues by expanding the definitions of “ x 0 A ” and “ x 0 B .”

通常,证明会继续扩展“x 0 ∈ A”和“x 0 ∈ B”的定义。

Sometimes, instead of proving a statement of the form A B , we already know that this statement is true. The following rule, which we might call a knowledge template , tells us how the known statement A B can be used to deduce further conclusions.

有时,我们无需证明 A ⊆ B 这样的命题,因为我们已经知道这个命题为真。以下规则(我们可以称之为知识模板)告诉我们,如何利用已知的命题 A ⊆ B 来推导出进一步的结论。

3.12. When A B is Already Known or Assumed:

3.12. 当 A ⊆ B 已知或假定时:

(a) If you know that x 0 A , you may deduce that x 0 B .

(a)如果你知道 x 0 ∈ A,你可以推断出 x 0 ∈ B。

(b) If you know that x 0 B , you may deduce that x 0 A .

(b)如果你知道 x0∉B,你就可以推断出 x0∉A。

To justify this, recall that A B means x , x A x B . By the Inference Rule for ALL, we may take x = x 0 here to deduce that “ x 0 A x 0 B ” is true. On one hand, if x 0 A is known, then x 0 B follows by the Inference Rule for IF. On the other hand, if x 0 B is known, then x 0 A follows by taking the contrapositive and using the Inference Rule for IF.

为了证明这一点,回顾一下 A ⊆ B 的含义:∀x,x∈A⇒x∈B。根据“全集推理规则”,我们可以取 x = x 0 来推导出“x 0 ∈ A ↠ x 0 ∈ B”为真。一方面,如果已知 x 0 ∈ A,那么根据“全集推理规则”,x 0 ∈ B 也成立。另一方面,如果已知 x0∉B,那么通过取逆否命题并运用“全集推理规则”,x0∉A 也成立。

The next few proofs illustrate both the subset proof template and the subset knowledge template. We are continuing to prove parts of the Theorem on Subsets, stated in the previous section. Throughout the discussion, we fix arbitrary sets A , B , and C . Comments about how we are generating the proofs appear in square brackets.

接下来的几个证明将同时展示子集证明模板和子集知识模板。我们将继续证明上一节中提出的子集定理的部分内容。在整个讨论过程中,我们固定任意集合 A、B 和 C。关于如何生成这些证明的注释将用方括号括起来。

3.13. Proof of Transitivity. We must prove: if A B and B C , then A C . [Try a direct proof of the IF-statement.] Assume A B and B C . Prove A C . [We are now proving a subset statement, so use the subset proof template.] Fix an arbitrary object x 0 . Assume x 0 A . Prove x 0 C . [To use the assumptions, invoke the subset knowledge template twice.] Since A B and x 0 A , we deduce x 0 B . Since B C and x 0 B , we deduce x 0 C . The proof is now done.

3.13. 传递性的证明。我们必须证明:如果 A ⊆ B 且 B ⊆ C,则 A ⊆ C。[尝试直接证明 IF 语句。] 假设 A ⊆ B 且 B ⊆ C。证明 A ⊆ C。[我们现在证明的是一个子集命题,因此使用子集证明模板。] 固定一个任意对象 x 0 。假设 x 0 ∈ A。证明 x 0 ∈ C。[要使用这些假设,请调用子集知识模板两次。] 因为 A ⊆ B 且 x 0 ∈ A,我们推导出 x 0 ∈ B。因为 B ⊆ C 且 x 0 ∈ B,我们推导出 x 0 ∈ C。证明完毕。

We can give some intuition for the result just proved using Venn diagrams. Here is a visual version of the transitivity of ⊆:

我们可以用维恩图来直观地理解刚才证明的结果。下图是⊆的传递性的可视化表示:

ufig3_7.jpg

Although these Venn diagrams provide valuable visual insight into why transitivity should be true, the diagrams by themselves are not a proof of this result. The preceding picture only shows that transitivity holds for the three specific sets shown (namely, the sets of points within the circles labeled A , B , and C ). In other words, the Venn diagrams prove the existential statement: “There exist three sets A , B , C , such that if A B and B C , then A C .” But the diagrams do not prove the stronger universal statement: “for all sets A , B , C , if A B and B C , then A C ” asserted in the theorem. Although the visual intuition conveyed by the diagrams is certainly compelling, we have no guarantee that this intuition extends to all abstract sets, which might be sets of numbers, or sets of functions, or sets of cows, or sets of other sets, not merely sets of points that we can draw. To get a rigorous proof of the universal version of the statement, we must rely on the formal definitions and the rules of logic, rather than invoking pictures. Similar comments apply to many other theorems about sets: Venn diagrams can prove existential statements, and they can supply intuition for other kinds of statements, but they cannot prove universal statements about all sets.

尽管这些维恩图为我们提供了关于传递性为何成立的宝贵视觉见解,但这些图本身并不能证明这一结论。前面的图仅仅表明,传递性对于图中所示的三个特定集合(即标记为 A、B 和 C 的圆圈内的点集)成立。换句话说,这些维恩图证明了存在性命题:“存在三个集合 A、B、C,使得如果 A ⊆ B 且 B ⊆ C,则 A ⊆ C。” 但这些图并不能证明定理中提出的更强的普遍性命题:“对于所有集合 A、B、C,如果 A ⊆ B 且 B ⊆ C,则 A ⊆ C。” 虽然这些图所传达的视觉直观性确实很有说服力,但我们无法保证这种直观性可以推广到所有抽象集合,这些抽象集合可能是数集、函数集、牛集或其他集合,而不仅仅是我们可以绘制的点集。为了对该命题的普遍形式进行严格证明,我们必须依赖形式定义和逻辑规则,而不是借助图像。类似的道理也适用于许多其他关于集合的定理:维恩图可以证明存在性命题,也可以为其他类型的命题提供直观的理解,但它们无法证明关于所有集合的普遍命题。

3.14. Proof of Inclusion Reversal. We must prove: if A B , then C B C A . [Try a direct proof of the IF-statement.] Assume A B . Prove C B C A . [We are proving a subset statement, so use that template.] Fix an arbitrary object x 0 . Assume x 0 C B . Prove x 0 C A . We have assumed x 0 C and x 0 B . We must prove x 0 C and x 0 A . First, prove x 0 C ; note this statement has already been assumed. Second, prove x 0 A . [Use the subset knowledge template, part (b).] Since we have assumed A B and x 0 B , we can deduce that x 0 A , as needed.

3.14. 包含关系反转的证明。我们必须证明:如果 A ⊆ B,则 C − B ⊆ C − A。[尝试直接证明 IF 语句。] 假设 A ⊆ B。证明 C − B ⊆ C − A。[我们正在证明一个子集命题,因此使用该模板。] 固定一个任意对象 x 0 。假设 x 0 ∈ C − B。证明 x 0 ∈ C − A。我们已经假设 x 0 ∈ C 且 x0∉B。我们必须证明 x 0 ∈ C 且 x0∉A。首先,证明 x 0 ∈ C;注意,此命题已被假设。其次,证明 x0∉A。 [使用子集知识模板,第 (b) 部分。] 由于我们假设 A ⊆ B 且 x0∉B,我们可以推断出 x0∉A,这是必要的。

The proof reveals why the inclusion of A within B gets reversed to become an inclusion of C B within C A . The following Venn diagrams provide additional intuition for why this reversal happens. The circle for A is nested within the circle for B , since we are assuming A B . The shaded part of the left diagram shows C B , whereas the shaded part of the right diagram shows C A . We see that the first shaded shape is contained within the second shaded shape.

该证明揭示了为什么 A 包含于 B 中会反转为 C − B 包含于 C − A 中。以下维恩图提供了这种反转发生原因的更多直观解释。由于我们假设 A ⊆ B,因此 A 的圆嵌套在 B 的圆内。左图的阴影部分表示 C − B,而右图的阴影部分表示 C − A。我们可以看到,第一个阴影部分包含于第二个阴影部分内。

ufig3_8.jpg

If you still think that these diagrams are enough to prove the result, note that the diagram makes hidden assumptions about how the set C intersects the other two sets. What if the circle for C is completely outside the circle for A , or completely inside the circle for B , or in some other position? What if A happens to be the empty set? We would need new Venn diagrams to cover these and other situations that might arise. On the other hand, the formal proof given earlier works for generic (arbitrary) sets A , B , and C .

如果你仍然认为这些图足以证明结论,请注意,这些图对集合 C 与其他两个集合的交集关系做了一些隐含的假设。如果集合 C 的圆完全位于集合 A 的圆之外,或者完全位于集合 B 的圆之内,或者处于其他位置呢?如果集合 A 恰好是空集呢?我们需要新的维恩图来涵盖这些以及其他可能出现的情况。另一方面,前面给出的形式化证明适用于一般的(任意的)集合 A、B 和 C。

Our next proof introduces another key ingredient in proof-writing: using previously proved results to shorten the proofs of new results.

我们的下一个证明引入了证明写作的另一个关键要素:利用先前已证明的结果来缩短新结果的证明。

3.15. Proof of Least Upper Bound. We must prove A B C iff A C and B C . [This is an IFF-statement, so we need a two-part proof.]

3.15. 最小上界的证明。我们必须证明 A∪B⊆C 当且仅当 A ⊆ C 且 B ⊆ C。[这是一个因果当且仅当命题,因此我们需要分两步证明。]

Part 1. Assume A B C . Prove A C and B C . [We must prove an AND-statement, so we need a two-part proof within part 1.]

第一部分:假设 A∪B⊆C。证明 A ⊆ C 且 B ⊆ C。[我们需要证明一个“与”命题,因此第一部分需要分两步证明。]

Part 1a. Prove A C . Fix x 0 . Assume x 0 A . Prove x 0 C . We already proved (Theorem on Subsets, part (e)) that A A B . Since x 0 A , it follows that x 0 A B . Since we assumed A B C , it follows that x 0 C , as needed.

第 1a 部分。证明 A ⊆ C。固定 x 0 。假设 x 0 ∈ A。证明 x 0 ∈ C。我们已经证明了(子集定理,第 (e) 部分)A⊆A∪B。由于 x 0 ∈ A,因此 x 0 ∈ A ∪ B。由于我们假设 A∪B⊆C,因此 x 0 ∈ C,满足要求。

Part 1b. Prove B C . [We could prove this as in part 1a, but here is a faster proof using known theorems.] We have already proved that B A B , and we have assumed that A B C in part 1. By transitivity (Theorem on Subsets, part (b)), B C follows. [Part 1a could have been proved in this way, as well. Observe that our proof of Part 1a essentially repeats the proof of transitivity.]

第 1b 部分:证明 B ⊆ C。[我们可以像第 1a 部分那样证明,但这里使用已知定理可以更快地证明。] 我们已经证明了 B ⊆ A ∪ B,并且在第 1 部分中假设了 A ∪ B ⊆ C。根据传递性(子集定理,第 (b) 部分),B ⊆ C 成立。[第 1a 部分也可以用这种方法证明。注意,我们对第 1a 部分的证明本质上重复了传递性的证明。]

Part 2. Assume A C and B C . Prove A B C . [Our goal is a subset statement, so we use that template.] Fix z 0 . Assume z 0 A B . Prove z 0 C . We have assumed z 0 A or z 0 B . [Having assumed an OR-statement, we are led to a proof by cases.]

第二部分。假设 A ⊆ C 且 B ⊆ C。证明 A∪B⊆C。[我们的目标是子集命题,因此我们使用子集命题模板。] 固定 z 0 。假设 z 0 ∈ A ∪ B。证明 z 0 ∈ C。我们假设 z 0 ∈ A 或 z 0 ∈ B。[假设了一个 OR 命题,我们就可以采用分情况的证明方法。]

Case 1. Assume z 0 A . Prove z 0 C . Since we have assumed A C and z 0 A , z 0 C follows.

情况 1. 假设 z 0 ∈ A。证明 z 0 ∈ C。由于我们假设 A ⊆ C 且 z 0 ∈ A,因此 z 0 ∈ C 成立。

Case 2. Assume z 0 B . Prove z 0 C . Since we have assumed B C and z 0 B , z 0 C follows.

情况 2. 假设 z 0 ∈ B。证明 z 0 ∈ C。由于我们假设 B ⊆ C 且 z 0 ∈ B,因此 z 0 ∈ C 成立。

Set Equality

集合相等

When are two sets equal? Informally, a set is a collection of objects called the members of the set. With this intuition, it seems reasonable to say that two sets should be equal iff they have precisely the same members. This leads to the following definition, which is sometimes called the Axiom of Extension .

两个集合何时相等?通俗地说,集合是称为集合成员的对象的集合。基于这种直觉,我们似乎可以合理地认为,两个集合相等当且仅当它们拥有完全相同的成员。由此可得出以下定义,该定义有时被称为外延公理。

3.16. Definition: Set Equality. For all sets A and B : A = B iff x , x A x B .

3.16. 定义:集合相等。对于所有集合 A 和 B:A=B 当且仅当 ∀x,x∈A⇔x∈B。

We can use logical transformations to relate this definition to the definition of subsets. First, x , x A x B is equivalent to

我们可以使用逻辑变换将此定义与子集的定义联系起来。首先,∀x,x∈A⇔x∈B 等价于

x x , ( ( x x A 一个 x x B B ) ( x x B B x x A 一个 ) )

by the Theorem on IFF. Second, by the quantifier property linking and ∧, the previous formula is equivalent to

根据 IFF 定理。其次,根据连接 ∀ 和 ∧ 的量词性质,前面的公式等价于

( y , y A 一个 y B B ) ( z z , z z B B z z A 一个 ) .

We recognize the definition of subset in two places here, so the last formula is equivalent to A B and B A . Chaining together these equivalences, we conclude:

这里我们两次看到了子集的定义,因此最后一个公式等价于 A ⊆ B 且 B ⊆ A。将这些等价关系串联起来,我们得出结论:

For all sets A and B , A = B iff A B and B A .

对于所有集合 A 和 B,A=B 当且仅当 A⊆B 且 B⊆A。

This leads to the following proof template for proving that two sets are equal.

由此可得出以下证明两个集合相等的证明模板。

3.17. Proof Template to Prove a Set Equality. A = B ( A and B are sets.)

3.17. 证明集合相等的证明模板。A = B(A 和 B 是集合。)

Part 1. Prove A B .

第一部分:证明 A ⊆ B。

Part 2. Prove B A .

第二部分。证明 B ⊆ A。

Typically, we would use the subset proof template to accomplish parts 1 and 2, though other methods can sometimes be used (e.g., invoking previously proved theorems). The next theorem lists many identities asserting the equality of various sets.

通常情况下,我们会使用子集证明模板来完成第 1 部分和第 2 部分,尽管有时也可以使用其他方法(例如,调用先前已证明的定理)。下一个定理列出了许多断言各种集合相等的恒等式。

3.18. Theorem on Set Equality. For all sets A , B , C , A ′, B ′:

3.18. 集合相等定理。对于所有集合 A、B、C、A′、B′:

(a) Reflexivity: A = A .

(a)自反性:A = A。

(b) Symmetry: If A = B , then B = A .

(b)对称性:如果 A = B,则 B = A。

(c) Transitivity: If A = B and B = C , then A = C .

(c)传递性:如果 A = B 且 B = C,则 A = C。

(d) Substitution Properties: If A = A ′ and B = B ′, then: A B = A B and A B = A B and A B = A ′ − B ′ and ( A B iff A ′⊆ B ′).

(d)替换性质:如果 A = A′ 且 B = B′,则:A∩B=A′∩B′ 且 A∪B=A′∪B′ 且 A − B = A′ − B′ 且(A ⊆ B 当且仅当 A′⊆ B′)。

(e) Commutativity: A B = B A and A B = B A .

(e)交换律:A∩B=B∩A 且 A∪B=B∪A。

(f) Associativity: ( A B ) C = A ( B C ) and ( A B ) C = A ( B C ) .

(f)结合律:(A∩B)∩C=A∩(B∩C)且(A∪B)∪C=A∪(B∪C)。

(g) Distributive Laws: A ( B C ) = ( A B ) ( A C ) and A ( B C ) = ( A B ) ( A C ) .

(g)分配律:A∩(B∪C)=(A∩B)∪(A∩C)且A∪(B∩C)=(A∪B)∩(A∪C)。

(h) Idempotent Laws: A A = A and A A = A .

(h)幂等定律:A∩A=A 和 A∪A=A。

(i) Absorption Laws: A ( A B ) = A and A ( A B ) = A .

(i)吸收定律:A∪(A∩B)=A 和 A∩(A∪B)=A。

(j) Detecting the Empty Set: ( B = ∅ iff z , z B ) and ( B iff z , z B ).

(j)检测空集:(B = ∅ 当且仅当 ∀z,z∉B)和(B≠∅ 当且仅当 ∃z,z∈B)。

(k) Properties of the Empty Set: A = A and A = and A = A and A = and A A = .

(k)空集的性质:A∪∅=A 且 A∩∅=∅ 且 A−∅=A 且 ∅−A=∅ 且 A−A=∅。

(l) de Morgan Laws for Sets: A ( B C ) = ( A B ) ( A C ) and A ( B C ) = ( A B ) ( A C ) .

(l)德摩根定律:A−(B∪C)=(A−B)∩(A−C) 和 A−(B∩C)=(A−B)∪(A−C)。

(m) Set Partition of One Set by Another: A = ( A B ) ( A B ) and ( A B ) ( A B ) = .

(m)一个集合对另一个集合的划分:A=(A−B)∪(A∩B)且(A−B)∩(A∩B)=∅。

(n) Set Partition of One Set by a Subset: If B A , then A − ( A B ) = B , A = B ( A B ) , and B ( A B ) = .

(n)一个集合被一个子集划分:如果 B ⊆ A,则 A − (A − B) = B,A=B∪(A−B),且 B∩(A−B)=∅。

(o) Characterizations of Subsets: The following conditions are equivalent: A B ; A B = A ; A B = B ; A B = .

(o)子集的特征:下列条件等价:A ⊆ B;A∩B=A;A∪B=B;A−B=∅。

The first four properties listed here may appear to be obvious facts about equality that do not require proof. In certain theories, where equality is taken as an undefined term, these facts might be adopted as axioms about the equality symbol. But in set theory, set equality is a defined term, so we can prove these facts from more basic principles. As a sample, we prove parts (a) and (c).

这里列出的前四个性质似乎是关于等式的显而易见的事实,无需证明。在某些理论中,等式被视为未定义术语,这些事实可能被视为关于等号的公理。但在集合论中,集合相等是一个已定义术语,因此我们可以从更基本的原理证明这些事实。作为示例,我们证明 (a) 和 (c)。

3.19. Proof of Reflexivity of Set Equality. Let A be a fixed but arbitrary set. We will prove A = A . [Follow the new template, replacing B by A .] Part 1. Prove A A . This is a known theorem (Theorem on Subsets, part (a)). Part 2. Prove A A . This is also a known theorem. So A = A holds.

3.19. 集合等式自反性的证明。设 A 为任意固定集合。我们将证明 A = A。[遵循新模板,将 B 替换为 A。] 第一部分:证明 A ⊆ A。这是一个已知定理(子集定理,(a))。第二部分:证明 A ⊆ A。这也是一个已知定理。因此,A = A 成立。

3.20. Proof of Transitivity of Set Equality. Let A , B , and C be fixed but arbitrary sets. We prove: if A = B and B = C , then A = C . Assume A = B and B = C ; we must prove A = C . [Expand the assumptions using the second version of the definition of set equality.] We have assumed that A B and B A and B C and C B . [Now attack the goal using the set equality template.]

3.20. 集合相等传递性的证明。设 A、B 和 C 为任意集合。我们证明:若 A = B 且 B = C,则 A = C。假设 A = B 且 B = C,则我们必须证明 A = C。[使用集合相等定义的第二种形式扩展假设。] 我们假设 A ⊆ B 且 B ⊆ A 且 B ⊆ C 且 C ⊆ B。[现在使用集合相等模板来证明目标。]

Part 1. Prove A C . Since A B and B C were assumed, A C follows by the previously proved transitivity of ⊆ (Theorem on Subsets, part (b)).

第 1 部分。证明 A ⊆ C。由于假设 A ⊆ B 且 B ⊆ C,因此根据先前证明的 ⊆ 的传递性(子集定理,第 (b) 部分),A ⊆ C 成立。

Part 2. Prove C A . Since C B and B A were assumed, C A follows by the known transitivity of ⊆.

第 2 部分。证明 C ⊆ A。由于假设 C ⊆ B 且 B ⊆ A,根据 ⊆ 的传递性,C ⊆ A 成立。

3.21. Remark (Optional). Our definition of set equality states that for all sets A and B , A = B iff x , x A x B . The forward direction of this IFF-statement can be viewed as a substitution property that allows us to replace a set A appearing on the right side of the membership symbol ∈ by an equal set B . We need one more axiom stating an analogous substitution property for sets appearing on the left side of the membership symbol. Here is the axiom: for all sets A , B , C , if A = B , then [( A C ) ⇔ ( B C )]. Various other substitution properties for equal sets (such as those in part (d) of the Theorem on Set Equality) need not be stated as axioms, since they can be proved from the definitions.

3.21. 备注(可选)。我们对集合相等的定义是:对于任意集合 A 和 B,A = B 当且仅当 ∀x,x∈A⇔x∈B。这个当且仅当语句的前向推导可以看作是一种替换性质,它允许我们将成员符号 ∈ 右侧的集合 A 替换为相等的集合 B。我们需要另一个公理来表述成员符号左侧集合的类似替换性质。该公理如下:对于任意集合 A、B、C,如果 A = B,则 [(A ∈ C) ⇔ (B ∈ C)]。其他一些相等集合的替换性质(例如集合相等定理 (d) 部分中的性质)无需作为公理表述,因为它们可以从定义中证明。

Set Equality Proofs

集合等式证明

We continue to illustrate the proof template for set equality by proving more items from the Theorem on Set Equality. Remember that it is no longer necessary to prove everything from the definitions alone; you can and should use previously known results such as the Theorem on Subsets. In the following, fix arbitrary sets A , B , C , A ′, and B ′.

我们继续用集合相等定理的更多内容来说明集合相等性的证明模板。请记住,不再需要仅凭定义来证明所有内容;您可以而且应该使用先前已知的结果,例如子集定理。在下面的例子中,固定任意集合 A、B、C、A′ 和 B′。

3.22. Proof of Idempotent Laws.

3.22. 幂等律的证明。

Part 1. We prove A A = A . Part 1a. We prove A A A . This follows from the known result A B A (Theorem on Subsets, part (c)) by replacing B by A . Part 1b. We prove A A A . Fix an arbitrary object x 0 . Assume x 0 A . Prove x 0 A A . We must prove “ x 0 A and x 0 A .” To do this, [following the AND-template] first prove x 0 A and then prove x 0 A . In each part, we already assumed the statement to be proved.

第一部分:证明 A∩A=A。第一部分a:证明 A ∩ A ⊆ A。这可由已知结果 A ∩ B ⊆ A(子集定理,第 (c) 部分)通过将 B 替换为 A 得出。第一部分b:证明 A⊆A∩A。固定任意对象 x 0 。假设 x 0 ∈ A。证明 x0∈A∩A。我们必须证明“x 0 ∈ A 且 x 0 ∈ A”。为此,[遵循 AND 模板] 首先证明 x 0 ∈ A,然后证明 x 0 ∈ A。在每个部分中,我们都已假设待证明的命题成立。

Part 2. We prove A A = A . Part 2a. We prove A A A . Fix an object y 0 . Assume y 0 A A . Prove y 0 A . We have assumed the truth of “ y 0 A or y 0 A .” By the truth table for OR, the truth of P P forces P to be true. So y 0 A must be true, as needed. [If we follow the templates mechanically, we could also have finished using a proof by cases. Case 1 assumes y 0 A and proves y 0 A ; case 2 assumes y 0 A and proves y 0 A . It seemed easier here to invoke the truth table instead.] Part 2b. We prove A A A . This follows from the known result A A B (Theorem on Subsets, part (e)) by replacing B by A .

第二部分:证明 A∪A=A。第二部分a:证明 A∪A⊆A。固定一个对象 y 0 。假设 y 0 ∈ A ∪ A。证明 y 0 ∈ A。我们假设“y 0 ∈ A 或 y 0 ∈ A”为真。根据 OR 的真值表,P∨P 为真则 P 为真。因此,y 0 ∈ A 也必须为真,符合要求。[如果我们机械地遵循模板,也可以使用分情况证明来完成。情况 1 假设 y 0 ∈ A 并证明 y 0 ∈ A;情况 2 假设 y 0 ∈ A,并证明 y 0 ∈ A。这里似乎更容易使用真值表。] 第 2b 部分。我们证明 A⊆A∪A。这可以从已知的结果 A⊆A∪B(子集定理,第 (e) 部分)中通过将 B 替换为 A 得出。

3.23 Proof that A = . Part 1. We prove A . This follows from the lower bound property (Theorem on Subsets, part (c)). Part 2. We prove A . This follows from the minimality of ∅ (Theorem on Subsets, part (g)).

3.23 证明 A∩∅=∅。第一部分:我们证明 A∩∅⊆∅。这由下界性质(子集定理,第 (c) 部分)得出。第二部分:我们证明 ∅⊆A∩∅。这由 ∅ 的极小性(子集定理,第 (g) 部分)得出。

The previous proof showed that a certain set was empty by using the standard set equality proof template. However, there is another way to prove that a set is empty that is much more frequently used. It is based on part (j) of the Theorem on Set Equality, which we now prove.

前面的证明使用了标准的集合相等性证明模板,证明了某个集合为空集。然而,还有一种更常用的证明集合为空集的方法。这种方法基于集合相等性定理的第(j)部分,我们现在就来证明它。

3.24. Detecting the Empty Set. We prove: B iff z , z B . The other assertion in part (j) of the theorem follows from this one by denying each side of the IFF-statement. [The following proof may be considered optional.]

3.24. 检测空集。我们证明:B≠∅ 当且仅当存在 z, z∈B。定理 (j) 部分的另一个断言由此断言通过否定 IFF 语句的每一边得出。[以下证明可视为可选。]

Part 1. Assume B . Prove z , z B . Negating the original definition of set equality, we have assumed that x , x B x . Let x 0 be a particular object that makes “ x 0 B XOR x 0 ” true. By definition of the empty set, x 0 is false. So, by definition of XOR, x 0 B is true. Now choose z to be x 0 to see that z , z B is true.

第一部分:假设 B≠∅。证明 ∃z,z∈B。否定集合相等的原始定义,我们假设 ∃x,x∈B⊕x∈∅。设 x 0 是一个特定的对象,使得“x 0 ∈ B XOR x0∈∅”为真。根据空集的定义,x0∈∅ 为假。因此,根据异或的定义,x 0 ∈ B 为真。现在选择 z 为 x 0 ,即可证明 ∃z,z∈B 为真。

Part 2. Assume z , z B . Prove B . We have assumed there is a fixed object z 0 with z 0 B . We must prove x , x B x . Choose x = z 0 . As in part 1, z 0 B is true and z 0 is false, making the required XOR-statement true.

第二部分:假设存在 z, z∈B。证明 B≠∅。我们假设存在一个固定对象 z 0 ,且 z 0 ∈ B。我们需要证明存在 x, x∈B ⊕ x∈∅。选择 x = z 0 。与第一部分类似,z 0 ∈ B 为真,z0∈∅ 为假,因此所需的异或语句为真。

The equivalence of the two boxed statements in 3.24 is used constantly in proofs involving the empty set, so it should be memorized carefully. (The equivalence should be intuitively evident as well: a set B is not empty iff there exists at least one thing that is a member of B .) We can use this equivalence to prove that a set is empty using proof by contradiction, as illustrated next.

3.24 节中两个方框语句的等价性在涉及空集的证明中经常用到,因此应该牢记。(这种等价性也应该在直觉上显而易见:集合 B 非空当且仅当存在至少一个元素属于 B。)我们可以利用这种等价性,通过反证法证明一个集合为空,如下例所示。

3.25 Second Proof that A = . Assume, to get a contradiction, that A . Our result above shows that there is an object z 0 A . By definition of ∩, we deduce z 0 A and z 0 . But also z 0 by definition of ∅. So we have the contradiction “ z 0 and z 0 .” This contradiction shows that A = is true.

3.25 证明 A∩∅=∅ 的第二个方法。假设 A∩∅≠∅,以得出矛盾。我们上面的结果表明存在对象 z0∈A∩∅。根据 ∩ 的定义,我们推导出 z 0 ∈ A 且 z0∈∅。但根据 ∅ 的定义,z0∉∅。因此,我们得到矛盾“z0∈∅ 且 z0∉∅”。这个矛盾证明了 A∩∅=∅ 为真。

We introduce the following terminology for sets that do not overlap.

我们引入以下术语来描述不重叠的集合。

3.26. Definition: Disjoint Sets. We say sets S and T are d i s j o i n t iff S T = . We say sets S 1 , S 2 , …, S n are pairwise disjoint iff for all i , j ∈ {1, 2, …, n } with i j , S i S j = .

3.26. 定义:不相交集。我们称集合 S 和 T 不相交,当且仅当 S∩T=∅。我们称集合 S 1 , S 2 , …, S n 两两不相交,当且仅当对于所有 i, j ∈ {1, 2, …, n} 且 i ≠ j,Si∩Sj=∅。

For example, part (m) of the Theorem on Set Equality says that A B and A B are disjoint sets whose union is A .

例如,集合相等定理的 (m) 部分指出 A − B 和 A ∩ B 是不相交的集合,它们的并集是 A。

Section Summary

章节概要

1. Proving A B vs. Knowing A B . To prove A B : fix x 0 ; assume x 0 A ; prove x 0 B . When you already know A B , knowing x 0 A allows you to deduce x 0 B ; and knowing y 0 B allows you to deduce y 0 A .

1.证明 A ⊆ B 与已知 A ⊆ B。要证明 A ⊆ B:固定 x 0 ;假设 x 0 ∈ A;证明 x 0 ∈ B。当你已经知道 A ⊆ B 时,知道 x 0 ∈ A 可以推导出 x 0 ∈ B;并且知道 y0∉B 可以推导出 y0∉A。

2. Set Equality. For all sets A , B , A = B iff x , x A x B .

2.集合相等。对于所有集合 A、B,A=B 当且仅当 ∀x,x∈A⇔x∈B。

Equivalently, A = B iff A B and B A .

等价地,A=B 当且仅当 A⊆B 且 B⊆A。

To prove two sets A and B are equal: 1. Prove A B . 2. Prove B A .

要证明两个集合 A 和 B 相等:1. 证明 A ⊆ B。2. 证明 B ⊆ A。

3. Theorem on Set Equality. Memorize the facts listed in the Theorem on Set Equality for later use. In particular, set equality is reflexive, symmetric, and transitive; equal sets may be substituted for one another in expressions; set intersection and set union obey the commutative, associative, distributive, idempotent, absorption, and de Morgan laws; any set A is the union of the disjoint sets A B and A B ; the four statements A B , A B = A , A B = B , A B = are all equivalent.

3.集合相等定理。记住集合相等定理中列出的事实,以便日后使用。特别地,集合相等具有自反性、对称性和传递性;相等的集合可以在表达式中相互替换;集合的交集和并集满足交换律、结合律、分配律、幂等律、吸收律和德摩根定律;任何集合 A 都是不相交集合 A − B 和 A ∩ B 的并集;四个命题 A ⊆ B、A∩B=A、A∪B=B、A−B=∅ 都是等价的。

4. Proofs Involving the Empty Set. For any set B , prove B = by proving x , x B . Prove B by proving x , x B . In general, statements involving the empty set may be easier to prove using negative logic (e.g., proof by contrapositive or proof by contradiction).

4.涉及空集的证明。对于任意集合 B,证明 B=∅ 的方法是证明 ∀x,x∉B。证明 B≠∅ 的方法是证明 ∃x,x∈B。一般来说,涉及空集的命题可能更容易用否定逻辑(例如,逆否命题证明或反证法)来证明。

Exercises

练习

1. Prove part (d) of the Theorem on Subsets: for all sets A , B , C , C A B iff ( C A and C B ).

1.证明子集定理的(d)部分:对于所有集合 A、B、C,C⊆A∩B 当且仅当(C ⊆ A 且 C ⊆ B)。

2. Prove part (b) of the Theorem on Set Equality: for all sets A , B , if A = B , then B = A .

2.证明集合相等定理的(b)部分:对于所有集合 A、B,如果 A = B,则 B = A。

3. Prove part (e) of the Theorem on Set Equality: for all sets A , B , A B = B A and A B = B A .

3.证明集合相等定理的(e)部分:对于所有集合 A、B,A∩B=B∩A 和 A∪B=B∪A。

4. Prove: for all sets A , A = A .

4.证明:对于所有集合A,A−∅=A。

5. (a) Prove: for all sets A , B , C , if C B , then A B and C are disjoint. (b) Use (a) to prove: for all sets A and B , A B and B A are disjoint.

5.(a) 证明:对于任意集合 A、B、C,如果 C ⊆ B,则 A − B 和 C 不相交。(b) 利用 (a) 证明:对于任意集合 A 和 B,A − B 和 B − A 不相交。

6. Prove: for all sets A and B , A B = A ( A B ) .

6.证明:对于所有集合 A 和 B,A−B=A−(A∩B)。

7. Give intuition for parts (l), (m), and (n) in the Theorem on Set Equality by drawing appropriate Venn diagrams. Do these diagrams prove these set identities?

7. 通过绘制适当的维恩图,对集合等式定理中的 (l)、(m) 和 (n) 部分给出直观解释。这些图能证明这些集合恒等式吗?

8. Prove part (m) of the Theorem on Set Equality: for all sets A and B , A = ( A B ) ( A B ) and ( A B ) ( A B ) = .

8.证明集合相等定理的(m)部分:对于所有集合 A 和 B,A=(A−B)∪(A∩B) 且 (A−B)∩(A∩B)=∅。

9. Prove part (n) of the Theorem on Set Equality: for all sets A , B , if B A , then A − ( A B ) = B and A = B ( A B ) and B ( A B ) = . (Use other parts of the theorem, such as parts (m) and (o), to obtain a shorter proof.)

9.证明集合相等定理的(n)部分:对于任意集合A和B,如果B⊆A,则A−(A−B)=B,且A=B∪(A−B),B∩(A−B)=∅。(利用定理的其他部分,例如(m)和(o)部分,可以得到更简洁的证明。)

10. Prove: for all sets R and S , the sets R S and S R and R S are pairwise disjoint, and the union of these sets is R S . Illustrate this result with a Venn diagram.

10.证明:对于任意集合R和S,集合R-S、S-R和R∩S两两不相交,且这些集合的并集为R∪S。用维恩图说明此结果。

11. Prove: for all sets A , B , C , D , if A B C D , then A B , B C , C D , D are pairwise disjoint, and the union of these sets is A . Illustrate this result with a Venn diagram.

11.证明:对于任意集合 A、B、C、D,如果 A⊇B⊇C⊇D,则 A − B、B − C、C − D、D 两两不相交,且这四个集合的并集为 A。用维恩图说明此结果。

12. Use a Venn diagram to express A B C as the union of: (a) two disjoint sets; (b) three disjoint sets; (c) seven disjoint sets.

12.用维恩图表示 A∪B∪C 为下列各组的并集:(a)两个不相交的集合;(b)三个不相交的集合;(c)七个不相交的集合。

13. Use the Theorem on Set Equality ( not definitions or proof templates) to prove the following identities hold for all sets. Justify each step by citing a part of the theorem.

13. 使用集合相等定理(而非定义或证明模板)证明下列恒等式对所有集合都成立。请引用定理的一部分来解释每一步。

(a) ( R S ) ( R S ) = R S .

(a)(R∪S)∪(R∩S)=R∪S。

(b) R ( ( S T ) U ) = ( ( R S ) ( R T ) ) ( R U ) .

(b)R∪((S∩T)∩U)=((R∪S)∩(R∪T))∩(R∪U)。

14. Use Venn diagrams to disprove each statement. Discuss why Venn diagrams are allowed here, despite earlier claims that set identities could not be proved via Venn diagrams.

14. 使用维恩图反驳每个陈述。讨论为什么这里允许使用维恩图,尽管之前有人声称集合恒等式不能用维恩图证明。

(a) For all sets A and C , C = A ( C A ) .

(a)对于所有集合 A 和 C,C=A∪(C−A)。

(b) For all sets A , B , C , A − ( B C ) = ( A B ) − C .

(b)对于所有集合 A、B、C,A − (B − C) = (A − B) − C。

(c) For all sets A , B , C , A ( B C ) = ( A B ) ( A C ) .

(c)对于所有集合 A、B、C,A∪(B∩C)=(A∩B)∪(A∩C)。

15. Disprove each statement in the preceding problem by constructing specific examples where every set is a subset of {1, 2, 3, 4}.

15.通过构造具体例子来反驳前面问题中的每个陈述,其中每个集合都是{1, 2, 3, 4}的子集。

16. Find and prove necessary and sufficient conditions on the sets A and C for C = A ( C A ) to be true.

16.找出并证明集合 A 和 C 满足 C=A∪(C−A) 的必要和充分条件。

17. Find and prove necessary and sufficient conditions on the sets A , B , and C for A − ( B C ) = ( A B ) − C to be true.

17.找出并证明集合 A、B 和 C 满足 A − (B − C) = (A − B) − C 的必要和充分条件。

18. (a) Prove: for all sets A , B , C , ( A B ) ( B C ) A C .

18.(a)证明:对于所有集合A、B、C,(A−B)∪(B−C)⊇A−C。

(b) Disprove: for all sets A , B , C , ( A B ) ( B C ) = A C .

(b)反证:对于所有集合 A、B、C,(A−B)∪(B−C)=A−C。

(c) Prove: for all sets A , B , C , if C B A , then ( A B ) ( B C ) = A C .

(c)证明:对于所有集合 A、B、C,如果 C ⊆ B ⊆ A,则 (A−B)∪(B−C)=A−C。

(d) Prove or disprove the converse of the statement in (c).

(d)证明或反驳(c)中陈述的逆命题。

19. (a) Prove: for all sets A , B , and C , if A C = B C , then A C = B C .

19.(a)证明:对于所有集合A、B和C,如果A − C = B − C,则A∪C=B∪C。

(b) Prove or disprove the converse of part (a).

(b)证明或反驳(a)部分的逆命题。

20. Prove or disprove: for all sets A , B , and C , if A C = B C and A C = B C , then A = B .

20.证明或反证:对于所有集合 A、B 和 C,如果 A − C = B − C 且 A∩C=B∩C,则 A = B。

21. Prove or disprove: for all sets A , B , and C , if A C = B C and C A = C B , then A = B .

21.证明或反证:对于所有集合 A、B 和 C,如果 A − C = B − C 且 C − A = C − B,则 A = B。

22. Find and prove necessary and sufficient conditions on the sets A , B , and C for A ( B C ) = ( A B ) C to be true.

22.找出并证明集合 A、B 和 C 满足 A∪(B∩C)=(A∪B)∩C 的必要和充分条件。

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3.3 Set Equality Proofs, Circle Proofs, and Chain Proofs

3.3 集合等式证明、圆证明和链式证明

Set equality proofs occur everywhere in mathematics, so we keep studying them in this section. A set equality is a particular kind of IFF-statement. In the second part of the section, we develop two new ways for proving IFF-statements — the circle proof method and the chain proof method — that can help prove set equalities and other biconditional statements.

集合等式的证明在数学中无处不在,因此本节将继续探讨它们。集合等式是一种特殊的因果关系式(IFF)命题。在本节的第二部分,我们将介绍两种证明因果关系式的新方法——圆证明法和链证明法——它们可以帮助证明集合等式和其他双条件命题。

Set Equality Proofs, Continued

集合相等性证明(续)

The most common method for proving A = B , where A and B are sets, is to use the set equality proof template:

证明集合 A 和 B 相等的最常用方法是使用集合相等性证明模板:

Part 1. Prove A B .

第一部分:证明 A ⊆ B。

Part 2. Prove B A .

第二部分。证明 B ⊆ A。

In this section, we continue practicing this proof method, but we also introduce several other ways of proving that two sets are equal. We first recall one such alternate method that is often used to prove that a given set equals the empty set.

在本节中,我们将继续练习这种证明方法,同时还会介绍几种证明两个集合相等的其他方法。首先,我们回顾一种常用的替代方法,这种方法通常用于证明给定的集合等于空集。

3.27 To Prove B = ∅ (where B is a set):

3.27 证明 B = ∅(其中 B 是一个集合):

Assume , to get a contradiction, that B .

为了得出矛盾,假设 B≠∅。

We have assumed that there is an object z 0 with z 0 B . Expand the definition of “ z 0 B ” to derive a contradiction.

我们假设存在一个对象 z 0 ,其中 z 0 ∈ B。扩展“z 0 ∈ B”的定义,得出矛盾。

Conclude that B = ∅.

得出结论:B = ∅。

3.28. Example. Prove that for all sets A , A A = .

3.28. 例。证明对于所有集合 A,A−A=∅。

Proof. Assume, to get a contradiction, that A A . Then there exists an object z 0 A A . By definition of set difference, z 0 A and z 0 A . This is a contradiction, so A A = .

证明。假设 A−A≠∅,以得出矛盾。那么存在一个对象 z 0 ∈ A − A。根据集合差的定义,z 0 ∈ A 且 z0∉A。这与假设矛盾,因此 A−A=∅。

3.29. Practice with Set Proofs. Try to prove the following three statements yourself, before reading the answers given below. Let A , B , and C be fixed, but arbitrary sets. Here, you may use any parts of the Theorem on Subsets except the result being proved.

3.29. 集合证明练习。在阅读下面的答案之前,请尝试自己证明以下三个命题。设 A、B 和 C 为任意集合。在这里,你可以使用子集定理的任何部分,但不能使用待证明的结论。

(a) Prove: If A B , then A C B C .

(a)证明:如果 A ⊆ B,则 A∪C⊆B∪C。

(b) Prove: A = and A = A .

(b)证明:∅−A=∅ 且 A=A∪∅。

(c) Prove: A ( B C ) = ( A B ) ( A C ) .

(c)证明:A−(B∪C)=(A−B)∩(A−C)。

3.30. Answer to Practice Proof. (a). [Use the IF-template to prove an IF-statement.] Assume A B . Prove A C B C . [Now use the subset template to prove the new goal.] Fix an arbitrary object x 0 . Assume x 0 A C . Prove x 0 B C . We have assumed x 0 A or x 0 C , and we must prove x 0 B or x 0 C . [Since an OR-statement was assumed, use cases.]

3.30. 练习证明答案。(a). [使用 IF 模板证明一个 IF 语句。] 假设 A ⊆ B。证明 A∪C⊆B∪C。[现在使用子集模板证明新的目标。] 固定一个任意对象 x 0 。假设 x 0 ∈ A ∪ C。证明 x 0 ∈ B ∪ C。我们假设 x 0 ∈ A 或 x 0 ∈ C,我们必须证明 x 0 ∈ B 或 x 0 ∈ C。[由于假设了一个 OR 语句,因此使用条件判断。]

Case 1. Assume x 0 A ; prove x 0 B or x 0 C . Since A B is known, we deduce that x 0 B , so “ x 0 B or x 0 C ” is true by definition of OR.

情况 1. 假设 x 0 ∈ A;证明 x 0 ∈ B 或 x 0 ∈ C。由于 A ⊆ B 已知,我们推断 x 0 ∈ B,因此根据 OR 的定义,“x 0 ∈ B 或 x 0 ∈ C”为真。

Case 2. Assume x 0 C ; prove x 0 B or x 0 C . This follows by definition of OR.

情况 2. 假设 x 0 ∈ C;证明 x 0 ∈ B 或 x 0 ∈ C。这由 OR 的定义得出。

3.31. Answer to Practice Proof. (b). Part 1. Prove A = . Assume, to get a contradiction, that A . Then there exists z 0 A . By definition of set difference, z 0 and z 0 A . But, we also know that z 0 . The contradiction “ z 0 and z 0 ” proves that A = .

3.31. 练习证明答案。(b) 第一部分。证明 ∅−A=∅。假设 ∅−A≠∅,以得出矛盾。那么存在 z0∈∅−A。根据集合差的定义,z0∈∅ 且 z0∉A。但是,我们也知道 z0∉∅。矛盾“z0∈∅ 且 z0∉∅”证明了 ∅−A=∅。

Part 2. Prove A = A . Part 2a. Prove A A . This is a known theorem (take B = ∅ in part (e) of the Theorem on Subsets). Part 2b. Prove A A . Since ∅ appears here, we try a contradiction proof. Assume, to get a contradiction, that A A . Then there exists z 0 with z 0 A and z 0 A . We know “ z 0 A or z 0 ” is true. However, z 0 A is false by assumption, and z 0 is false by definition of the empty set. So, “ z 0 A or z 0 ” is false. This contradiction proves part 2b.

第二部分:证明 A=A∪∅。第二部分a:证明 A⊆A∪∅。这是一个已知定理(在子集定理的(e)部分中取 B = ∅)。第二部分b:证明 A∪∅⊆A。由于 ∅ 出现在这里,我们尝试反证法。假设 A∪∅⊈A,为了得到矛盾。那么存在 z 0 ,使得 z0∈A∪∅ 且 z0∉A。我们知道“z 0 ∈ A 或 z0∈∅”为真。然而,根据假设,z 0 ∈ A 为假,并且根据空集的定义,z0∈∅ 也为假。因此,“z 0 ∈ A 或 z0∈∅”为假。这一矛盾证明了第 2b 部分。

3.32. Answer to Practice Proof. (c). Part 1. Prove A ( B C ) ( A B ) ( A C ) . Fix x 0 ; assume x 0 A ( B C ) ; prove x 0 ( A B ) ( A C ) . We have assumed x 0 A and x 0 B C ; we must prove x 0 A B and x 0 A C . Continuing to expand definitions, we have assumed x 0 A and ( x 0 B AND x 0 C ); we must prove x 0 A and x 0 B and x 0 A and x 0 C . All four parts of the new goal have already been assumed, so part 1 is done.

3.32. 练习证明答案。(c). 第 1 部分。证明 A−(B∪C)⊆(A−B)∩(A−C)。固定 x 0 ;假设 x0∈A−(B∪C);证明 x0∈(A−B)∩(A−C)。我们假设 x 0 ∈ A 且 x0∉B∪C;我们必须证明 x 0 ∈ A − B 且 x 0 ∈ A − C。继续扩展定义,我们假设 x 0 ∈ A 且 (x0∉B 且 x0∉C);我们必须证明 x 0 ∈ A 且 x0∉B 且 x 0 ∈ A 且 x0∉C。新目标的四个部分都已经确定,所以第一部分已经完成。

Part 2. Prove ( A B ) ( A C ) A ( B C ) . Fix y 0 ; assume y 0 ( A B ) ( A C ) ; prove y 0 A ( B C ) . Expanding definitions as in part 1, we eventually see that we have assumed ( y 0 A y 0 B ) ( y 0 A y 0 C ) , and we must prove ( y 0 A ( y 0 B y 0 C ) ) . All three parts of this AND-statement have already been assumed, so part 2 is done.

第二部分:证明 (A−B)∩(A−C)⊆A−(B∪C)。固定 y 0 ;假设 y0∈(A−B)∩(A−C);证明 y0∈A−(B∪C)。展开定义,如同第一部分,我们最终发现我们已经假设了 (y0∈A∧y0∉B)∧(y0∈A∧y0∉C),因此我们必须证明 (y0∈A∧(y0∉B∧y0∉C))。这个 AND 语句的所有三个部分都已经假设,因此第二部分完成。

Circle Proof Method

圆证明法

Consider part (o) of the Theorem on Set Equality. This item asserts that four propositions (call them P , Q , R , and S ) are all equivalent. More explicitly, part (o) says that

考虑集合等式定理的第 (o) 部分。该部分断言四个命题(分别称为 P、Q、R 和 S)都是等价的。更具体地说,第 (o) 部分指出:

( P P Q ) ( Q R ) ( R S S ) ( S S P P ) ( P P R ) ( Q S S ) .

Using the AND-template and IFF-template to prove this, we would need to perform twelve independent subproofs:

使用 AND 模板和 IFF 模板来证明这一点,我们需要执行十二个独立的子证明:

1a. Prove P Q .      1b. Prove Q P .

1a. 证明 P ↠ Q。 1b. 证明 Q ↠ P。

2a. Prove Q R .      2b. Prove R Q .

2a. 证明 Q ↠ R。2b. 证明 R ↠ Q。

3a. Prove R S .      3b. Prove S R .

3a. 证明 R ↠ S。3b. 证明 S ↠ R。

4a. Prove S P .      4b. Prove P S .

4a.证明 S ↠ P. 4b。证明 P ↠ S。

5a. Prove P R .      5b. Prove R P .

5a. 证明 P ↠ R。5b. 证明 R ↠ P。

6a. Prove Q S .      6b. Prove S Q .

6a. 证明 Q ↠ S。6b. 证明 S ↠ Q。

That is a lot of work! It turns out, however, that there is a shorter way. Suppose we give proofs for Steps 1a, 2a, 3a, and 4a, so we know P Q and Q R and R S and S P . You can check that

这工作量可不小!不过,实际上还有更简便的方法。假设我们给出了步骤 1a、2a、3a 和 4a 的证明,那么我们知道 P ↠ Q 且 Q ↠ R 且 R ↠ S 且 S ↠ P。你可以验证这一点。

( ( P P Q ) ( Q R ) ) ( P P R )

is a tautology (called “Transitivity of ↠”). So, once we complete Steps 1a and 2a, the Inference Rule for IF tells us that P R is true, completing Step 5a. Using this fact and Step 3a, the Inference Rule for IF tells us that P S is true, completing Step 4b. Repeatedly using Transitivity of IF and the Inference Rule for IF, we see that Steps 2a and 3a and 4a yield Q S and Q P , which takes care of Step 6a and Step 1b. Continuing in this way, we see that after proving Steps 1a, 2a, 3a, and 4a, the eight remaining statements all follow automatically. Thus it suffices to prove the four statements in 1a, 2a, 3a, and 4a. This is called the circle method for proving the equivalence of four statements, because of the following picture:

这是一个重言式(称为“↠的传递性”)。因此,一旦我们完成步骤1a和2a,IF推理规则告诉我们P ↠ R为真,从而完成步骤5a。利用这一事实和步骤3a,IF推理规则告诉我们P ↠ S为真,从而完成步骤4b。反复使用IF的传递性和IF推理规则,我们发现步骤2a、3a和4a得出Q ↠ S和Q ↠ P,这就满足了步骤6a和步骤1b的要求。以此类推,我们发现,在证明步骤1a、2a、3a和4a之后,剩余的八个语句都自动成立。因此,只需证明步骤1a、2a、3a和4a中的四个语句即可。这被称为证明四个命题等价性的圆法,因为如下图所示:

P P Q P P Q implies 暗示 S S R S S R .

Following the single arrows (known IF-statements) around the left circle and using Transitivity of IF, we deduce the bidirectional arrows (IFF-statements to be proved) appearing in the right circle, together with the diagonals P R and Q S . The circle method generalizes to any number of statements, as follows.

沿着左侧圆圈上的单箭头(已知的 IF 语句)并利用 IF 的传递性,我们可以推导出右侧圆圈中出现的双向箭头(待证明的 IFF 语句),以及对角线 P ⇔ R 和 Q ⇔ S。圆圈法可以推广到任意数量的语句,如下所示。

3.33 Circle Method to Prove P 1 , P 2 , …, P n are Logically Equivalent.

3.33 圆法证明 P 1 , P 2 , …, P n 在逻辑上等价。

Part 1. Prove P 1 P 2 (by any method).

第 1 部分。证明 P 1 ↠ P 2 (用任何方法)。

Part 2. Prove P 2 P 3 (by any method).

第 2 部分。证明 P 2 ↠ P 3 (用任何方法)。

Part i . Prove P i P i +1 (by any method).

第一部分。证明 P i ↠ P i +1 (用任何方法)。

Part n . Prove P n P 1 (by any method).

第 n 部分。证明 P n ↠ P 1 (用任何方法)。

Let us illustrate the circle method by proving part (o) of the Set Equality Theorem.

让我们通过证明集合等式定理的第(o)部分来说明圆的方法。

3.34. Proof of Subset Characterization. Let A and B be fixed, but arbitrary sets.

3.34. 子集刻画的证明。设 A 和 B 为固定的任意集合。

Part 1. Prove: A B A B = A .

第一部分。证明:A⊆B⇒A∩B=A。

Part 2. Prove: A B = A A B = B .

第二部分。证明:A∩B=A⇒A∪B=B。

Part 3. Prove: A B = B A B = .

第三部分。证明:A∪B=B⇒A−B=∅。

Part 4. Prove: A B = A B .

第四部分。证明:A−B=∅⇒A⊆B。

[Stop reading here, and try to fill in the proofs of each part yourself. Hints: Use the Theorem on Subsets when you can; try proof by contradiction in part 3; try a contrapositive proof in part 4. Remember that sets already known to be equal have the same members.]

【请在此处停止阅读,并尝试自行完成各部分的证明。提示:尽可能使用子集定理;尝试在第 3 部分使用反证法;尝试在第 4 部分使用逆否命题法。记住,已知相等的集合具有相同的元素。】

Proof of Part 1. Assume A B . Prove A B = A .

第一部分证明。假设 A ⊆ B。证明 A∩B=A。

Part 1a. Prove A B A . This is already known (part (c) of the Theorem on Subsets).

第 1a 部分。证明 A ∩ B ⊆ A。这已经是已知的(子集定理的 (c) 部分)。

Part 1b. Prove A A B . Fix an arbitrary object x 0 . Assume x 0 A . Prove x 0 A B . We must prove x 0 A and x 0 B . First, prove x 0 A ; note we have already assumed this. Second, prove x 0 B . To do so, recall we have assumed A B and x 0 A , from which x 0 B follows [by the knowledge template for subsets].

第 1b 部分。证明 A⊆A∩B。固定一个任意对象 x 0 。假设 x 0 ∈ A。证明 x0∈A∩B。我们必须证明 x 0 ∈ A 且 x 0 ∈ B。首先,证明 x 0 ∈ A;注意我们已经假设了这一点。其次,证明 x 0 ∈ B。为此,回想一下我们已经假设了 A ⊆ B 且 x 0 ∈ A,由此可知 x 0 ∈ B [根据子集的知识模板]。

Proof of Part 2. Assume A B = A . Prove A B = B .

第二部分证明。假设 A∩B=A。证明 A∪B=B。

Part 2a. Prove A B B . Fix an arbitrary object y 0 . Assume y 0 A B . Prove y 0 B . We know y 0 A or y 0 B , so use cases within part 2a.

第 2a 部分。证明 A∪B⊆B。固定任意对象 y 0 。假设 y 0 ∈ A ∪ B。证明 y 0 ∈ B。我们知道 y 0 ∈ A 或 y 0 ∈ B,因此使用第 2a 部分中的例子。

Case 1. Assume y 0 A ; prove y 0 B . Since y 0 A and A B = A is known, we deduce that y 0 A B . Since A B B is known, we deduce that y 0 B as needed.

情况 1. 假设 y 0 ∈ A;证明 y 0 ∈ B。由于 y 0 ∈ A 且 A∩B=A 已知,我们推断 y0∈A∩B。由于 A ∩ B ⊆ B 已知,我们推断 y 0 ∈ B,满足要求。

Case 2. Assume y 0 B ; prove y 0 B . This is immediate.

情况 2. 假设 y 0 ∈ B;证明 y 0 ∈ B。这是显而易见的。

Part 2b. Prove B A B . This is known from part (e) of the Theorem on Subsets.

第 2b 部分。证明 B⊆A∪B。这可由子集定理的第 (e) 部分得出。

Proof of Part 3. Assume, to get a contradiction, that A B = B and A B . The second assumption tells us that there is an object z 0 with z 0 A B . This means z 0 A and z 0 B . On one hand, since z 0 A and A A B , we see that z 0 A B . We have also assumed A B = B , so z 0 B follows. We now have the contradiction “ z 0 B and z 0 B ,” which completes the proof of part 3.

第三部分的证明。为了得出矛盾,我们假设 A∪B=B 且 A−B≠∅。第二个假设告诉我们存在一个对象 z 0 ,使得 z 0 ∈ A − B。这意味着 z 0 ∈ A 且 z0∉B。一方面,由于 z 0 ∈ A 且 A⊆A∪B,我们可知 z 0 ∈ A ∪ B。我们还假设 A∪B=B,因此 z 0 ∈ B。现在我们得到了矛盾“z 0 ∈ B 且 z0∉B”,至此第三部分的证明完成。

Proof of Part 4. Using the contrapositive method, assume A B ; prove A B . We have assumed there exists an object z 0 such that z 0 A z 0 B . We must prove w , w A B , which means w , w A w B . Choosing w = z 0 completes the proof.

第四部分证明。利用逆否命题法,假设 A⊈B;证明 A−B≠∅。我们假设存在一个对象 z 0 使得 z0∈A∧z0∉B。我们需要证明 ∃w,w∈A−B,即 ∃w,w∈A∧w∉B。选择 w = z 0 即可完成证明。

Having proved part (o) of the Theorem on Set Equality, we can quickly deduce the absorption laws (part (i)) as follows.

证明了集合等式定理的第(o)部分之后,我们可以迅速推导出吸收定律(第(i)部分)如下。

3.35. Proof of Absorption Laws. Let X and Y be arbitrary sets; we prove X ( X Y ) = X and X ( X Y ) = X . Since X Y X is already known, part (o) of the theorem (with A there replaced by X Y , and with B there replaced by X ) tells us that ( X Y ) X = X . Assuming commutativity of union has already been proved, we conclude that X ( X Y ) = X . Similarly, since X X Y is already known, part (o) gives X ( X Y ) = X , which is the second absorption law.

3.35. 吸收律的证明。设 X 和 Y 为任意集合;我们证明 X∪(X∩Y)=X 且 X∩(X∪Y)=X。由于 X ∩ Y ⊆ X 已知,定理 (o) 部分(将 A 替换为 X ∩ Y,将 B 替换为 X)得出 (X∩Y)∪X=X。假设并集的交换律已证明,我们得出 X∪(X∩Y)=X。类似地,由于 X⊆X∪Y 已知,定理 (o) 部分得出 X∩(X∪Y)=X,即第二吸收律。

Chain Proof Method

链式验证方法

Suppose we are trying to prove an algebraic identity of the form a = c , where a and c are expressions denoting real numbers. A common way to proceed is to write down a sequence of known identities that looks like this:

假设我们要证明一个形如 a = c 的代数恒等式,其中 a 和 c 是表示实数的表达式。一种常见的方法是列出一系列已知的恒等式,如下所示:

We know 我们知道 a 一个 = b b 1 b b 1 = b b 2 b b 2 = b b 3 b b 3 = b b 4 b b 4 = c c .

By repeatedly using the theorem “ x , y , z R , if x = y and y = z , then x = z ,” we can conclude that a = c . This theorem is almost never mentioned explicitly; instead, we abbreviate the above discussion by presenting a single chain of known equalities

通过反复使用定理“∀x,y,z∈R,如果x=y且y=z,则x=z”,我们可以得出a=c。这个定理几乎从未被明确提及;相反,我们通过给出一个已知的等式链来简化上述讨论。

a 一个 = b b 1 = b b 2 = b b 3 = b b 4 = c c

that begins with a (the left side of the equality to be proved) and ends with c (the right side of the equality to be proved). We state this method as a new proof template.

它以 a(待证明等式的左侧)开头,以 c(待证明等式的右侧)结尾。我们将此方法表述为一个新的证明模板。

3.36. Chain Proof Template to Prove a = c . To prove a = c , we can present a chain of known equalities

3.36. 链式证明模板:证明 a = c。为了证明 a = c,我们可以给出一个已知等式的链。

a 一个 = b b 1 = b b 2 = b b 3 = = b b n n 1 = b b n n = c c

starting at a and ending at c . The letters a , b 1 , …, b n , c represent any expressions denoting objects of the same type (e.g., sets, numbers, functions, etc.).

从 a 开始,到 c 结束。字母 a、b 1 、…、b n 、c 表示表示相同类型对象(例如,集合、数字、函数等)的任何表达式。

3.37. Example. We use a chain proof to prove x , y R , ( x y ) ( x + y ) = x 2 y 2 . Fix arbitrary x , y ∈ ℝ. We compute

3.37. 示例。我们使用链式证明来证明 ∀x,y∈R,(x−y)(x+y)=x²−y²。固定任意 x, y ∈ ℝ。我们计算

( x x y ) ( x x + y ) = x x ( x x + y ) y ( x x + y ) = x x x x + x x y y x x y y = x x 2 + x x y x x y y 2 = x x 2 + 0 y 2 = x x 2 y 2 .

Each equality in the chain is true because of a law of algebra; for instance, the first two equalities hold by the distributive law, and the third holds by commutativity of multiplication and the definitions x 2 = xx and y 2 = yy . We conclude that ( x y )( x + y ) = x 2 y 2 , as needed.

链中的每个等式都成立,因为存在代数定律;例如,前两个等式由分配律成立,第三个等式由乘法交换律以及定义 x 2 = xx 和 y 2 = yy 成立。我们得出结论:(x − y)(x + y) = x 2 − y 2 ,符合要求。

Now let us return to the problem of proving IFF-statements. You can check that

现在让我们回到证明逆正则表达式的问题。你可以验证……

( ( P P Q ) ( Q R ) ) ( P P R )

is a tautology (called “Transitivity of ⇔”). By analogy with what we did above for equality proofs, we can formulate the following chain method for proving IFF-statements.

这是一个重言式(称为“⇔的传递性”)。类比我们上面对等式证明所做的工作,我们可以为证明IFF语句制定以下链式方法。

3.38. Chain Proof Template to Prove P R . To prove the IFF-statement P R , we can present a chain of known IFF-statements

3.38. 证明 P ⇔ R 的链式证明模板。为了证明 IFF 语句 P ⇔ R,我们可以给出一个已知的 IFF 语句链。

P P Q 1 Q 2 Q 3 Q n n 1 Q n n R

starting at P and ending at R .

从 P 点开始,到 R 点结束。

Note that the displayed formula is really an abbreviation for

请注意,显示的公式实际上是以下公式的缩写:

( P P Q 1 ) ( Q 1 Q 2 ) ( Q n n R ) .

For instance, P Q R (with no parentheses) is being used to abbreviate ( P Q ) ( Q R ) . This abbreviating convention is dangerous, since (for example) the propositional form ( P Q ) ⇔ R is not logically equivalent to ( P Q ) ( Q R ) . However, expressions such as ( P Q ) ⇔ R seldom arise in practice, and context frequently tells us when this abbreviation is being used. To illustrate the chain method, we prove part (f) of the Theorem on Set Equality.

例如,P ⇔ Q ⇔ R(不带括号)常被用来简写 (P⇔Q)∧(Q⇔R)。这种简写约定是危险的,因为(例如)命题形式 (P ⇔ Q) ⇔ R 在逻辑上并不等价于 (P⇔Q)∧(Q⇔R)。然而,像 (P ⇔ Q) ⇔ R 这样的表达式在实践中很少出现,而且上下文通常可以告诉我们何时使用了这种简写。为了说明链式方法,我们证明集合等式定理的第 (f) 部分。

3.39. Proof of Associativity of ∩ and ∪. Let A , B , C be fixed sets. We must prove ( A B ) C = A ( B C ) . By the original definition of set equality, we must prove:

3.39. ∩ 和 ∪ 结合律的证明。设 A、B、C 为固定的集合。我们需要证明 (A∩B)∩C=A∩(B∩C)。根据集合相等的原始定义,我们需要证明:

x x , x x ( A 一个 B B ) C C x x A 一个 ( B B C C ) .

Let x 0 be a fixed object. To prove the IFF-statement for x 0 , we present the following chain of known equivalences:

设 x 0 为一个固定对象。为了证明 x 0 的 IFF 语句成立,我们给出以下已知的等价关系链:

x x 0 ( A 一个 B B ) C C ( x x 0 A 一个 B B ) x x 0 C C ( x x 0 A 一个 x x 0 B B ) x x 0 C C x x 0 A 一个 ( x x 0 B B x x 0 C C ) x x 0 A 一个 ( x x 0 B B C C ) x x 0 A 一个 ( B B C C ) .

The third equivalence is true by the known associativity of AND (see the Theorem on Logical Equivalences). All the other equivalences are instances of the definition of ∩. We can prove associativity of ∪ in the same way, by replacing ∩ by ∪ and ∧ by ∨ everywhere in the proof above.

第三个等价关系成立,是因为已知“与”运算具有结合律(参见逻辑等价定理)。所有其他等价关系都是 ∩ 定义的实例。我们可以用同样的方法证明 ∪ 的结合律,只需将上述证明中的 ∩ 替换为 ∪,将 ∧ 替换为 ∨ 即可。

We can give similar chain proofs for several other parts of the Set Equality Theorem (including commutativity, the distributive laws, and the absorption laws), by reducing the identities for sets to corresponding known equivalences from the Theorem on Logical Equivalences. This proof technique vividly reveals how properties of the logical operators lead directly to analogous properties of set operations.

我们可以通过将集合恒等式简化为逻辑等价定理中相应的已知等价关系,对集合等价定理的其他几个部分(包括交换律、分配律和吸收律)给出类似的链式证明。这种证明技巧生动地揭示了逻辑运算符的性质如何直接导出集合运算的类似性质。

Transitivity and Chain Proofs (Optional)

传递性和链式证明(可选)

At this point, we have encountered several mathematical relations that have a property called transitivity . We give an abstract definition of this property later, when we study relations (see §6.1 ). We now list some specific examples of the transitive law occurring in algebra, set theory, and logic.

至此,我们已经接触到几个具有传递性这一性质的数学关系。我们将在稍后学习关系时(参见§6.1)给出这一性质的抽象定义。现在,我们列举一些在代数、集合论和逻辑中出现的传递律的具体例子。

3.40. Transitivity Properties.

3.40. 传递性性质。

(a) Transitivity of Equality: For all objects a , b , c , if a = b and b = c , then a = c .

(a)等式的传递性:对于所有对象 a、b、c,如果 a = b 且 b = c,则 a = c。

(b) Transitivity of ≤: For all a , b , c ∈ ℝ, if a b and b c , then a c .

(b)≤ 的传递性:对于所有 a、b、c ∈ ℝ,如果 a ≤ b 且 b ≤ c,则 a ≤ c。

(c) Transitivity of >: For all a , b , c ∈ ℝ, if a > b and b > c , then a > c .

(c)> 的传递性:对于所有 a、b、c ∈ ℝ,如果 a > b 且 b > c,则 a > c。

(d) Transitivity of Subsets: For all sets A , B , C , if A B and B C , then A C .

(d)子集的传递性:对于所有集合 A、B、C,如果 A ⊆ B 且 B ⊆ C,则 A ⊆ C。

(e) Transitivity of IF: For all propositions P , Q , R , if P Q and Q R , then P R .

(e)IF 的传递性:对于所有命题 P、Q、R,如果 P ↠ Q 且 Q ↠ R,则 P ↠ R。

(f) Transitivity of IFF: For all propositions P , Q , R , if P Q and Q R , then P R .

(f)IFF 的传递性:对于所有命题 P、Q、R,如果 P ⇔ Q 且 Q ⇔ R,则 P ⇔ R。

Whenever we have a transitive relation, we can use the chain proof technique to prove that this relation holds between two given objects. We have already discussed chain proofs of equalities and IFF-statements. To give a chain proof of an inequality such as a > c , we present a chain of known inequalities

每当我们遇到传递关系时,我们都可以使用链式证明技巧来证明该关系在两个给定对象之间成立。我们已经讨论过等式和不等式正则表达式的链式证明。为了给出诸如 a > c 之类的不等式的链式证明,我们首先给出一个已知的不等式链。

a 一个 > b b 1 > b b 2 > b b 3 > > b b n n > c c ,

which abbreviates “ a > b 1 and b 1 > b 2 and b 2 > b 3 and ... and b n > c .” Using Transitivity of > repeatedly, we successively deduce a > b 2 , then a > b 3 , and so on, until finally we obtain the needed conclusion a > c . To prove A C by the chain proof method, we find a chain of known set inclusions

这可以简写为“a > b 1 且 b 1 > b 2 且 b 2 > b 3 且 ... 且 b n > c”。利用“>”的传递性,我们依次推导出 a > b 2 ,然后是 a > b 3 ,依此类推,直到最终得到所需的结论 a > c。为了用链式证明法证明 A ⊆ C,我们找到一个已知的集合包含链。

A 一个 B B 1 B B 2 B B 3 B B n n C C .

To prove P R by the chain proof method, we find a chain of known IF-statements

为了用链式证明法证明 P ⇒ R,我们找到一个已知的 IF 语句链。

P P Q 1 Q 2 Q 3 Q n n R .

Beware: the chain proof technique is very convenient when it succeeds, but this proof method cannot be used in all situations. When trying to prove a complicated set inclusion or set equality, it is often safer to use the subset proof template or set equality template rather than attempting a chain proof.

注意:链式证明法在成功时非常方便,但这种证明方法并非适用于所有情况。在尝试证明复杂的集合包含关系或集合相等关系时,通常使用子集证明模板或集合相等性模板比尝试链式证明更为安全。

Section Summary

章节概要

1. Circle Proof Method. To prove that statements P 1 , P 2 , …, P n are all logically equivalent, we can prove P 1 P 2 and P 2 P 3 and ... and P n −1 P n and P n P 1 . Reordering the original list of statements is allowed and may make the implications easier to prove.

1. 循环证明法。为了证明语句 P 1 , P 2 , …, P n 在逻辑上等价,我们可以证明 P 1 ↠ P 2 且 P 2 ↠ P 3 且 ... 且 P n −1 ↠ P n 且 P n ↠ P 1 。允许对原始语句列表进行重新排序,这可能会使蕴含关系更容易证明。

2. Chain Proof Method for Equalities. To prove an equality a = c , find a chain of known equalities a = b 1 = b 2 = · · · = b n = c starting at a and ending at c .

2.等式链证明法。要证明等式 a = c,找到一个从 a 开始到 c 结束的已知等式链:a = b 1 = b 2 = · · · = b n = c。

3. Chain Proof Method for IFF Statements. To prove an IFF statement P R , find a chain of known equivalences P Q 1 Q 2 ⇔ · · · ⇔ Q n R starting at P and ending at R .

3. 因果命题语句的链式证明法。要证明因果命题语句 P ⇔ R,找到一条从 P 开始到 R 结束的已知等价链 P ⇔ Q 1 ⇔ Q 2 ⇔ · · · ⇔ Q n ⇔ R。

4. Chain Proof Method for Transitive Relations. Whenever we have a transitive relation (such as =, ≤, <, ≥, >, ⊆, ↠, or ⇔), we can prove the relation holds between given objects A and B by finding a finite chain of known relations leading from A to B through intermediate objects. Each step in the chain should be explained by citing known results.

4.传递关系的链式证明法。对于任何传递关系(例如 =、≤、<、≥、>、⊆、↠ 或 ⇔),我们可以通过找到一条有限的已知关系链来证明给定对象 A 和 B 之间的关系成立。这条链从 A 出发,经过中间对象,最终到达 B。链中的每一步都应该用已知的结论来解释。

5. Advice for Set Proofs. When asked to prove a complicated statement involving sets, steadfastly keep applying the proof templates for subsets, set equality, set operations, and the logical operators as they arise within the statement. Remember the special templates involving the empty set, and be ready to try contrapositive proofs or proof by contradiction. Some set identities can be reduced to logical identities by using a chain proof.

5.集合证明的建议。当被要求证明一个涉及集合的复杂命题时,要始终如一地运用子集、集合相等、集合运算以及逻辑运算符的证明模板,直到命题完成。记住涉及空集的特殊模板,并准备好尝试逆否命题证明或反证法。一些集合恒等式可以通过链式证明转化为逻辑恒等式。

Exercises

练习

1. Outline a proof of the following statement using the circle method. For all sets X ⊆ ℝ n , the following statements are equivalent: X is compact; X is sequentially compact; X is closed and bounded; X is complete and totally bounded.

1. 使用圆法证明以下命题。对于所有集合 X ⊆ ℝ n ,以下命题等价:X 是紧集;X 是序列紧集;X 是闭集且有界集;X 是完备集且完全有界集。

2. Given statements P , Q , R , S , T , suppose we have proved P Q and Q S and S P and T R and R P . Have we proved that all five statements are logically equivalent? If not, find one additional IF-statement that will complete the proof; give all possible answers.

2. 给定命题 P、Q、R、S、T,假设我们已经证明了 P ↠ Q 且 Q ↠ S 且 S ↠ P 且 T ↠ R 且 R ↠ P。我们是否证明了这五个命题逻辑等价?如果不是,请找出一条额外的 IF 语句来完成证明;并给出所有可能的答案。

3. Repeat the previous question assuming we have proved P R and Q R and S R and T S .

3.假设我们已经证明了 P ⇔ R 且 Q ⇔ R 且 S ⇔ R 且 T ⇔ S,重复上一个问题。

4. Give a chain proof of the commutative laws for set union and intersection (Theorem on Set Equality, part (e)).

4.给出集合并集和交集交换律的链式证明(集合相等定理,第(e)部分)。

5. Give a chain proof of the distributive laws for set union and intersection (Theorem on Set Equality, part (g)).

5.给出集合并集和交集分配律的链式证明(集合相等定理,第(g)部分)。

6. Give chain proofs of the following algebraic identities. (You will need the distributive law and similar facts.)

6. 给出下列代数恒等式的链式证明。(你需要用到分配律及类似性质。)

(a) x , y R , ( x + y ) 2 = x 2 + 2 x y + y 2 .

(a)∀x,y∈R,(x+y)2=x2+2xy+y2。

(b) x , y R , ( x y ) ( x 2 + x y + y 2 ) = x 3 y 3 .

(b)∀x,y∈R,(x−y)(x2+xy+y2)=x3−y3。

7. Let x be a fixed positive real number. Prove the following statements are equivalent using a circle proof: x > 1; x 2 > x ; 1/ x 2 < 1/ x ; 1/ x < 1.

7.设 x 为一个固定的正实数。用圆证明法证明下列语句等价:x > 1;x > x;1/x < 1/x;1/x < 1。

8. Let x be a fixed negative real number. Which pairs of statements in the previous exercise are logically equivalent?

8.设 x 为一个固定的负实数。上题中哪些语句对在逻辑上是等价的?

9. Using any method, prove this part of the Theorem on Set Equality: for all sets A , B , C , A ( B C ) = ( A B ) ( A C ) .

9.使用任何方法证明集合相等定理的这一部分:对于所有集合 A、B、C,A−(B∩C)=(A−B)∪(A−C)。

10. Give direct proofs of the implications A B A B = B and A B = A A B in part (o) of the Theorem on Set Equality.

10.给出集合等式定理(o)部分中蕴含式 A⊆B⇒A∪B=B 和 A∩B=A⇒A⊆B 的直接证明。

11. Prove: for all sets A , B , C , ( A B ) C = A ( B C ) = ( A C ) B .

11.证明:对于所有集合 A、B、C,(A−B)−C=A−(B∪C)=(A−C)−B。

12. Prove part (d) of the Theorem on Set Equality (substitution properties).

12.证明集合等式定理(代换性质)的(d)部分。

13. Prove: for all sets A and B , the following statements are equivalent: A B = ; A B = A ; B A = B ; C A , D B , C D = .

13.证明:对于所有集合 A 和 B,下列命题等价:A∩B=∅;A − B = A;B − A = B;∀C⊆A,∀D⊆B,C∩D=∅。

14. Prove: for all x ∈ ℝ ≠0 , the following statements are equivalent: x is not rational; x /3 is not rational; 3/ x is not rational; (3/ x ) + 2 is not rational; 2 x + 3 is not rational.

14.证明:对于所有 x ∈ ℝ ≠0 ,下列陈述等价:x 不是有理数;x/3 不是有理数;3/x 不是有理数;(3/x) + 2 不是有理数;2x + 3 不是有理数。

15. Define the symmetric difference of two sets A and B as follows:

15.定义两个集合A和B的对称差如下:

x x , x x A 一个 B B x x A 一个 x x B B .

Informally, A B consists of all objects that belong to exactly one of the sets A and B . Prove the following identities hold for all sets A , B , and C .

通俗地说,A△B 由恰好属于集合 A 和 B 之一的所有对象组成。证明下列恒等式对所有集合 A、B 和 C 都成立。

(a) A B = B A (commutativity).

(a)A△B=B△A(交换律)。

(b) ( A B ) C = A ( B C ) (associativity).

(b)(A△B)△C=A△(B△C)(结合律)。

(c) A = A = A (identity).

(c)A△∅=A=∅△A(恒等式)。

(d) ( A B ) C = ( A C ) ( B C ) (distributive law).

(d)(A△B)∩C=(A∩C)△(B∩C)(分配律)。

16. Illustrate the identities in the previous exercise (as best you can) using Venn diagrams.

16.用维恩图尽可能地说明上一个练习中的各个恒等式。

17. Let A and B be arbitrary sets.

17.设 A 和 B 为任意集合。

(a) Prove: A B = ( A B ) ( B A ) .

(a)证明:A△B=(A−B)∪(B−A)。

(b) Prove: A B = ( A B ) ( A B ) .

(b)证明:A△B=(A∪B)−(A∩B)。

18. Prove: A , ! B , A B = = B A .

18.证明:∀A,∃!B,A△B=∅=B△A。

19. Prove or disprove: For all A , B , C , x , x ( A B ) C iff x is a member of exactly one of the sets A , B , C .

19.证明或反证:对于所有 A、B、C、x,x∈(A△B)△C 当且仅当 x 属于集合 A、B、C 中的一个。

20. (a) Is ∨ (OR) transitive? In other words, is it always true that if A B and B C , then A C ? Prove your answer.

20.(a) ∨(或)是否具有传递性?换句话说,如果 A∨B 且 B∨C,那么 A∨C 是否总是成立?请证明你的答案。

(b) Decide (with proof) whether ∧ (AND) is transitive.

(b)判断(并证明)∧(与)是否具有传递性。

(c) Decide (with proof) whether (XOR) is transitive.

(c)判断(并证明)⊕(异或)是否具有传递性。

3.4 Small Sets and Power Sets

3.4 小型套装和动力套装

In our initial informal discussion of set theory, we used the notation {1, 4, 5, 7} to describe a set whose members were 1, 4, 5, and 7. In this section, we give a precise formal definition of this notation. This enables us to understand several key technical issues involving sets — whether order and repetition matter in sets, how sets may appear as members of other sets, and the distinction between ∈ (the set membership relation) and ⊆ (the subset relation). We also discuss the power set of a set X , which is the set of all subsets of X .

在我们最初对集合论的非正式讨论中,我们使用符号 {1, 4, 5, 7} 来描述一个包含元素 1、4、5 和 7 的集合。在本节中,我们将给出该符号的精确形式化定义。这使我们能够理解集合相关的几个关键技术问题——集合中元素的顺序和重复是否重要,集合如何作为其他集合的成员出现,以及 ∈(集合成员关系)和 ⊆(子集关系)之间的区别。我们还将讨论集合 X 的幂集,即 X 的所有子集构成的集合。

Small Sets

小套装

We have already introduced the informal notation U = { z 1 , z 2 , …, z n } to represent a finite set whose members are z 1 , z 2 , …, z n (and nothing else). We now give more formal definitions of what this notation means, for specific small values of n . To see why this formal definition is necessary, consider the following questions.

我们之前已经引入了非正式符号 U = {z 1 , z 2 , …, z n 来表示一个有限集合,其成员为 z 1 , z 2 , …, z n (除此之外没有其他元素)。现在,我们将针对特定的较小 n 值,给出该符号的更正式定义。为了理解为什么需要这个正式定义,请考虑以下问题。

(a) Is {1, 2} = {2, 1}? (Does order matter in a set?)

(a){1, 2} 是否等于 {2, 1}?(集合中的元素顺序重要吗?)

(b) Is {3} = {3, 3, 3}? (Does repetition matter in a set?)

(b) {3} = {3, 3, 3} 吗?(集合中元素的重复重要吗?)

(c) Is { } = ? (Is a set containing the empty set the same as the empty set?)

(c) {∅}=∅吗?(包含空集的集合是否等同于空集?)

(d) Is {1, {2, 3}} = {1, 2, 3}? (Do nested braces change the set?)

(d) {1, {2, 3}} = {1, 2, 3} 吗?(嵌套的大括号会改变集合吗?)

To give incontrovertible answers to these questions, we need formal definitions to work with. As before, we define new sets by specifying precisely which objects are members of these sets.

为了对这些问题给出无可辩驳的答案,我们需要形式化的定义。和之前一样,我们通过精确指定哪些对象是这些集合的成员来定义新的集合。

3.41. Definitions of Small Sets.

3.41. 小集合的定义。

(a) Empty Set: For all z , z . Using curly brace notation: for all z , z { } .

(a)空集:对于所有z,z∉∅。使用花括号表示法:对于所有z,z∉{}。

(b) Singleton Sets: For all a , z , z { a } iff z = a .

(b)单元素集:对于所有 a、z、z∈{a} 当且仅当 z=a。

(c) Unordered Pairs: For all a , b , z { a , b } iff z = a or z = b .

(c)无序对:对于所有 a、b、z∈{a,b} 当且仅当 z=a 或 z=b。

(d) Unordered Triples: For all a , b , c , z , z { a , b , c } iff z = a or z = b or z = c .

(d)无序三元组:对于所有 a、b、c、z,z∈{a,b,c} 当且仅当 z=a 或 z=borz=c。

We can give analogous definitions for { a , b , c , d }, { a , b , c , d , e }, etc.

我们可以对 {a, b, c, d}、{a, b, c, d, e} 等给出类似的定义。

We have called { a , b } the unordered pair with members a and b , suggesting that order does not matter in a set. We can prove this fact now.

我们将集合中的元素 a 和 b 称为无序对 {a, b},这表明集合中的顺序无关紧要。现在我们可以证明这一点。

3.42. Theorem. For all a , b , { a , b } = { b , a }.

3.42. 定理。对于所有 a, b,{a, b} = {b, a}。

Proof. Fix arbitrary objects a , b . We must prove z , z { a , b } z { b , a } . We use a chain proof. Fix an arbitrary object z 0 . We know

证明。固定任意对象 a 和 b。我们需要证明 ∀z, z∈{a, b} ⇔ z∈{b, a}。我们使用链式证明。固定任意对象 z 0 。我们知道

z z 0 { a 一个 , b b } ( z z 0 = a 一个 z z 0 = b b ) ( z z 0 = b b z z 0 = a 一个 ) z z 0 { b b , a 一个 } .

The first and third equivalences come from the definition of unordered pair, and the middle equivalence is an instance of the tautology ( P Q ) ⇔ ( Q P ).

第一个和第三个等价关系来自无序对的定义,中间的等价关系是重言式 (P∨Q) ⇔ (Q∨P) 的一个实例。

An analogous proof shows that for all objects a , b , c ,

类似的证明表明,对于所有对象 a、b、c,

{ a 一个 , b b , c c } = { a 一个 , c c , b b } = { b b , a 一个 , c c } = { b b , c c , a 一个 } = { c c , a 一个 , b b } = { c c , b b , a 一个 } ,

and similarly for larger sets. In general, displaying the members of a set in a different order does not change the set . Later, we discuss concepts (such as ordered pairs, ordered triples, and sequences) in which the order does make a difference.

对于更大的集合,情况也类似。一般来说,改变集合中元素的排列顺序并不会改变集合本身。稍后,我们将讨论一些概念(例如有序对、有序三元组和序列),在这些概念中,元素的排列顺序确实会产生影响。

Next consider the issue of repetition of elements. The definition of unordered pair allows the possibility that a = b . In this case, we can prove that the unordered pair is the same as a singleton set.

接下来考虑元素重复的问题。无序对的定义允许 a = b 的情况。在这种情况下,我们可以证明该无序对等同于单元素集。

3.43. Theorem. For all a , { a , a } = { a }.

3.43. 定理。对于所有 a,{a, a} = {a}。

Proof. Fix an arbitrary object a . We must prove z , z { a , a } z { a } . We again use a chain proof. Fix an arbitrary object z 0 . We know

证明。固定任意对象 a。我们必须证明 ∀z,z∈{a,a}⇔z∈{a}。我们再次使用链式证明。固定任意对象 z 0 。我们知道

z z 0 { a 一个 , a 一个 } ( z z 0 = a 一个 z z 0 = a 一个 ) ( z z 0 = a 一个 ) z z 0 { a 一个 } .

The equivalences follow by definition of unordered pair, by the tautology ( P P ) ⇔ P , and by definition of singleton sets.

这些等价关系可以通过无序对的定义、重言式 (P∨P) ⇔ P 以及单元素集的定义得出。

Similarly, we can prove that for all a , b , c , { a } = { a , a , a , a }, { a , b , c } = { a , b , b , c , c , c }, and so on. In general, listing the members of a set with repetitions does not change the set . Combining this with the previous observation, we can summarize by saying that

类似地,我们可以证明对于所有 a、b、c,{a} = {a, a, a, a},{a, b, c} = {a, b, b, c, c, c},依此类推。一般来说,列出集合中可重复出现的元素并不会改变该集合。结合之前的观察,我们可以总结如下:

Order and repetition do not matter for sets;

对于组来说,顺序和重复并不重要;

only membership in the set matters when deciding set equality.

判断集合是否相等,只有集合的成员关系才重要。

Next, let us turn to the nuances of the curly brace notation for small sets. Our first result shows that adding new braces to a set can change the set.

接下来,我们来探讨一下小集合的花括号表示法的细微差别。我们的第一个结果表明,向集合中添加新的花括号会改变该集合。

3.44. Theorem. { } and { { } } { } .

3.44. 定理。{∅}≠∅ 且 {{∅}}≠{∅}。

Proof. We first prove { } . To do so, we prove z , z { } . Choose z = ∅; we must prove { } . By definition of singleton, we must prove = , which is a known fact (reflexivity of set equality).

证明。我们首先证明 {∅}≠∅。为此,我们证明存在 z,z∈{∅}。取 z = ∅;我们必须证明 ∅∈{∅}。根据单元素集合的定义,我们必须证明 ∅=∅,这是一个已知事实(集合等式的自反性)。

Next we prove { { } } { } . Note that we can prove a set inequality A B by finding a member of A that is not a member of B . In this case, A is { { } } and B is {∅}. Consider z = { } . On one hand, by definition of a singleton set, z A (i.e., { } { { } } ) because { } = { } . On the other hand, using the result proved in the last paragraph, z B (i.e., { } { } ) since { } .

接下来我们证明 {{∅}}≠{∅}。注意,我们可以通过找到集合 A 中不属于集合 B 的一个元素来证明集合不等式 A ≠ B。在本例中,A 为 {{∅}},B 为 {∅}。考虑 z={∅}。一方面,根据单元素集合的定义,z ∈ A(即,{∅}∈{{∅}}),因为 {∅}={∅}。另一方面,利用上一段证明的结果,z∉B(即,{∅}∉{∅}),因为 {∅}≠∅。

If you find the proof just given to be cryptic, here is some intuition that may help. A set (collection of objects) should not be confused with the members of the set. In particular, a singleton set { b } containing a single object b is not the same thing as the object b itself. For example, the set {George Washington} is not the same object as George Washington (a set is not a person). We might think of a set as a box that contains its members (in no particular order). Thus, {1, 2, 3} is a box containing the three numbers 1, 2, 3 and nothing else. On the other hand, S = {1, {2, 3}} is a box containing two objects: the number 1, and a second box that itself contains the numbers 2 and 3. Question: is 2 ∈ S ? The answer may be unclear using the box analogy, so we refer to the formal definition. We know 2 ∈ {1, {2, 3}} iff (2 = 1 or 2 = {2, 3}). Neither alternative is true, so 2 is not a member of S . However, 2 is a member of one of the members of S .

如果你觉得刚才的证明晦涩难懂,这里有一些直观的解释或许会有帮助。集合(对象的集合)不应与集合中的元素混淆。特别地,包含单个对象 b 的单元素集合 {b} 与对象 b 本身并非同一事物。例如,集合 {George Washington} 与乔治·华盛顿并非同一对象(集合并非人)。我们可以将集合想象成一个盒子,里面装着它的元素(顺序不限)。因此,{1, 2, 3} 就是一个只包含数字 1、2、3 的盒子,除此之外别无其他。另一方面,S = {1, {2, 3}} 是一个包含两个对象的盒子:数字 1,以及另一个包含数字 2 和 3 的盒子。问题:2 是否属于 S?使用盒子类比可能难以得出明确答案,因此我们参考正式定义。我们知道 2 ∈ {1, {2, 3}} 当且仅当 2 = 1 或 2 = {2, 3}。这两种情况都不成立,所以 2 不是 S 的成员。然而,2 是 S 中某个成员的成员。

Similarly, T = {{1, 2, 3}} is a set with just one member, namely the set {1, 2, 3}. We have 1 T , 2 T , 3 T , but {1, 2, 3} ∈ T . We think of T as a box containing another box containing the numbers 1, 2, and 3. Now, the empty set (which can be denoted by ∅ or {}) can be viewed as an empty box — a box that contains nothing. This set is not the same as {∅}, which is a box containing an empty box. Similarly, { { } } = { { { } } } is a box containing a box containing an empty box. The set { , { } } is a box containing two things: an empty box, and a box containing an empty box. And so on. In summary, nested braces are important when determining set membership; only items at the outermost level are considered members of the outermost set. Sets may contain other sets as members.

类似地,T = {{1, 2, 3}} 是一个只有一个元素的集合,即集合 {1, 2, 3}。我们有 1∉T,2∉T,3∉T,但 {1, 2, 3} ∈ T。我们可以把 T 想象成一个盒子,里面装着另一个盒子,盒子里装着数字 1、2 和 3。现在,空集(可以用 ∅ 或 {} 表示)可以看作是一个空盒子——一个什么都不包含的盒子。这个集合与 {∅} 不同,{∅} 是一个盒子,里面装着一个空盒子。类似地,{{∅}}={{{}}} 是一个盒子,里面装着一个盒子,盒子里装着一个空盒子。集合 {∅,{∅}} 是一个盒子,里面装着两个东西:一个空盒子,以及一个盒子,盒子里装着一个空盒子。以此类推。总之,嵌套的大括号在确定集合成员关系时非常重要;只有最外层的元素才被认为是该最外层集合的成员。集合可以包含其他集合作为成员。

3.45. Example. Let U = {5.3, {7}, {8, 9, {10, 11}}, ℤ}. U is a finite set with four members. One member of U is the real number 5.3. Another member of U is a set whose only member is 7. Another member of U is an unordered triple whose three members are 8, 9, and the unordered pair {10, 11}. Another member of U is the infinite set ℤ. We could write

3.45. 示例。设 U = {5.3, {7}, {8, 9, {10, 11}}, ℤ}。U 是一个有限集,包含四个元素。U 的一个元素是实数 5.3。U 的另一个元素是一个集合,该集合只有一个元素 7。U 的另一个元素是一个无序三元组,其三个元素分别是 8、9 和无序对 {10, 11}。U 的另一个元素是无限集 ℤ。我们可以写出

U U = { { 7 } , Z Z , 5.3 , { 8 , { 11 , 10 } , 9 } , Z Z , 53 / 10 } , but U U { 5.3 , { 7 } , { 8 , 9 , 10 , 11 } , Z Z } .

Although 8 ∈ {8, 9, {10, 11}} and 8 ∈ ℤ, 8 U .

虽然 8 ∈ {8, 9, {10, 11}} 且 8 ∈ ℤ,但 8∉U。

Power Sets

力量集

Informally, for any set X , the power set P ( X ) is the set of all subsets of X . Formally, we define the power set by specifying its members, as follows.

通俗地说,对于任意集合 X,幂集 P(X) 是 X 的所有子集的集合。正式地说,我们通过指定其成员来定义幂集,如下所示。

3.46. Definition: Power Set. For all objects S and all sets X , S P X iff S X . So, the members of P ( X ) are precisely the subsets of X .

3.46. 定义:幂集。对于所有对象 S 和所有集合 X,S∈PX 当且仅当 S⊆X。因此,P(X) 的元素恰好是 X 的子集。

3.47. Example. You can check that

3.47. 示例。您可以检查一下

P P ( { 7 , 8 , 9 } ) = { , { 7 } , { 8 } , { 9 } , { 7 , 8 } , { 7 , 9 } , { 8 , 9 } , { 7 , 8 , 9 } } .

The right side is an eight-element set with members ∅, {7}, {8}, and so on. Changing the braces used here produces an incorrect statement. For instance, we know { 7 , 8 , 9 } by the Theorem on Subsets, and so P ( { 7 , 8 , 9 } ) . On the other hand, { } { 7 , 8 , 9 } , since ∅ is a member of {∅} that is not a member of {7, 8, 9}. So { } P ( { 7 , 8 , 9 } ) . Similarly, 7 { 7 , 8 , 9 } , since the number 7 is not even a set 1 . But {7}⊆{7, 8, 9}, since every member of {7} (namely, the number 7) is a member of {7, 8, 9}.

右侧是一个包含八个元素的集合,其成员包括 ∅、{7}、{8} 等等。改变此处使用的花括号会导致语句错误。例如,根据子集定理,我们知道 ∅⊆{7,8,9},因此 ∅∈P({7,8,9})。另一方面,{∅}⊈{7,8,9},因为 ∅ 是 {∅} 的一个成员,但它不是 {7, 8, 9} 的成员。所以 {∅}∉P({7,8,9})。类似地,7⊈{7,8,9},因为数字 7 甚至不是一个集合。但是 {7}⊆{7, 8, 9},因为 {7} 的每个成员(即数字 7)都是 {7, 8, 9} 的成员。

3.48. Example. What is P ( { , { } } ) ? In general, for any unordered pair { a , b }, we can see that

3.48. 示例。P({∅,{∅}}) 是什么?一般来说,对于任意无序对 {a, b},我们可以看到

P P ( { a 一个 , b b } ) = { , { a 一个 } , { b b } , { a 一个 , b b } } .

Making the replacements a = ∅ and b = { } , we conclude that

令 a = ∅ 和 b = {∅},我们得出以下结论:

P P ( { , { } } ) = { , { } , { { } } , { , { } } } .

Similary, since P ( { a } ) = { , { a } } , we see that P ( { } ) = { , { } } , P ( { { } } ) = { , { { } } } , and so on. Finally, what is P ( ) ? Answer: {∅}, not ∅.

类似地,由于 P({a})={∅,{a}},我们看到 P({∅})={∅,{∅}},P({{∅}})={∅,{{∅}}},依此类推。最后,P(∅) 是什么?答案:{∅},而不是 ∅。

We must take care not to confuse the meaning of T S and T S . The statement T S says that the object T (which might or might not be a set) is a member of the set S . The statement T S says that every member of the set T is also a member of the set S , i.e., z , z T z S . By definition, T S means the same thing as T P ( S ) , but in general, T S does not mean the same thing as T S .

我们必须注意不要混淆 T ∈ S 和 T ⊆ S 的含义。语句 T ∈ S 表示对象 T(它可能是集合,也可能不是)是集合 S 的元素。语句 T ⊆ S 表示集合 T 的每个元素也都是集合 S 的元素,即 ∀z,z∈T⇒z∈S。根据定义,T ⊆ S 等同于 T∈P(S),但一般来说,T ⊆ S 并不等同于 T ∈ S。

3.49. Example. Let U = { 5 , 6 , 7 , { 5 , 7 } , { 6 } , { { 5 } } , { { 5 , 7 } } , { 5 , { 6 } , 7 } , { 4 } , } . What is U P ( U ) ? Solution. For each fixed z , we know z U P ( U ) iff ( z U and z P ( U )) iff ( z U and z U ). Thus, we are looking for all objects z that are simultaneously members of U and subsets of U . Let us test each of the ten members of U to see which of them are also subsets of U . The first three displayed members of U , namely 5, 6 and 7, are numbers, not subsets. The next member of U is {5, 7}, which is also a subset of U because 5 ∈ U and 7 ∈ U . Similarly, {6} ∈ U and {6}⊆ U (because 6 ∈ U ). On the other hand, {{5}} ∈ U but { { 5 } } U because {5} is a member of {{5}} that is not a member of U . Next, {{5, 7}} ∈ U and also {{5, 7}}⊆ U , where the second fact holds because {5, 7} ∈ U . We see that {5, {6}, 7}⊆ U since each of the three members of the set on the left is a member of U . But { 4 } U since 4 U . Finally, ∅ ⊆ U . In summary,

3.49. 例。设 U={5,6,7,{5,7},{6},{{5}},{{5,7}},{5,{6},7},{4},∅}。U∩P(U) 是什么?解:对于每个固定的 z,我们知道 z∈U∩P(U) 当且仅当 (z ∈ U 且 z∈P(U)) 当且仅当 (z ∈ U 且 z ⊆ U)。因此,我们寻找所有既是 U 的成员又是 U 的子集的对象 z。让我们测试 U 的十个成员,看看哪些也是 U 的子集。U 的前三个成员,即 5、6 和 7,是数字,而不是子集。U 的下一个成员是 {5, 7},它也是 U 的子集,因为 5 ∈ U 且 7 ∈ U。类似地,{6} ∈ U 且 {6}⊆ U(因为 6 ∈ U)。另一方面,{{5}} ∈ U,但 {{5}}⊈U,因为 {5} 是 {{5}} 的元素,而 {{5}} 不是 U 的元素。接下来,{{5, 7}} ∈ U,并且 {{5, 7}}⊆ U,其中第二个结论成立是因为 {5, 7} ∈ U。我们看到 {5, {6}, 7}⊆ U,因为左侧集合中的每个元素都是 U 的元素。但是 {4}⊈U,因为 4∉U。最后,∅ ⊆ U。综上所述,

U U P P ( U U ) = { { 5 , 7 } , { 6 } , { { 5 , 7 } } , { 5 , { 6 } , 7 } , } .

3.50. Theorem on Power Sets (Optional). For all sets A and B :

3.50. 幂集定理(可选)。对于所有集合 A 和 B:

(a) P ( A ) and A P ( A ) .

(a)∅∈P(A)且A∈P(A)。

(b) A B iff P ( A ) P ( B ) .

(b)A ⊆ B 当且仅当 P(A)⊆P(B)。

(c) P ( A B ) = P ( A ) P ( B ) .

(c)P(A∩B)=P(A)∩P(B)。

(d) P ( A ) P ( B ) P ( A B ) .

(d)P(A)∪P(B)⊆P(A∪B)。

(e) P ( A B ) { } P ( A ) P ( B ) .

(e)P(A−B)−{∅}⊆P(A)−P(B)。

This theorem can be proved more quickly if we remember facts proved earlier in the Theorem on Subsets. We illustrate by proving parts (a), (b), and (d). Let A and B be fixed, but arbitrary sets. To prove part (a), we must prove ∅ ⊆ A and A A . Both parts are already known from the Theorem on Subsets. For part (b), we first prove that A B implies P ( A ) P ( B ) . Assume A B ; prove P ( A ) P ( B ) . [Continue with the subset template.] Fix an arbitrary object S ; assume S P ( A ) ; prove S P ( B ) . We have assumed that S A ; we must prove S B . Now, from S A and A B , we deduce that S B (by transitivity of ⊆, part of the Theorem on Subsets).

如果我们回顾一下子集定理中已证明的事实,就可以更快地证明这个定理。我们通过证明 (a)、(b) 和 (d) 部分来说明这一点。设 A 和 B 是任意集合。为了证明 (a) 部分,我们必须证明 ∅ ⊆ A 和 A ⊆ A。这两部分在子集定理中已经成立。对于 (b) 部分,我们首先证明 A ⊆ B 蕴含 P(A) ⊆ P(B)。假设 A ⊆ B;证明 P(A) ⊆ P(B)。[继续使用子集模板。] 固定一个任意对象 S;假设 S ∈ P(A);证明 S ∈ P(B)。我们已经假设 S ⊆ A;我们必须证明 S ⊆ B。现在,由 S ⊆ A 和 A ⊆ B,我们可以推断出 S ⊆ B(由 ⊆ 的传递性得出,这是子集定理的一部分)。

For the converse in part (b), we now assume that P ( A ) P ( B ) , and prove that A B . Fix x ; assume x A ; prove x B . [Here we need a creative step — how can we use our assumption about the power sets to gain information about x ? The key is to realize that { x } is a subset of A , hence is a member of P ( A ) .] We claim that { x }⊆ A . To prove this claim, fix y , assume y ∈ { x }, and prove y A . Our assumption tells us that y = x , so that y A by assumption on x . It follows from the claim and the definition of power set that { x } P ( A ) . We have assumed P ( A ) P ( B ) , so we deduce that { x } P ( B ) . This means { x }⊆ B . Since we know x ∈ { x } (as x = x ), we conclude that x B as needed.

对于 (b) 部分的逆命题,我们现在假设 P(A)⊆P(B),并证明 A ⊆ B。固定 x;假设 x ∈ A;证明 x ∈ B。[这里我们需要一个巧妙的步骤——如何利用关于幂集的假设来获取关于 x 的信息?关键在于认识到 {x} 是 A 的子集,因此它是 P(A) 的成员。] 我们断言 {x}⊆ A。为了证明这个断言,固定 y,假设 y ∈ {x},并证明 y ∈ A。我们的假设告诉我们 y = x,因此根据关于 x 的假设,y ∈ A。由断言和幂集的定义可知,{x}∈P(A)。我们已经假设 P(A)⊆P(B),所以我们推断 {x}∈P(B)。这意味着 {x}⊆ B。由于我们知道 x ∈ {x}(因为 x = x),我们得出 x ∈ B,这正是我们所需要的结论。

We continue by proving part (d). Fix an arbitrary object S ; assume S P ( A ) P ( B ) ; prove S P ( A B ) . We have assumed S P ( A ) or S P ( B ) , so use cases.

我们继续证明 (d) 部分。固定一个任意对象 S;假设 S∈P(A)∪P(B);证明 S∈P(A∪B)。我们已经假设 S∈P(A) 或 S∈P(B),因此使用个案。

Case 1. Assume S P ( A ) ; prove S P ( A B ) . Since A A B is known, we see from part (b) of the theorem that P ( A ) P ( A B ) . Now, because S P ( A ) , we deduce that S P ( A B ) .

情形 1. 假设 S∈P(A);证明 S∈P(A∪B)。由于 A⊆A∪B 已知,由定理的 (b) 部分可知 P(A)⊆P(A∪B)。现在,因为 S∈P(A),我们推断 S∈P(A∪B)。

Case 2. Assume S P ( B ) ; prove S P ( A B ) . Since B A B is known, we see from part (b) of the theorem that P ( B ) P ( A B ) . Now, because S P ( B ) , we deduce that S P ( A B ) .

情形 2. 假设 S∈P(B);证明 S∈P(A∪B)。由于 B⊆A∪B 已知,由定理的 (b) 部分可知 P(B)⊆P(A∪B)。现在,因为 S∈P(B),我们推断 S∈P(A∪B)。

This completes the proof of part (d). The next example shows that we cannot replace ⊆ by = in part (d).

至此,(d) 部分的证明完毕。下一个例子表明,在 (d) 部分中,我们不能用 = 代替 ⊆。

3.51. Example. Disprove: for all sets A and B , P ( A ) P ( B ) = P ( A B ) .

3.51. 示例。反证:对于所有集合 A 和 B,P(A)∪P(B)=P(A∪B)。

Solution. We must prove there exist sets A and B with P ( A ) P ( B ) P ( A B ) . Choose A = {1} and B = {2}. We know P ( A ) = { , { 1 } } and P ( B ) = { , { 2 } } , so

解:我们需要证明存在集合 A 和 B,使得 P(A)∪P(B)≠P(A∪B)。取 A = {1} 和 B = {2}。我们知道 P(A)={∅,{1}} 且 P(B)={∅,{2}},所以

P P ( A 一个 ) P P ( B B ) = { , { 1 } , { 2 } } .

On the other hand, A B = { 1 , 2 } , so

另一方面,A∪B={1,2},所以

P P ( A 一个 B B ) = { , { 1 } , { 2 } , { 1 , 2 } } .

We see that {1, 2} is a member of P ( A B ) that is not a member of P ( A ) P ( B ) , so these two sets are unequal.

我们看到 {1, 2} 是 P(A∪B) 的成员,但不是 P(A)∪P(B) 的成员,因此这两个集合不相等。

We conclude with one last example of the distinction between ∈ and ⊆. We have proved that for all sets X , X X . Using an axiom of set theory called the Axiom of Foundation , it can be proved that for all sets X , X X . (For details, see § 3.7 .) So, although every set is a subset of itself, no set is a member of itself. On the other hand, for all sets X , X ∈ { X } and X P ( X ) . Similarly, “for all sets X , ∅ ⊆ X ” is true, but “for all sets X , X ” is false.

最后,我们用一个例子来说明 ∈ 和 ⊆ 的区别。我们已经证明,对于所有集合 X,都有 X ⊆ X。利用集合论中的基础公理,可以证明对于所有集合 X,都有 X ∉ X。(详情参见 §3.7。)因此,虽然每个集合都是自身的子集,但没有哪个集合是自身的成员。另一方面,对于所有集合 X,都有 X ∈ {X} 且 X ∈ P(X)。类似地,“对于所有集合 X,∅ ⊆ X”为真,但“对于所有集合 X,∅ ∈ X”为假。

Section Summary

章节概要

1. Memorize these definitions:

1. 记住这些定义:

Empty Set: z , z .

空集:∀z,z∉∅。

Singleton Sets: For all a , z , z { a } iff z = a .

单元素集:对于所有 a、z、z∈{a} 当且仅当 z=a。

Unordered Pairs: For all a , b , z , z { a , b } iff z = a or z = b .

无序对:对于所有 a、b、z,z∈{a,b} 当且仅当 z=a 或 z=b。

Unordered Triples: For all a , b , c , z { a , b , c } iff z = a or z = b or z = c .

无序三元组:对于所有 a、b、c、z∈{a,b,c} 当且仅当 z=a 或 z=borz=c。

Power Set: For all S , X , S P X iff S X .

幂集:对于所有 S、X,S∈PX 当且仅当 S⊆X。

2. Order and repetition do not matter for sets; only membership in the set matters when deciding set equality. For example, {1, 2} = {2, 1} and {3} = {3, 3} and {4, 4, 4, 5, 5, 6} = {6, 6, 5, 4, 4}.

2.集合中元素的顺序和重复并不重要;判断集合是否相等时,只有元素是否属于该集合才重要。例如,{1, 2} = {2, 1},{3} = {3, 3},{4, 4, 4, 5, 5, 6} = {6, 6, 5, 4, 4}。

3. Sets may contain other sets as members. Nested braces are important when determining set membership; only items at the outermost level are considered members of the outermost set. Note carefully that ∅ and { } denote the empty set, but {∅} is a set with one member (namely, the empty set).

3. 集合可以包含其他集合作为成员。嵌套的大括号在确定集合成员关系时非常重要;只有最外层元素才被视为最外层集合的成员。请注意,∅ 和 { } 都表示空集,但 {∅} 是一个包含一个成员(即空集)的集合。

4. Take care not to confuse ∈ (the set membership relation) with ⊆ (the subset relation). Note x B means x is a member of B , whereas A B means every member of the set A is a member of the set B . In particular, A A is always false although A A is always true. For the set B = {4, 5, {6, 7}}, we have 4 ∈ B and 5 ∈ B and {4, 5}⊆ B and { 4 , 5 } B , whereas {6, 7} ∈ B and 6 B and 7 B and { 6 , 7 } B .

4. 注意不要将 ∈(集合成员关系)与 ⊆(子集关系)混淆。注意 x ∈ B 表示 x 是 B 的成员,而 A ⊆ B 表示集合 A 中的每个元素都是集合 B 的成员。特别地,A ∈ A 始终为假,而 A ⊆ A 始终为真。对于集合 B = {4, 5, {6, 7}},我们有 4 ∈ B 且 5 ∈ B 且 {4, 5} ⊆ B 且 {4, 5} ∉ B,而 {6, 7} ∈ B 且 6 ∉ B 且 7 ∉ B 且 {6, 7} ⊈ B。

Exercises

练习

1. (a) Carefully prove that for all a , b , c , { a , b , c } = { b , a , c } and { a , b , c } = { a , c , b }.

1.(a)仔细证明对于所有a、b、c,{a、b、c} = {b、a、c} 且 {a、b、c} = {a、c、b}。

(b) Use (a) to deduce that for all x , y , z , { x , y , z } = { x , z , y } = { y , x , z } = { y , z , x } = { z , x , y } = { z , y , x }.

(b)利用(a)推断出对于所有x、y、z,{x、y、z} = {x、z、y} = {y、x、z} = {y、z、x} = {z、x、y} = {z、y、x}。

2. (a) Prove: for all a , { a } = { a , a , a , a }.

2.(a)证明:对于所有a,{a} = {a, a, a, a}。

(b) Prove: for all a , b , c , { a , b , c } = { a , b , b , c , c , c }.

(b)证明:对于所有 a、b、c,{a、b、c} = {a、b、b、c、c、c}。

3. Is the set equality { a , { b }} = {{ a }, b } true for all objects a , b , for some objects a , b , or for no objects a , b ? Explain carefully.

3.集合等式{a, {b}} = {{a}, b}对于所有对象a、b、某些对象a、b,还是对于任何对象a、b都成立?请详细解释。

4. Let S = { , 8 , 7 / 5 , 2 , Z , Q , { R } , { 3 , 8 } , P ( Z ) } . (Assume numbers are not sets here.) (a) Find all x , y S such that x y . (b) Find all x , y S such that x y .

4.设 S={∅,8,7/5,2,Z,Q,{R},{3,8},P(Z)}。(假设这里的数字不是集合。)(a)求所有满足 x ∈ y 的 x, y ∈ S。(b)求所有满足 x ⊆ y 的 x, y ∈ S。

5. True or false? Proofs are not required, but expanding definitions may help you reach the correct answer.

5.判断题:对还是错?不需要证明,但扩展定义可能有助于你得出正确答案。

(a) {1, 2, 3, 4, 5} ≠ {2, 4, 5, 3, 1}.

(a){1, 2, 3, 4, 5} ≠ {2, 4, 5, 3, 1}。

(b) {1, 2, 3, 4} = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4}.

#

(b){1, 2, 3, 4} = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4}。

(c) 5 ∈ {5}.

(c)5 ∈ {5}。

(d) 5 ∈ {{5}}.

(d)5 ∈ {{5}}。

(e) {5} ∈ {5}.

(e){5} ∈ {5}。

(f) {5} ∈ {{5}}.

(f){5} ∈ {{5}}。

(g) 5 ∈ {5, {{5}}}.

(g)5 ∈ {5, {{5}}}。

(h) {5} ∈ {5, {{5}}}.

(h){5} ∈ {5, {{5}}}。

(i) {{5}} ∈ {5, {{5}}}.

(i){{5}} ∈ {5, {{5}}}。

(j) 5⊆{5, {{5}}}.

(j)5⊆{5, {{5}}}。

(k) {5}⊆{5, {{5}}}.

(k){5}⊆{5, {{5}}}。

(l) {{5}}⊆{5, {{5}}}.

(l){{5}}⊆{5, {{5}}}。

(m) {{{5}}}⊆{5, {{5}}}.

(m){{{5}}}⊆{5, {{5}}}。

(n) { 5 } P ( { 5 , { { 5 } } } ) .

(n){5}∈P({5,{{5}}})。

(o) { { 5 } } P ( { 5 , { { 5 } } } ) .

(o){{5}}∈P({5,{{5}}})。

(p) { { { 5 } } } P ( { 5 , { { 5 } } } ) .

(p){{{5}}}∈P({5,{{5}}}).

(q) { { 5 } } P ( { 5 , { { 5 } } } ) .

(q){{5}}⊆P({5,{{5}}}).

(r) { { { 5 } } } P ( { 5 , { { 5 } } } ) .

(r){{{5}}}⊆P({5,{{5}}})。

6. Compute each of the following power sets; be sure to use braces correctly. (a) P ( { 5 , 6 } ) (b) P ( { 5 , { 6 } } ) (c) P ( { { 5 , 6 } } ) (d) P ( { 5 , { 5 } , { 5 , { 5 } } } )

6. 计算下列各幂集;务必正确使用花括号。(a) P({5,6}) (b) P({5,{6}}) (c) P({{5,6}}) (d) P({5,{5},{5,{5}}})

7. (a) If S has four members, how many members does P ( S ) have?

7.(a)如果 S 有 4 个成员,那么 P(S) 有多少个成员?

(b) If S has five members, how many members does P ( S ) have?

(b)如果 S 有 5 个成员,那么 P(S) 有多少个成员?

8. Someone claims that { , } { } because a box containing two empty boxes is not the same as a box containing one empty box. Is this claim correct? Explain carefully.

8. 有人声称 {∅,∅}≠{∅},因为装有两个空盒子的盒子与装有一个空盒子的盒子并不相同。这种说法正确吗?请详细解释。

9. Let a , b , c be distinct objects that are not sets. Define

9.设 a、b、c 为不同的对象,且它们不是集合。定义

S S = { a 一个 , { b b } , c c , { { b b } , c c } , { a 一个 , c c } , { { a 一个 , c c } } , { { { a 一个 , c c } } } , { { a 一个 , c c } , b b } , { { b b } , a 一个 , { c c } } , { a 一个 , { b b } , c c } } .

(a) How many members does S have? How many members of S are sets?

(a)集合 S 有多少个成员?集合 S 中有多少个成员是集合?

(b) List all members of S P ( S ) .

(b)列出S∩P(S)的所有成员。

(c) List all members of S P ( S ) .

(c)列出 S∩P(S) 的所有成员。

(d) How many members does P ( S ) P ( S ) have?

(d)P(S)∩P(S)有多少个成员?

10. True or false? Explain each answer. Assume the variable X ranges over sets.

10.判断正误并解释每个答案。假设变量 X 取值于若干集合。

(a) X , X . (b) X , X . (c) ! X , X . (d) ! X , X . (e) ! X , X { } . (f) X , X P ( X ) . (g) ! X , X P ( X ) . (h) X , ! y , y P ( X ) . (i) ! y , X , y P ( X ) .

(a) ∃X,∅⊆X。(b) ∃X,∅∈X。(c) ∃!X,X⊆∅。(d) ∃!X,X∈∅。(e) ∃!X,X∈{∅}。(f) ∀X,X⊆P(X)。(g) ∃!X,X∈P(X)。(h) ∃X,∃!y,y∈P(X)。(i) ∃!y,∀X,y∈P(X)。

11. Prove: X , ( { X } X ) .

11.证明:∀X,∼({X}⊆X)。

12. Prove part (c) of the Theorem on Power Sets: for all sets A and B , P ( A B ) = P ( A ) P ( B ) . (This can be done very quickly using part of the Theorem on Subsets.)

12.证明幂集定理的(c)部分:对于任意集合A和B,P(A∩B)=P(A)∩P(B)。(利用子集定理的一部分可以很快地证明这一点。)

13. (a) Prove part (e) of the Theorem on Power Sets:

13.(a)证明幂集定理的(e)部分:

for all sets A and B , P ( A B ) { } P ( A ) P ( B ) .

对于所有集合 A 和 B,P(A−B)−{∅}⊆P(A)−P(B)。

(b) Disprove: for all sets A and B , P ( A B ) P ( A ) P ( B ) .

(b)反证:对于所有集合 A 和 B,P(A−B)−∅⊆P(A)−P(B)。

(c) Prove or disprove: for all sets A and B , P ( A ) P ( B ) P ( A B ) .

(c)证明或反证:对于所有集合 A 和 B,P(A)−P(B)⊆P(A−B)。

14. Prove from the definitions: for all sets X and all objects a , a X iff { a } P ( X ) .

14.根据定义证明:对于所有集合 X 和所有对象 a,a ∈ X 当且仅当 {a}∈P(X)。

15. Let S be an arbitrary set. Find all members of P ( S ) P ( { S } ) .

15.设 S 为任意集合。求 P(S)∩P({S}) 的所有元素。

16. (a) Prove x , y , { x } = { y } x = y .

16.(a)证明∀x,∀y,{x}={y}⇔x=y。

(b) Prove x , y , z , { x , y } = { z } ( z = x z = y ) .

(b)证明 ∀x,∀y,∀z,{x,y}={z}⇔(z=x∧z=y)。

(c) Prove x , y , z , w , { x , y } = { z , w } ( ( x = z y = w ) ( x = w y = z ) ) .

(c)证明 ∀x,∀y,∀z,∀w,{x,y}={z,w}⇔((x=z∧y=w)∨(x=w∧y=z))。

17. Prove: a , b , c , d , { { a } , { a , b } } = { { c } , { c , d } } [ ( a = c ) ( b = d ) ] .

17.证明:∀a,b,c,d,{{a},{a,b}}={{c},{c,d}}⇔[(a=c)∧(b=d)].

18. Give a fully detailed proof of the following fact, used in this section: for all a , b , P ( { a , b } ) = { , { a } , { b } , { a , b } } . (You may need proof by cases. Note that for any set S , you know a S or a S ; and you know b S or b S .)

18. 请给出本节中使用的以下事实的完整详细证明:对于所有 a 和 b,P({a,b})={∅,{a},{b},{a,b}}。(你可能需要分情况证明。注意,对于任意集合 S,已知 a ∈ S 或 a∉S;并且已知 b ∈ S 或 b∉S。)

19. (a) Prove:

19.(a)证明:

S S , x x S S , T T , T T P P ( S S { x x } ) iff 当……时 [ T T P P ( S S ) ( x x T T T T { x x } P P ( S S ) ) ] .

(b) Explain informally what the statement proved in part (a) means. In particular, for finite S , how does the size of P ( S { x } ) compare to the size of P ( S ) ?

(b) 用通俗易懂的方式解释(a)部分证明的命题的含义。特别是,对于有限的集合S,P(S∪{x})的大小与P(S)的大小相比如何?

20. A set S is called ∈-transitive iff for all x , y , if x y and y S , then x S .

20.集合 S 被称为 ∈-传递的,当且仅当对于所有 x, y,如果 x ∈ y 且 y ∈ S,则 x ∈ S。

(a) Prove: For all sets S , S is ∈-transitive iff ( y , y S y S ) iff S P ( S ) .

(a)证明:对于所有集合 S,S 是 ∈-传递的当且仅当(∀y,y∈S⇒y⊆S)当且仅当 S⊆P(S)。

(b) Is ∅ ∈-transitive? Explain.

(b)∅ ∈ 是否具有传递性?请解释。

(c) Is {∅} ∈-transitive? Explain.

(c){∅} 是否具有 ∈-传递性?请解释。

(d) Is { { } } ∈-transitive? Explain.

(d){{∅}} 是否具有 ∈-传递性?请解释。

21. (a) Prove: for all S , T , if S and T are ∈-transitive, then S T is ∈-transitive.

21.(a)证明:对于所有 S、T,如果 S 和 T 是 ∈-传递的,则 S ∩ T 是 ∈-传递的。

(b) Prove or disprove: for all S , T , if S and T are ∈-transitive, then S T is ∈-transitive.

(b)证明或反证:对于所有 S、T,如果 S 和 T 是 ∈-传递的,则 S∪T 是 ∈-传递的。

(c) Prove: for all ∈-transitive sets S , S { S } is ∈-transitive.

(c)证明:对于所有 ∈-传递集 S,S∪{S} 是 ∈-传递的。

22. Give an explicit example of an ∈-transitive set S with exactly five distinct members.

22.请给出一个具有五个不同元素的∈传递集合S的具体例子。

23. Prove or disprove: for all ∈-transitive sets S , P ( S ) is ∈-transitive.

23.证明或反证:对于所有∈-传递集S,P(S)是∈-传递的。

3.5 Ordered Pairs and Product Sets

3.5 有序对和产品集

This section introduces the concept of an ordered pair ( a , b ). In contrast to the unordered pairs studied earlier, ( a , b ) is not the same object as ( b , a ) when a b . Ordered pairs are used in analytic geometry and calculus to represent points in the plane. We also define a product operation for sets to build sets of ordered pairs such as the xy -plane ℝ 2 = ℝ × ℝ.

本节介绍有序对 (a, b) 的概念。与之前学习的无序对不同,当 a ≠ b 时,(a, b) 与 (b, a) 不是同一个对象。有序对在解析几何和微积分中用于表示平面上的点。我们还定义了集合的乘积运算,以构建有序对集合,例如 xy 平面 ℝ 2 = ℝ × ℝ。

Ordered Pairs and Intervals

有序数对和区间

Recall that the unordered pair { a , b } is a set with a and b as members. We proved that { a , b } is the same set as { b , a }; order does not matter in a set. In many situations, however, order does matter. We now introduce a new undefined term, called an ordered pair, for such situations.

回想一下,无序对 {a, b} 是一个包含 a 和 b 的集合。我们已经证明 {a, b} 与 {b, a} 是同一个集合;在集合中,元素的顺序无关紧要。然而,在许多情况下,元素的顺序却很重要。现在,我们引入一个新的未定义术语,称为有序对,来描述这种情况。

3.52 Undefined Term: Ordered Pair For any objects a and b , the notation ( a , b ) is called the ordered pair with first component a and second component b .

3.52 未定义术语:有序对 对于任意对象 a 和 b,符号 (a, b) 称为以 a 为第一分量、b 为第二分量的有序对。

We always use round parentheses for ordered pairs and curly braces for unordered pairs. The following axiom states the main property of ordered pairs.

我们总是用圆括号表示有序数对,用花括号表示无序数对。以下公理阐述了有序数对的主要性质。

3.53 Ordered Pair Axiom For all objects a , b , c , d , ( a , b ) = ( c , d ) iff ( a = c and b = d ) .

3.53 有序对公理 对于所有对象 a、b、c、d,(a,b)=(c,d) 当且仅当 (a=c,b=d)。

In words, two ordered pairs are equal if and only if their first coordinates agree and their second coordinates agree. Thus, for instance, (3, 5) ≠ (5, 3), but (1 + 1, 2 + 2) = (3 − 1, 2 × 2). The components of an ordered pair can be any objects — integers, real numbers, sets, or even other ordered pairs. For instance, ({1, 2}, (2, 1)) is an ordered pair whose first component is the set {1, 2} and whose second component is the ordered pair (2, 1). For now, we postulate that ordered pairs, sets, and numbers are different kinds of objects. So, for instance, ((1, 2), 3) ≠ (1, (2, 3)) because “(1, 2) = 1 and 3 = (2, 3)” is false.

换句话说,两个有序对相等当且仅当它们的第一个坐标相等且第二个坐标也相等。例如,(3, 5) ≠ (5, 3),但 (1 + 1, 2 + 2) = (3 − 1, 2 × 2)。有序对的组成部分可以是任何对象——整数、实数、集合,甚至是其他有序对。例如,({1, 2}, (2, 1)) 就是一个有序对,它的第一个组成部分是集合 {1, 2},第二个组成部分是有序对 (2, 1)。目前,我们假设有序对、集合和数是不同类型的对象。因此,例如,((1, 2), 3) ≠ (1, (2, 3)),因为“(1, 2) = 1 且 3 = (2, 3)”不成立。

Remarkably, it is possible to define ordered pairs in terms of the previously introduced concepts of unordered pairs and singleton sets. Then ordered pairs are certain sets, and the Ordered Pair Axiom can be proved as a theorem. We give the details of this process in an optional subsection at the end of this section.

值得注意的是,我们可以利用前面介绍的无序对和单元素集的概念来定义有序对。这样,有序对就是确定性集合,有序对公理就可以被证明为一个定理。我们将在本节末尾的可选小节中详细介绍这一过程。

We now introduce interval notation for certain subsets of ℝ. Intuitively, for given a , b ∈ ℝ, the closed interval [ a , b ] consists of all real numbers x satisfying a x b . The open interval ( a , b ) consists of all real numbers x satisfying a < x < b . Half-open intervals and intervals extending to ±∞ are defined similarly. The formal definition is as follows.

现在我们引入区间表示法来描述 ℝ 的某些子集。直观地说,对于给定的 a, b ∈ ℝ,闭区间 [a, b] 由所有满足 a ≤ x ≤ b 的实数 x 组成。开区间 (a, b) 由所有满足 a < x < b 的实数 x 组成。半开区间和延拓至 ±∞ 的区间定义类似。其正式定义如下。

3.54. Definition: Intervals. Let a , b be fixed real numbers.

3.54. 定义:区间。设 a、b 为固定的实数。

(a) Closed Intervals: For all z , z [ a , b ] iff z R and a z b .

(a)闭区间:对于所有 z,z∈[a,b] 当且仅当 z∈Randa≤z≤b。

(b) Open Intervals: For all z , z ( a , b ) iff z R and a < z < b .

(b)开区间:对于所有 z,z∈(a,b) 当且仅当 z∈Randa<z<b。

(c) Half-Open Intervals: For all z , z ( a , b ] iff z R and a < z b ; and z [ a , b ) iff $ z R and a z < b .

(c)半开区间:对于所有 z,z∈(a,b] 当且仅当 z∈Randa<z≤b;并且 z∈[a,b) 当且仅当 z∈Randa≤z<b。

(d) Infinite Intervals: For all z , z ( , a ] iff z R and z a . We define ( − ∞, a ), [ a , ∞), and ( a , ∞) analogously. Finally, ( − ∞, ∞) is the set ℝ. We remark that the symbols ∞ and −∞ are not real numbers, and we do not define intervals such as [− ∞, ∞].

(d) 无限区间:对于所有 z,z∈(−∞,a] 当且仅当 z∈R 且 z≤a。我们类似地定义 (−∞,a)、[a,∞) 和 (a,∞)。最后,(−∞,∞) 是集合 ℝ。我们注意到符号 ∞ 和 −∞ 不是实数,并且我们不定义诸如 [−∞,∞] 之类的区间。

Note that the notation for open intervals conflicts with the notation for ordered pairs ; but this notation conflict seldom causes problems, since context often tells us what is intended. Some authors avoid this conflict by writing ] a , b [ or 〈 a , b 〉 to denote an open interval, but we will not use this convention. We stress that curly braces, round parentheses, and square brackets are not interchangeable in mathematics: { a , b } denotes an unordered pair, ( a , b ) denotes an ordered pair or open interval, and [ a , b ] denotes a closed interval. The three pictures below illustrate the set {1.5, 4}, the open interval (1.5, 4), and the closed interval [1.5, 4], respectively. Note that open circles are used to indicate endpoints that are excluded, while closed (filled-in) circles indicate endpoints that are included.

请注意,开区间的表示方法与有序对的表示方法存在冲突;但这种表示冲突很少会造成问题,因为上下文通常能告诉我们其含义。一些作者为了避免这种冲突,使用 ]a, b[ 或 〈a, b〉 来表示开区间,但我们不会采用这种约定。我们强调,在数学中,花括号、圆括号和方括号不能互换使用:{a, b} 表示无序对,(a, b) 表示有序对或开区间,而 [a, b] 表示闭区间。下图分别展示了集合 {1.5, 4}、开区间 (1.5, 4) 和闭区间 [1.5, 4]。请注意,空心圆表示不包含的端点,而实心圆表示包含的端点。

ufig3_9.jpg

Product Sets

产品套装

Given two sets A and B , the product set A × B is the set of all ordered pairs ( x , y ) where x A and y B . We state this formally as follows.

给定两个集合 A 和 B,乘积集 A × B 是所有有序对 (x, y) 的集合,其中 x ∈ A 且 y ∈ B。我们将其正式表述如下。

3.55. Definition: Product Set. For all sets A and B and all objects z : z A × B iff x , y , ( x A y B ) z = ( x , y ) .

3.55. 定义:乘积集。对于所有集合 A 和 B 以及所有对象 z:z∈A×B 当且仅当 ∃x,∃y,(x∈A∧y∈B)∧z=(x,y)。

It can be proved from this definition and the Ordered Pair Axiom that for all sets A and B and all objects x , y ,

根据这个定义和有序对公理可以证明,对于所有集合 A 和 B 以及所有对象 x, y,

( x x , y ) A 一个 × × B B iff 当……时 x x A 一个 and y B B

This version of the definition is frequently used in proofs.

该定义的这个版本经常用于证明中。

3.56 Example Given A = {1, 2, 3} and B = {3, 5},

3.56 例 已知 A = {1, 2, 3} 且 B = {3, 5},

A 一个 × × B B = { ( 1 , 3 ) , ( 2 , 3 ) , ( 3 , 3 ) , ( 1 , 5 ) , ( 2 , 5 ) , ( 3 , 5 ) } .

On the other hand,

另一方面,

B B × × A 一个 = { ( 3 , 1 ) , ( 3 , 2 ) , ( 3 , 3 ) , ( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) } A 一个 × × B B .

We have ( A × B ) ( B × A ) = { ( 3 , 3 ) } . Note that A B = { 3 } , and ( A B ) × ( A B ) = { ( 3 , 3 ) } . We have ( A B ) × ( B A ) = {1, 2} × {5} = {(1, 5), (2, 5)}. On the other hand, ( A × B ) − ( B × A ) = {(1, 3), (2, 3), (1, 5), (2, 5), (3, 5)}.

我们有 (A×B)∩(B×A)={(3,3)}。注意 A∩B={3},且 (A∩B)×(A∩B)={(3,3)}。我们有 (A − B) × (B − A) = {1, 2} × {5} = {(1, 5), (2, 5)}。另一方面,(A × B) − (B × A) = {(1, 3), (2, 3), (1, 5), (2, 5), (3, 5)}。

3.57 Example Given A = {{1, 2}, (3, 4)} and B = { , { 3 } } ,

3.57 示例 给定 A = {{1, 2}, (3, 4)} 和 B={∅,{3}},

A 一个 × × B B = { ( { 1 , 2 } , ) , ( { 1 , 2 } , { 3 } ) , ( ( 3 , 4 ) , ) , ( ( 3 , 4 ) , { 3 } ) } .

3.58 Example Given C = {0, 1}, we have C × C = {(0, 0), (0, 1), (1, 0), (1, 1)}. We can then compute

3.58 例 给定 C = {0, 1},我们有 C × C = {(0, 0), (0, 1), (1, 0), (1, 1)}。然后我们可以计算

( C C × × C C ) × × C C = { ( ( 0 , 0 ) , 0 ) , ( ( 0 , 1 ) , 0 ) , ( ( 1 , 0 ) , 0 ) , ( ( 1 , 1 ) , 0 ) , ( ( 0 , 0 ) , 1 ) , ( ( 0 , 1 ) , 1 ) , ( ( 1 , 0 ) , 1 ) , ( ( 1 , 1 ) , 1 ) } ; C C × × ( C C × × C C ) = { ( 0 , ( 0 , 0 ) ) , ( 0 , ( 1 , 0 ) ) , ( 1 , ( 0 , 0 ) ) , ( 1 , ( 1 , 0 ) ) , ( 0 , ( 0 , 1 ) ) , ( 0 , ( 1 , 1 ) ) , ( 1 , ( 0 , 1 ) ) , ( 1 , ( 1 , 1 ) ) } .

These two sets are not equal; in fact, [ ( C × C ) × C ] [ C × ( C × C ) ] = .

这两个集合不相等;事实上,[(C×C)×C]∩[C×(C×C)]=∅。

The product set ℝ × ℝ consists of all ordered pairs ( x , y ) with x ∈ ℝ and y ∈ ℝ. We write ℝ 2 = ℝ × ℝ; more generally, define A 2 = A × A for any set A . We visualize ℝ 2 as the xy -plane by dr two perpendicular axes intersecting at a point called the origin , and identifying the ordered pair ( x , y ) with the point reached by traveling from the origin a directed distance of x horizontally and a directed distance of y vertically. Then, as seen in the next example, the product set [ a , b ] × [ c , d ] is a solid rectangle in the plane.

乘积集 ℝ × ℝ 由所有满足 x ∈ ℝ 且 y ∈ ℝ 的有序对 (x, y) 组成。我们记 ℝ 2 = ℝ × ℝ;更一般地,对于任意集合 A,定义 A2 = A×A。我们将 ℝ 2 可视化为 xy 平面,其中两条互相垂直的轴相交于一点,称为原点。我们将有序对 (x, y) 等同于从原点出发,水平方向移动距离 x,垂直方向移动距离 y 所到达的点。然后,如下例所示,乘积集 [a, b] × [c, d] 是平面上的一个实心矩形。

3.59 Example Draw the following sets in the xy -plane.

3.59 例 在 xy 平面上画出下列集合。

(a) R = [1, 3] × [ − 2, 2]; (b) S = ( − 2, 2) × (1, 3); (c) R S ; (d) R S ; (e) {1, 3, 4} × { − 1, 0, 2}; (f) ( ( 1 , 3 ) × { 2 , 2 } ) ( { 1 , 3 } × ( 2 , 2 ] ) .

(a) R = [1, 3] × [ − 2, 2]; (b) S = ( − 2, 2) × (1, 3); (c) R∪S; (d) R ∩ S; (e) {1, 3, 4} × { − 1, 0, 2}; (f) ((1,3)×{−2,2})∪({1,3}×(−2,2]).

Solution. See Figure 3.1 . The four edges of the rectangle are part of the set R in (a), but not part of the set S in (b). We use dashed lines and open circles to indicate excluded regions, as in (c) and (d). Part (e) is a finite set of nine ordered pairs, each of which is drawn as a filled dot. In part (f), the graph of (1, 3) × { − 2, 2} consists of two horizontal line segments with both endpoints excluded; whereas the graph of {1, 3} × ( − 2, 2] consists of two vertical line segments with the top endpoints included. The union of these sets is shown in the figure; the interior of the rectangle is not part of the union.

解答。参见图 3.1。矩形的四条边在 (a) 中属于集合 R,但在 (b) 中不属于集合 S。我们用虚线和空心圆来表示排除区域,如图 (c) 和 (d) 所示。(e) 部分是一个包含九个有序对的有限集合,每个有序对用一个实心圆点表示。在 (f) 部分中,集合 (1, 3) × { − 2, 2} 由两条水平线段组成,且不包含两个端点;而集合 {1, 3} × { − 2, 2} 由两条垂直线段组成,且包含两个端点。图中显示了这些集合的并集;矩形的内部不属于该并集。

fig3_1

Figure 3.1

图 3.1

Solution to Example 3.59 .

例 3.59 的解答。

Properties of Product Sets

乘积集的性质

The next theorem lists general properties of product sets, which can all be proved from the definitions using the proof templates.

下一个定理列出了乘积集的一般性质,这些性质都可以使用证明模板从定义中证明出来。

3.60 Theorem on Product Sets For all sets A , B , C , D :

3.60 关于乘积集的定理 对于所有集合 A、B、C、D:

(a) Monotonicity: If A C and B D , then A × B C × D .

(a)单调性:如果 A ⊆ C 且 B ⊆ D,则 A × B ⊆ C × D。

(b) Distributive Laws:

(b)分配法则:

A × ( B C ) = ( A × B ) ( A × C ) and ( A B ) × C = ( A × C ) ( B × C ) and

A×(B∪C)=(A×B)∪(A×C) 且 (A∪B)×C=(A×C)∪(B×C) 且

A × ( B C ) = ( A × B ) ( A × C ) and ( A B ) × C = ( A × C ) ( B × C ) and

A×(B∩C)=(A×B)∩(A×C) 且 (A∩B)×C=(A×C)∩(B×C) 且

A × ( B C ) = ( A × B ) − ( A × C ) and ( A B ) × C = ( A × C ) − ( B × C ).

A × (B − C) = (A × B) − (A × C) 且 (A − B) × C = (A × C) − (B × C)。

(c) Intersection of Products: ( A × B ) ( C × D ) = ( A C ) × ( B D ) .

(c)乘积的交集:(A×B)∩(C×D)=(A∩C)×(B∩D)。

(d) Union of Products: ( A × B ) ( C × D ) ( A C ) × ( B D ) , but equality does not always hold.

(d)产品并集:(A×B)∪(C×D)⊆(A∪C)×(B∪D),但等式并不总是成立。

(e) Product of Empty Sets: A × = and × B = .

(e)空集的乘积:A×∅=∅ 且 ∅×B=∅。

(f) Criterion for Commutativity: A × B = B × A iff ( A = B or A = ∅ or B = ∅).

(f)交换律的判据:A × B = B × A 当且仅当(A = B 或 A = ∅ 或 B = ∅)。

We prove a few parts of this theorem to illustrate the techniques, leaving the remaining parts as exercises.

我们证明该定理的几个部分来说明其中的技巧,其余部分留给读者作为练习。

3.61. Proof of Monotonicity. Let A , B , C , D be fixed, but arbitrary sets. We prove: if A C and B D , then A × B C × D . [Start with the IF-template.] Assume A C and B D . Prove A × B C × D . [Continue with the subset template.] Fix an arbitrary object z ; assume z A × B ; prove z C × D . Our assumption means that z = ( a , b ) for some a A and b B . We must prove ( a , b ) ∈ C × D , which means a C and b D . On one hand, since a A and A C , we deduce a C . On the other hand, since b B and B D , we deduce b D .

3.61. 单调性证明。设 A、B、C、D 为任意固定集合。我们证明:若 A ⊆ C 且 B ⊆ D,则 A × B ⊆ C × D。[从 IF 模板开始。] 假设 A ⊆ C 且 B ⊆ D。证明 A × B ⊆ C × D。[继续使用子集模板。] 固定任意对象 z;假设 z ∈ A × B;证明 z ∈ C × D。我们的假设意味着对于某些 a ∈ A 和 b ∈ B,有 z = (a, b)。我们必须证明 (a, b) ∈ C × D,这意味着 a ∈ C 且 b ∈ D。一方面,由于 a ∈ A 且 A ⊆ C,我们推导出 a ∈ C。另一方面,由于 b ∈ B 且 B ⊆ D,我们推导出 b ∈ D。

The following figure gives some intuition for why monotonicity holds, though the figure cannot replace the formal proof.

下图从直观上解释了为什么单调性成立,但该图不能代替正式证明。

ufig3_10.jpg

3.62 Proof that × B = . Let B be an arbitrary set. [To show a set is empty, use proof by contradiction.] Assume, to get a contradiction, that × B . This assumption means z , z × B . In turn, this means that x , y , ( x y B ) z = ( x , y ) . In particular, x . But we know x also, which is a contradiction. So × B = , after all.

3.62 证明 ∅×B=∅。设 B 为任意集合。[要证明一个集合为空集,可以使用反证法。] 为了得出矛盾,假设 ∅×B≠∅。这个假设意味着 ∃z,z∈∅×B。反过来,这意味着 ∃x,∃y,(x∈∅∧y∈B)∧z=(x,y)。特别地,x∈∅。但我们知道 x∉∅,这与假设矛盾。所以,∅×B=∅。

Before proceeding to the next proof, we describe a modification of the subset proof template that is often convenient when dealing with product sets.

在进行下一个证明之前,我们描述子集证明模板的一种修改,这种修改在处理乘积集时通常很方便。

3.63. Subset Proof Template to Prove A × B T .

3.63. 证明 A × B ⊆ T 的子集证明模板。

Fix an arbitrary ordered pair ( x , y ).

固定任意有序对 (x, y)。

Assume ( x , y ) ∈ A × B , which means x A and y B .

假设 (x, y) ∈ A × B,这意味着 x ∈ A 且 y ∈ B。

Prove ( x , y ) ∈ T .

证明 (x, y) ∈ T。

The template works since the only objects that can belong to the product set A × B are ordered pairs. So we may as well fix an ordered pair at the outset, rather than fixing an object z and then saying x , y , ( x A y B ) z = ( x , y ) . We illustrate this template in the next proof.

该模板之所以有效,是因为乘积集 A × B 中只能包含有序对。因此,我们不妨一开始就指定一个有序对,而不是先指定一个对象 z,然后再说 ∃x,∃y,(x∈A∧y∈B)∧z=(x,y)。我们将在下一个证明中演示该模板。

3.64 Proof that A × ( B C ) = ( A × B ) − ( A × C ). Let A , B , C be arbitrary sets.

3.64 证明 A × (B − C) = (A × B) − (A × C)。设 A、B、C 为任意集合。

Part 1. Assume ( x , y ) is an arbitrary member of A × ( B C ); prove ( x , y ) ∈ ( A × B ) − ( A × C ). We have assumed that x A and y B C , so y B and y C . We must prove that ( x , y ) ∈ A × B and ( x , y ) A × C . The first part of what we must prove means that x A and y B , and both statements are known from our current assumptions. The second part of what we must prove is that “ x A or y C .” Since y C was assumed, this OR-statement is true.

第一部分:假设 (x, y) 是 A × (B − C) 中的任意元素;证明 (x, y) ∈ (A × B) − (A × C)。我们假设 x ∈ A 且 y ∈ B − C,因此 y ∈ B 且 y∉C。我们需要证明 (x, y) ∈ A × B 且 (x, y)∉A × C。我们需要证明的第一部分是 x ∈ A 且 y ∈ B,这两个命题都由我们当前的假设得出。我们需要证明的第二部分是“x∉A 或 y∉C”。由于 y∉C 已被假设,因此这个“或”命题成立。

Part 2. [We are about to prove ( A × B ) − ( A × C ) ⊆ A × ( B C ). We observe that every member of the left set belongs to A × B , hence must be an ordered pair. Thus we may begin part 2 by fixing an ordered pair rather than an arbitrary object.] Assume ( x , y ) is an arbitrary member of ( A × B ) − ( A × C ); prove ( x , y ) ∈ A × ( B C ). We have assumed that ( x , y ) ∈ A × B and ( x , y ) A × C , which means ( x A and y B ) and ( x A or y C ). We must prove x A and y B C ; i.e., we must prove x A and y B and y C . The first two statements follow from the first parenthesized statement in our assumption. On the other hand, since x A is true and “ x A or y C ” is also true, we deduce y C by definition of OR.

第二部分。[我们即将证明 (A × B) − (A × C) ⊆ A × (B − C)。我们观察到左侧集合中的每个元素都属于 A × B,因此它们必定是一个有序对。因此,我们可以从固定一个有序对而不是任意对象开始第二部分。] 假设 (x, y) 是 (A × B) − (A × C) 中的任意元素;证明 (x, y) ∈ A × (B − C)。我们假设 (x, y) ∈ A × B 且 (x, y) ∉ A × C,这意味着 (x ∈ A 且 y ∈ B) 且 (x ∉ A 或 y ∉ C)。我们必须证明 x ∈ A 且 y ∈ B − C;也就是说,我们必须证明 x ∈ A 且 y ∈ B 且 y ∉ C。前两个结论由我们假设中第一个括号内的结论得出。另一方面,由于 x ∈ A 为真,且“x∉A 或 y∉C”也为真,根据 OR 的定义,我们推导出 y∉C。

3.65. Proof of Commutativity Criterion. Let A and B be arbitrary sets; we prove A × B = B × A iff ( A = B or A = ∅ or B = ∅).

3.65. 交换律准则的证明。设 A 和 B 为任意集合;我们证明 A × B = B × A 当且仅当(A = B 或 A = ∅ 或 B = ∅)。

Part 1. We prove: “if A × B = B × A , then A = B or A = ∅ or B = ∅” using a contrapositive proof. Assume A B and A and B ; prove A × B B × A . Since A and B are nonempty, we know there exist objects a 0 A and b 0 B . We also know A B or B A , so use cases.

第一部分:我们使用逆否命题证明:“如果 A × B = B × A,则 A = B 或 A = ∅ 或 B = ∅”。假设 A ≠ B 且 A ≠ ∅ 且 B ≠ ∅;证明 A × B ≠ B × A。由于 A 和 B 非空,我们知道存在对象 a 0 ∈ A 和 b 0 ∈ B。我们也知道 A⊈B 或 B⊈A,因此可以使用这些情况。

Case 1. Assume A B , which means there exists c with c A and c B . Consider the ordered pair ( c , b 0 ). On one hand, ( c , b 0 ) ∈ A × B because c A and b 0 B . On the other hand, ( c , b 0 ) B × A since c B . Thus, A × B B × A since we have found a member of A × B that is not a member of B × A .

情况 1. 假设 A⊈B,这意味着存在 c 使得 c ∈ A 且 c∉B。考虑有序对 (c, b 0 )。一方面,(c, b 0 ) ∈ A × B,因为 c ∈ A 且 b 0 ∈ B。另一方面,(c,b0)∉B×A,因为 c∉B。因此,A×B⊈B×A,因为我们找到了 A × B 中的一个元素,它不属于 B × A。

Case 2. Assume B A . The proof is similar to the proof of Case 1, so we let you fill in the details of Case 2.

情况 2. 假设 B⊈A。证明与情况 1 的证明类似,因此我们留给您填写情况 2 的细节。

Part 2. We prove: “if A = B or A = ∅ or B = ∅, then A × B = B × A .” Assume A = B or A = ∅ or B = ∅. Prove A × B = B × A . [We have assumed an OR-statement, so use cases.]

第二部分。我们证明:“如果 A = B 或 A = ∅ 或 B = ∅,则 A × B = B × A。” 假设 A = B 或 A = ∅ 或 B = ∅。证明 A × B = B × A。[我们假设这是一个 OR 语句,因此使用条件判断。]

Case 1. Assume A = B . Then A × B = A × A = B × A .

情况 1. 假设 A = B。则 A × B = A × A = B × A。

Case 2. Assume A = ∅. Then both A × B and B × A equal the empty set, by part (e) of the theorem, hence A × B = B × A .

情况 2. 假设 A = ∅。则根据定理的 (e) 部分,A × B 和 B × A 都等于空集,因此 A × B = B × A。

Case 3. Assume B = ∅. Then both A × B and B × A equal the empty set, by part (e) of the theorem, hence A × B = B × A .

情况 3. 假设 B = ∅。则根据定理的 (e) 部分,A × B 和 B × A 都等于空集,因此 A × B = B × A。

Formal Definition of Ordered Pairs (Optional)

有序对的正式定义(可选)

We mentioned earlier that ordered pairs could be defined in terms of the previously introduced concepts of singleton sets and unordered pairs. Here is the relevant definition.

我们之前提到过,有序对可以根据之前介绍的单元素集和无序对的概念来定义。以下是相关的定义。

3.66. Definition: Ordered Pairs. For any objects a , b , define ( a , b ) = { { a } , { a , b } } .

3.66. 定义:有序对。对于任意对象 a、b,定义 (a,b)={{a},{a,b}}。

3.67 Example We have (3, 5) = {{3}, {3, 5}} and (5, 3) = {{5}, {5, 3}} = {{5}, {3, 5}}; note that the set (3, 5) does not equal the set (5, 3). The ordered pair (4, 4) is the set {{4}, {4, 4}}. Since {4, 4} = {4}, we can also write (4, 4) = {{4}, {4}} = {{4}}. On the other hand, (4, {4}) = {{4}, {4, {4}}}, whereas ({4}, 4) = {{{4}}, {{4}, 4}} = {{{4}}, {4, {4}}}. Thus, the three sets (4, 4) and (4, {4}) and ({4}, 4) are all distinct.

3.67 例 我们有 (3, 5) = {{3}, {3, 5}} 和 (5, 3) = {{5}, {5, 3}} = {{5}, {3, 5}};注意集合 (3, 5) 不等于集合 (5, 3)。有序对 (4, 4) 是集合 {{4}, {4, 4}}。由于 {4, 4} = {4},我们也可以写成 (4, 4) = {{4}, {4}} = {{4}}。另一方面,(4, {4}) = {{4}, {4, {4}}},而 ({4}, 4) = {{{4}}, {{4}, 4}} = {{{4}}, {4, {4}}}。因此,集合 (4, 4)、(4, {4}) 和 ({4}, 4) 都是不同的。

We can now prove the Ordered Pair Axiom as a theorem.

现在我们可以将有序对公理证明为一个定理。

3.68 Ordered Pair Theorem For all objects a , b , c , d , ( a , b ) = ( c , d ) iff ( a = c and b = d ).

3.68 有序对定理 对于所有对象 a、b、c、d,(a, b) = (c, d) 当且仅当 (a = c 且 b = d)。

Proof . Our proof uses the following three facts, stated in the exercises of § 3.4 . For all objects x , y , z , w : (1) { x } = { y } ⇔ x = y ; (2) { x , y } = { z } ( z = x z = y ) ; and (3) { x , y } = { z , w } ( ( x = z y = w ) ( x = w y = z ) ) . Fix arbitrary objects a , b , c , d .

证明。我们的证明使用了以下三个事实,这些事实在 §3.4 的练习中给出。对于所有对象 x、y、z、w:(1) {x} = {y} ⇔ x = y;(2) {x,y}={z}⇔(z=x∧z=y);以及 (3) {x,y}={z,w}⇔((x=z∧y=w)∨(x=w∧y=z))。固定任意对象 a、b、c、d。

Part 1. Assume ( a , b ) = ( c , d ). Prove a = c and b = d . By definition of ordered pairs, we have assumed {{ a }, { a , b }} = {{ c }, { c , d }}. Using Fact (3) with x = { a }, y = { a , b }, z = { c }, and w = { c , d }, we deduce that ({ a } = { c } and { a , b } = { c , d }) or ({ a } = { c , d } and { a , b } = { c }). [We have just written a known OR-statement, so use proof by cases.]

第一部分:假设 (a, b) = (c, d)。证明 a = c 且 b = d。根据有序对的定义,我们假设 {{a}, {a, b}} = {{c}, {c, d}}。利用事实 (3),令 x = {a},y = {a, b},z = {c},w = {c, d},我们推导出 ({a} = {c} 且 {a, b} = {c, d}) 或 ({a} = {c, d} 且 {a, b} = {c})。[我们刚刚写出了一个已知的 OR 语句,因此使用分情况证明。]

Case 1. Assume { a } = { c } and { a , b } = { c , d }; prove a = c and b = d . Since { a } = { c }, Fact (1) gives a = c . Since { a , b } = { c , d }, Fact (3) gives ( a = c and b = d ) or ( a = d and b = c ). [This is another OR-statement, so use cases within Case 1.]

情况 1. 假设 {a} = {c} 且 {a, b} = {c, d};证明 a = c 且 b = d。由于 {a} = {c},根据事实 (1) 可得 a = c。由于 {a, b} = {c, d},根据事实 (3) 可得 (a = c 且 b = d) 或 (a = d 且 b = c)。[这是另一个 OR 语句,因此请在情况 1 中使用其他情况。]

Case 1a. Assume a = c and b = d ; prove a = c and b = d . The two statements to be proved have already been assumed.

情况 1a. 假设 a = c 且 b = d;证明 a = c 且 b = d。待证明的两个命题已经假设成立。

Case 1b. Assume a = d and b = c ; prove a = c and b = d . We already know a = c . Using the assumptions, we see b = c = a = d , so b = d .

情况 1b. 假设 a = d 且 b = c;证明 a = c 且 b = d。我们已经知道 a = c。利用假设,我们看到 b = c = a = d,所以 b = d。

Case 2. Assume { a } = { c , d } and { a , b } = { c }; prove a = c and b = d . Since { a } = { c , d }, Fact (2) gives a = c and a = d . Since { a , b } = { c }, Fact (2) gives c = a and c = b . So, a = c and b = c = a = d , as needed. [We might initially think Case 2 cannot occur, since { a } is a one-element set and { c , d } is a two-element set. But when c = d , { c , d } = { c } is a one-element set after all! Compare to the example (4, 4) = {{4}}.]

情况 2. 假设 {a} = {c, d} 且 {a, b} = {c};证明 a = c 且 b = d。因为 {a} = {c, d},根据事实 (2) 可知 a = c 且 a = d。因为 {a, b} = {c},根据事实 (2) 可知 c = a 且 c = b。所以,a = c 且 b = c = a = d,符合要求。[我们最初可能认为情况 2 不可能发生,因为 {a} 是一个单元素集合,而 {c, d} 是一个双元素集合。但是当 c = d 时,{c, d} = {c} 最终也是一个单元素集合!与示例 (4, 4) = {{4}} 进行比较。]

Part 2. Assume a = c and b = d . Prove ( a , b ) = ( c , d ). By definition of ordered pairs, we must prove {{ a }, { a , b }} = {{ c }, { c , d }}. Since a = c , Fact (1) shows that { a } = { c }. Similarly, since a = c and b = d , Fact (3) gives { a , b } = { c , d }. Using Fact (3) again taking x = { a }, y = { a , b }, z = { c }, and w = { c , d }, we conclude that {{ a }, { a , b }} = {{ c }, { c , d }}, as needed.

第二部分。假设 a = c 且 b = d。证明 (a, b) = (c, d)。根据有序对的定义,我们必须证明 {{a}, {a, b}} = {{c}, {c, d}}。由于 a = c,事实 (1) 表明 {a} = {c}。类似地,由于 a = c 且 b = d,事实 (3) 得出 {a, b} = {c, d}。再次利用事实 (3),令 x = {a},y = {a, b},z = {c},w = {c, d},我们得出 {{a}, {a, b}} = {{c}, {c, d}},符合要求。

We should not get too distracted by the odd-looking definition of ( a , b ) or the logical subtleties of the preceding proof. The definition ( a , b ) = {{ a }, { a , b }} has various bizarre consequences, such as ( a , b ) P ( { a , b } ) . It is theoretically interesting that we can construct ordered objects from sets, which are intrinsically unordered. However, in applications of set theory to the rest of mathematics, the only important feature of this construction is the criterion for equality of ordered pairs stated in the Ordered Pair Theorem. That is why we can safely ignore the details of this definition, as we do in the main text, treating ( a , b ) as an undefined concept obeying the Ordered Pair Axiom.

我们不应过于纠结于 (a, b) 看似奇怪的定义或前述证明的逻辑细节。定义 (a, b) = {{a}, {a, b}} 会产生各种奇特的推论,例如 (a,b)⊆P({a,b})。理论上,我们可以从本质上无序的集合构造有序对象,这固然有趣。然而,在集合论应用于其他数学领域时,这种构造的唯一重要特征是有序对定理中给出的有序对相等性准则。因此,我们可以像正文中那样,忽略该定义的细节,将 (a, b) 视为一个满足有序对公理的未定义概念。

Section Summary

章节概要

1. Intervals in ℝ. For all real numbers a , b and all x :

1. ℝ 中的区间。对于所有实数 a、b 和所有 x:

Open Interval : x ( a , b ) iff x R and a < x < b .

开区间:x∈(a,b) 当且仅当 x∈Randa<x<b。

Closed Interval : x [ a , b ] iff x R and a x b

闭区间:x∈[a,b] 当且仅当 x∈Randa≤x≤b

Half-Open Interval : x [ a , b ) iff x R and a x < b .

半开区间:x∈[a,b) 当且仅当 x∈Randa≤x<b。

We define ( a , b ] and intervals extending to ±∞ analogously.

我们以类似的方式定义 (a, b] 和延伸到 ±∞ 的区间。

2. Ordered Pairs. For all objects a , b , c , d , ( a , b ) = ( c , d ) iff ( a = c and b = d ). The round parentheses in ( a , b ) denote an ordered pair (or an open interval); the curly braces in { a , b } denote an unordered pair; the square brackets in [ a , b ] denote a closed interval.

2.有序对。对于任意对象 a、b、c、d,(a, b) = (c, d) 当且仅当 (a = c 且 b = d)。(a, b) 中的圆括号表示一个有序对(或一个开区间);{a, b} 中的花括号表示一个无序对;[a, b] 中的方括号表示一个闭区间。

3. Product Sets. For all objects x , y , ( x , y ) A × B iff x A and y B . For all sets A , B and all objects z , z A × B iff x , y , x A y B z = ( x , y ) . To prove that A × B T , we may fix an ordered pair ( x , y ) satisfying x A and y B , and prove that ( x , y ) ∈ T .

3. 乘积集。对于任意对象 x, y,(x, y)∈A×B 当且仅当 x∈A 且 y∈B。对于任意集合 A, B 和任意对象 z,z∈A×B 当且仅当存在 x, y,x∈A ∧ y∈B ∧ z=(x, y)。为了证明 A × B ⊆ T,我们可以固定一个有序对 (x, y),满足 x ∈ A 且 y ∈ B,并证明 (x, y) ∈ T。

4. Theorem on Product Sets. The product set construction has properties such as monotonicity, distributive laws, ( A × B ) ( C × D ) = ( A C ) × ( B D ) , ( A × B ) ( C × D ) ( A C ) × ( B D ) , and A × = = × A . However, A × B does not equal B × A , except when A = B or one of the sets is empty.

4.关于乘积集的定理。乘积集构造具有单调性、分配律、(A×B)∩(C×D)=(A∩C)×(B∩D)、(A×B)∪(C×D)⊆(A∪C)×(B∪D)以及A×∅=∅=∅×A等性质。然而,除非A=B或其中一个集合为空,否则A×B不等于B×A。

5. Formal Definition of Ordered Pairs (Optional). For all objects a , b , we can define ( a , b ) = {{ a }, { a , b }} and use this definition to prove the Ordered Pair Axiom from facts about unordered pairs.

5.有序对的正式定义(可选)。对于任意对象 a 和 b,我们可以定义 (a, b) = {{a}, {a, b}},并利用此定义从无序对的性质证明有序对公理。

Exercises

练习

1. Let A = {2, 4, 5}, B = {4, 5, 6}, and C = {1, 4, 6}. Compute:

1.设 A = {2, 4, 5},B = {4, 5, 6},C = {1, 4, 6}。计算:

(a) A × B (b) B × A (c) ( A × B ) ( B × A ) (d) ( A × B ) ( B × C ) (e) ( A B ) × ( B C ) (f) ( B C ) × ( C B ) (g) ( B × A ) − ( C × B ) (h) [ ( B A ) × C ] [ C × ( A B ) ] .

(a) A × B (b) B × A (c) (A×B)∩(B×A) (d) (A×B)∩(B×C) (e) (A∩B)×(B∩C) (f) (B − C) × (C − B) (g) (B × A) − (C × B) (h) [(B−A)×C]∪[C×(A−B)].

2. Let A = {1, 2} and B = {3, 4}. Compute:

2.设 A = {1, 2},B = {3, 4}。计算:

(a) A × B (b) P ( A ) (c) P ( B ) (d) P ( A × B ) (e) P ( A ) × P ( B ) (f) ( A × B ) × A (g) A × ( B × A ) (h) Are the sets found in (d) and (e) equal? Why? (i) Are the sets found in (f) and (g) equal? Why?

(a) A × B (b) P(A) (c) P(B) (d) P(A×B) (e) P(A)×P(B) (f) (A × B) × A (g) A × (B × A) (h) (d) 和 (e) 中的集合相等吗?为什么? (i) (f) 和 (g) 中的集合相等吗?为什么?

3. Given A = {[0, 3], ℤ, {2}} and B = {(1, {3}), {1, 3}}, compute A × B and B × A .

3.给定 A = {[0, 3], ℤ, {2}} 和 B = {(1, {3}), {1, 3}},计算 A × B 和 B × A。

4. Let A = [ − 1, 2], B = [ − 1, 3], C = [1, 4], and D = [2, 5].

4.设 A = [ − 1, 2],B = [ − 1, 3],C = [1, 4],D = [2, 5]。

(a) Draw pictures of A , B , C , D , A C , and B D on a number line. Express A C and B D in interval notation.

(a)在数轴上画出A、B、C、D、A-C和B-D的图像。用区间表示法表示A-C和B-D。

(b) Draw pictures in ℝ 2 showing A × B , C × D , ( A × B ) − ( C × D ), and ( A C ) × ( B D ).

(b)在 ℝ 2 中画出图形,表示 A × B、C × D、(A × B)−(C × D)和(A − C)×(B − D)。

5. Define intervals A = (0, 5), B = [2, 3), C = ( − 1, 2], and D = (1, 7).

5.定义区间A=(0,5),B=[2,3],C=(-1,2],D=(1,7)。

(a) Draw pictures of A , B , C , D , A B , A C , C D , and D B on a number line. Express A B and D B as unions of intervals.

(a)在数轴上画出 A、B、C、D、A − B、A − C、C∪D 和 D − B 的图像。将 A − B 和 D − B 表示为区间的并集。

(b) Draw pictures in ℝ 2 showing ( A B ) × ( D B ), ( D B ) × ( A B ), ( A × D ) − ( B × B ), and ( D × A ) − ( C × C ).

(b)在 ℝ 2 中画出图形,表示 (A − B) × (D − B)、(D − B) × (A − B)、(A × D) − (B × B) 和 (D × A) − (C × C)。

6. Define A = [ − 3, 3], B = ( − 2.5, 1.5), and C = { − 1, 1, 2}. Draw pictures in ℝ 2 of the following sets: (a) A × B (b) C × C (c) ( A × B ) ( Z × Z ) (d) ( A C ) × ( B C ) (e) ℝ × A (f) B × ℤ (g) C × ℝ (h) ℤ × C .

6.定义 A = [ − 3, 3],B = ( − 2.5, 1.5),C = { − 1, 1, 2}。在 ℝ 2 中画出下列集合的图像:(a) A × B (b) C × C (c) (A×B)∩(Z×Z) (d) (A − C) × (B − C) (e) ℝ × A (f) B × ℤ (g) C × ℝ (h) ℤ × C。

7. (a) Let S be the boundary of the rectangle in ℝ 2 with corners (1, 2), (5, 2), (5, 7), and (1, 7). Describe S as a union of product sets. (b) Give another description of the set S in part (a) using only interval notation, product sets, and set difference.

7.(a)设 S 为 ℝ 2 中顶点分别为 (1, 2)、(5, 2)、(5, 7) 和 (1, 7) 的矩形的边界。将 S 描述为若干个乘积集的并集。(b) 仅使用区间表示法、乘积集和集合差集,对 (a) 部分中的集合 S 给出另一种描述。

8. Let T be the cube [0, 2] × [0, 2] × [1, 3].

8.设 T 为 [0, 2] × [0, 2] × [1, 3] 的立方。

(a) What is the geometric significance of the set {0, 2} × {0, 2} × {1, 3}? What about [0, 2] × {2} × [1, 3]? (b) Use set operations such as × and ∪ to describe the set of points in ℝ 3 lying on one of the six faces of T . (c) Similarly, describe the set of points in ℝ 3 lying on one of the 12 edges of T .

(a)集合{0, 2} × {0, 2} × {1, 3}的几何意义是什么?[0, 2] × {2} × [1, 3]的几何意义又是什么?(b)使用集合运算(例如×和∪)描述ℝ 3 中位于T的六个面上的点集。(c)类似地,描述ℝ 3 中位于T的12条边上的点集。

9. Assume a , b , c , d are fixed real numbers. Prove the following properties of intervals in ℝ, and illustrate with sketches.

9. 假设 a、b、c、d 为固定的实数。证明 ℝ 中区间的下列性质,并用示意图说明。

(a) If a < b , then [ a , b ] − ( a , b ) = { a , b }.

(a)如果 a < b,则 [a, b] − (a, b) = {a, b}。

(b) If a < b < c , then [ a , b ] ( b , c ] = [ a , c ] .

(b)如果 a < b < c,则 [a,b]∪(b,c]=[a,c].

(c) If a < b < c < d , then [ a , d ] [ b , c ] = [ a , b ) ( c , d ] .

(c)如果 a < b < c < d,则 [a,d]−[b,c]=[a,b)∪(c,d)。

(d) If a < b , then ( a , b ] = ( , b ] ( a , ) .

(d)如果 a < b,则(a,b]=(−∞,b]∩(a,∞)。

10. (a) Prove: for all sets A , B , C , D , ( A × B ) ( C × D ) = ( A C ) × ( B D ) .

10.(a)证明:对于所有集合A、B、C、D,(A×B)∩(C×D)=(A∩C)×(B∩D)。

(b) Draw pictures illustrating the identity in (a), taking the sets A , B , C , D to be specific closed intervals.

(b)画图说明(a)中的恒等式,其中集合 A、B、C、D 为特定的闭区间。

11. Prove: for all sets A , A × = .

11.证明:对于所有集合A,A×∅=∅。

12. Fix arbitrary sets A , B , C .

12. 固定任意集合 A、B、C。

(a) Prove the distributive law ( A B ) × C = ( A × C ) ( B × C ) .

(a)证明分配律(A∪B)×C=(A×C)∪(B×C)。

(b) Prove the distributive law A × ( B C ) = ( A × B ) ( A × C ) .

(b)证明分配律 A×(B∩C)=(A×B)∩(A×C)。

(c) Prove the distributive law ( A B ) × C = ( A × C ) − ( B × C ).

(c)证明分配律(A − B)× C =(A × C)−(B × C)。

13. Draw pictures in ℝ 2 illustrating each identity in the preceding problem.

13.在ℝ 2 中画图,说明前面问题中的每个身份。

14. (a) Prove: for all sets A , B , C , D , ( A × B ) ( C × D ) ( A C ) × ( B D ) .

14.(a)证明:对于所有集合A、B、C、D,(A×B)∪(C×D)⊆(A∪C)×(B∪D)。

(b) Draw a picture in ℝ 2 to show that equality does not always hold in (a).

(b)在 ℝ 2 中画一幅图,以表明(a)中的等式并不总是成立。

(c) Based on your picture, guess a formula expressing ( A C ) × ( B D ) as a union of other sets. Prove your formula.

(c) 根据你的图,猜出一个公式,将 (A∪C)×(B∪D) 表示为其他集合的并集。证明你的公式。

15. (a) Let A = [0, 5], B = [1, 2], C = [0, 6], and D = [3, 4]. Draw a picture of ( A B ) × ( C D ).

15.(a)设 A = [0, 5],B = [1, 2],C = [0, 6],D = [3, 4]。画出 (A − B) × (C − D) 的图像。

(b) Repeat part (a) with A = C = [0, 3], B = [2, 4], and D = [ − 1, 1].

(b)重复(a)部分,其中 A = C = [0, 3],B = [2, 4],D = [ − 1, 1]。

(c) Repeat part (a) with A = C = {1, 2, 3, 4, 5}, B = {1, 2, 4}, and D = {4, 5, 7}.

(c)重复(a)部分,其中 A = C = {1, 2, 3, 4, 5},B = {1, 2, 4},D = {4, 5, 7}。

(d) For arbitrary sets A , B , C , D , guess a formula expressing ( A B ) × ( C D ) in terms of A × C and other sets. Prove your formula.

(d)对于任意集合A、B、C、D,猜出一个用A×C和其他集合表示(A-B)×(C-D)的公式。证明你的公式。

16. Carefully sketch the set A B C , where: A = ( [ 1 , 8 ] Z ) × ( 1 , 3 ] , B = ( ( 4 , 5 ) ( 7 , 8 ) ) × { 2 } , and C = (([1, 6.5] − (3, 4)) − (5, 5.5)) × {3}.

16.仔细绘制集合 A∪B∪C,其中:A=([1,8]∩Z)×(1,3],B=((4,5)∪(7,8))×{2},C=(([1, 6.5] − (3, 4)) − (5, 5.5)) × {3}。

17. (a) Prove: for any closed intervals I and J in ℝ, I J is a closed interval or the empty set. (b) Prove: for any open intervals I and J in ℝ, if I J , then I J is an open interval. (c) Prove or disprove the converse of the statement in (b).

17.(a)证明:对于ℝ中的任意闭区间I和J,I∩J要么是闭区间,要么是空集。(b)证明:对于ℝ中的任意开区间I和J,如果I∩J≠∅,则I∪J是开区间。(c)证明或反证(b)的逆命题。

18. For any objects a , b , c , define the ordered triple ( a , b , c ) to be (( a , b ), c ). Use the Ordered Pair Axiom to prove:

18. 对于任意对象 a、b、c,定义有序三元组 (a, b, c) 为 ((a, b), c)。利用有序对公理证明:

a 一个 , b b , c c , d d , e e , f f , ( a 一个 , b b , c c ) = ( d d , e e , f f ) [ ( a 一个 = d d ) ( b b = e e ) ( c c = f f ) ] .

19. For any objects a , b , c , we give an alternate definition of the ordered triple by letting ( a , b , c ) be the set {(1, a ), (2, b ), (3, c )}. Use this definition and properties of ordered pairs to prove:

19. 对于任意对象 a、b、c,我们给出有序三元组的另一种定义,令 (a, b, c) 为集合 {(1, a), (2, b), (3, c)}。利用此定义和有序对的性质证明:

a 一个 , b b , c c , d d , e e , f f , ( a 一个 , b b , c c ) = ( d d , e e , f f ) [ ( a 一个 = d d ) ( b b = e e ) ( c c = f f ) ] .

20. Use Definition 3.66 to prove: (a) a , b , ( a , b ) P ( { a , b } ) . (b) A , B , A × B P ( A B ) . (c) x , y , x = y ! z , z ( x , y ) .

20. 利用定义 3.66 证明:(a) ∀a,b,(a,b)⊆P({a,b})。(b) ∀A,B,A×B⊆P(A∪B)。(c) ∀x,y,x=y⇔∃!z,z∈(x,y)。

21. Consider the following alternate definition of ordered pairs: for all objects a , b , define ( a , b ) = {{1, a }, {2, b }}. Prove the Ordered Pair Axiom holds using this definition of ( a , b ).

21.考虑以下有序对的替代定义:对于所有对象 a、b,定义 (a, b) = {{1, a}, {2, b}}。使用 (a, b) 的这个定义证明有序对公理成立。

22. Suppose we try to define ordered triples by letting ( a , b , c ) = {{1, a }, {2, b }, {3, c }} for all objects a , b , c . Disprove:

22. 假设我们尝试定义有序三元组,令 (a, b, c) = {{1, a}, {2, b}, {3, c}},其中 a、b、c 为任意对象。反证:

a 一个 , b b , c c , d d , e e , f f , ( a 一个 , b b , c c ) = ( d d , e e , f f ) [ ( a 一个 = d d ) ( b b = e e ) ( c c = f f ) ] .

23. Suppose we try to define ordered pairs by letting ( a , b ) = { a , b , { a }} for all objects a and b . Prove or disprove the Ordered Pair Axiom using this definition of ( a , b ).

23.假设我们尝试定义有序对,令 (a, b) = {a, b, {a}},其中 a 和 b 为所有对象。使用 (a, b) 的这个定义证明或反驳有序对公理。

3.6 General Unions and Intersections

3.6 一般联合体和交叉口

Given two sets A and B , we have defined the union A B and the intersection A B of these sets. To take the union or intersection of finitely many sets, one way to proceed is to iterate the binary version of the operation. For instance, the union of five sets A , B , C , D , E could be defined to be ( ( ( A B ) C ) D ) E . Because of the commutative and associative properties of union and intersection, the order of sets and placement of parentheses in this expression do not affect the answer. However, we cannot use this method to define the union or intersection of infinitely many sets. This section describes how to handle this situation, using the idea of an indexed collection of sets. We also introduce set-builder notation, which gives a concise way to define new sets.

给定两个集合 A 和 B,我们已经定义了它们的并集 A ∪ B 和交集 A ∩ B。要计算有限个集合的并集或交集,一种方法是迭代执行该运算的二进制版本。例如,五个集合 A、B、C、D、E 的并集可以定义为 (((A∪B)∪C)∪D)∪E。由于并集和交集满足交换律和结合律,因此该表达式中集合的顺序和括号的位置不会影响结果。然而,我们不能使用这种方法来定义无限个集合的并集或交集。本节将介绍如何使用索引集合的概念来处理这种情况。我们还将介绍集合构造器符号,它提供了一种简洁的方式来定义新的集合。

Unions and Intersections of Indexed Collections of Sets

集合索引集合的并集和交集

Let I be any nonempty set, called an index set . Each member i of I is called an index . Suppose that for each index i I , we have a set A i . Informally, the union of the sets A i , as i ranges over I , is the set obtained by joining together all the members of all the sets A i into one giant set. The intersection of the sets A i is the set of all objects z (if any) that belong to all of the A i . Formally, we have the following definitions.

设 I 为任意非空集合,称为索引集。I 中的每个元素 i 称为索引。假设对于每个索引 i ∈ I,我们有一个集合 A i 。通俗地说,当 i 取遍 I 时,集合 A i 的并集是将所有集合 A i 的所有元素合并成一个大集合。集合 A i 的交集是所有属于 A i 的对象 z(如果有)的集合。正式地说,我们有以下定义。

3.69. Definition: Union and Intersection of an Indexed Family of Sets. Given a nonempty set I and a set A i for each i I :

3.69. 定义:索引集合族的并集和交集。给定一个非空集合 I 和一个集合 A i ,其中 i ∈ I:

(a) General Union: For all z , z i I A i iff i I , z A i .

(a)一般联盟:对于所有 z,z∈⋃i∈IAi 当且仅当 ∃i∈I,z∈Ai。

(b) General Intersection: For all z , z i I A i iff i I , z A i .

(b)一般交集:对于所有 z,z∈∩i∈IAi 当且仅当 ∀i∈I,z∈Ai。

For certain index sets I , variations of this notation are used. Specifically, if I = {1, 2, …, n } for some positive integer n , we write

对于某些索引集 I,可以使用这种符号的变体。具体来说,如果 I = {1, 2, …, n},其中 n 为某个正整数,我们写出

i I A 一个 i = i = 1 n n A 一个 i = A 一个 1 A 一个 2 A 一个 n n ; i I A 一个 i = i = 1 n n A 一个 i = A 一个 1 A 一个 2 A 一个 n n .

If I = ℤ >0 is the set of all positive integers, we write

如果 I = ℤ >0 是所有正整数的集合,我们写

i I A 一个 i = i = 1 A 一个 i , i I A 一个 i = i = 1 A 一个 i .

We use similar notation for I = ℤ > a , I = { b , b + 1, …, c }, etc. We can also change the name of the index i to any unused letter; for instance, ∪ i I A i = ∪ k I A k = ∪ n I A n .

我们对 I = ℤ > a 、I = {b, b + 1, …, c} 等使用类似的符号。我们还可以将索引 i 的名称更改为任何未使用的字母;例如,∪ i I A i = ∪ k I A k = ∪ n I A n

3.70 Example Let the index set I be ℝ >0 , the set of positive real numbers. For each r ∈ ℝ >0 , let A r = { − r , 0, r }. Then r R > 0 A r = R , whereas r R > 0 A r = { 0 } . If we shrink the index set from ℝ >0 to ℤ >0 , then n Z > 0 A n = Z .

3.70 例 设索引集 I 为 ℝ >0 ,即正实数集。对于每个 r ∈ ℝ >0 ,令 A r = { − r, 0, r}。则 ⋃r∈R>0Ar=R,而 ∩r∈R>0Ar={0}。如果我们将索引集从 ℝ >0 缩小到 ℤ >0 ,则 ⋃n∈Z>0An=Z。

3.71 Example Suppose I = ℤ >0 , and for each n I , let A n = [ − n , 1/ n ]. For instance, A 1 = [ − 1, 1], A 2 = [ − 2, 1/2], A 3 = [ − 3, 1/3], and so on, as shown in this picture:

3.71 示例 假设 I = ℤ >0 ,对于每个 n ∈ I,令 A n = [ − n, 1/n]。例如,A 1 = [ − 1, 1],A 2 = [ − 2, 1/2],A 3 = [ − 3, 1/3],依此类推,如下图所示:

ufig3_11.jpg

It is visually apparent from this picture that n = 1 A n = ( , 1 ] and n = 1 A n = [ 1 , 0 ] . In particular, for any positive real number r , there is an integer n 0 with 1/ n 0 < r , so that r A n 0 and hence r n = 1 A n .

从图中可以明显看出,⋃n=1∞An=(−∞,1] 且 ∩n=1∞An=[−1,0]。特别地,对于任意正实数 r,存在整数 n 0 使得 1/n 0 < r,因此 r∉An0,进而 r∉∩n=1∞An。

3.72 Example Let I be the set of rational numbers larger than π . For each q I , let A q = [ q , ∞). On one hand, q I A q = , so that the intersection of infinitely many infinite sets can be the empty set. On the other hand, ∪ q I A q = ( π , ∞), so that the union of infinitely many closed intervals can be an open interval.

3.72 例 设 I 为大于 π 的有理数集合。对于每个 q ∈ I,令 A q = [q, ∞)。一方面,∩q∈IAq=∅,因此无穷多个无限集合的交集可以是空集。另一方面,∪ q I A q = (π, ∞),因此无穷多个闭区间的并集可以是开区间。

Let us check the claimed set equalities in this example in more detail. First, assume to get a contradiction that q I A q . Then there exists z 0 such that z 0 ∈ ∩ q I A q . This means that for all q I , z 0 A q . So for every rational q > π , we have q z 0 < ∞. However, if we let q be the smallest integer larger than z 0 (such an integer exists and is a rational number larger than π ), we have z 0 < q , which is a contradiction. So the intersection is indeed empty.

让我们更详细地检验一下这个例子中所声称的集合等式。首先,为了得出矛盾,假设∩q∈IAq≠∅。那么存在z 0 ,使得z 0 ∈ ∩ q I A q 。这意味着对于所有q ∈ I,z 0 ∈ A q 。因此,对于每个大于π的有理数q,我们有q ≤ z 0 < ∞。然而,如果我们令q为大于z 0 的最小整数(这样的整数存在且是大于π的有理数),则有z 0 < q,这与假设矛盾。所以交集确实为空集。

To check ∪ q I A q = ( π , ∞), we prove set inclusions in both directions. To prove ⊆ , fix an arbitrary z 0 ∈ ∪ q I A q , and prove z 0 ∈ ( π , ∞). We know there exists q 0 I with z 0 A q 0 . This means that q 0 ∈ ℚ, q 0 > π , z 0 ∈ ℝ, and q 0 z 0 . Since π < q 0 z 0 , we see that π < z 0 , proving z 0 ∈ ( π , ∞). To prove , fix an arbitrary y 0 ∈ ( π , ∞). Then y 0 ∈ ℝ and π < y 0 . Assume we already know the theorem that between any two distinct real numbers there exists a rational number. So, we can find q 0 ∈ ℚ with π < q 0 < y 0 . Then y 0 A q 0 , and so y 0 ∈ ∪ q I A q , as needed.

为了验证 ∪ q I A q = (π, ∞),我们需要证明双向集合包含关系。为了证明 ⊆,固定任意 z 0 ∈ ∪ q I A q ,并证明 z 0 ∈ (π, ∞)。我们知道存在 q 0 ∈ I 使得 z0∈Aq0。这意味着 q 0 ∈ ℚ,q 0 > π,z 0 ∈ ℝ,且 q 0 ≤ z 0 。由于 π < q 0 ≤ z 0 ,我们可知 π < z 0 ,从而证明 z 0 ∈ (π, ∞)。为了证明 ⊇,固定任意 y 0 ∈ (π, ∞)。则 y 0 ∈ ℝ 且 π < y 0 。假设我们已经知道任意两个不同的实数之间存在一个有理数。因此,我们可以找到 q 0 ∈ ℚ,使得 π < q 0 < y 0 。那么 y0∈Aq0,因此 y 0 ∈ ∪ q I A q ,如有必要。

We remark that if we replace I by a finite subset I ′ = { q 1 < q 2 < · · · < q n }, then q I A q = ( q 1 , ) = A q 1 and q I A q = ( q n , ) = A q n .

我们注意到,如果我们用有限子集 I′ = {q 1 < q 2 < · · · < q n 替换 I,则 ⋃q∈I′Aq=(q1,∞)=Aq1 且 ∩q∈I′Aq=(qn,∞)=Aqn。

3.73 Example Take the index set to be I = ℤ >0 . For fixed n I , let A n be the set of real numbers of the form k / n for some k ∈ ℤ. Thus, A n is the set of real numbers that can be written as fractions where the numerator is any integer and the denominator is n . Then n = 1 A n = Q , whereas n = 1 A n = Z . When checking the latter equality in detail (see Exercise 4), it helps to first prove that ℤ ⊆ A n for every n I , and ℤ = A 1 .

3.73 例 取索引集为 I = ℤ >0 。对于固定的 n ∈ I,令 A n 为形如 k/n 的实数集,其中 k ∈ ℤ。因此,A n 为可以写成分子为任意整数、分母为 n 的分数形式的实数集。那么 ⋃n=1∞An=Q,而 ∩n=1∞An=Z。在详细检验后一个等式时(参见练习 4),首先证明对于每个 n ∈ I,ℤ ⊆ A n ,且 ℤ = A 1 会有所帮助。

Properties of General Unions and Intersections

一般并集和交集的性质

The next theorem states identities satisfied by the union and intersection operations for indexed collections of sets. Many of these properties are analogous to properties of binary union and binary intersection, which were given earlier in the Theorem on Subsets, the Theorem on Set Equality, and the Theorem on Product Sets.

下一个定理阐述了带索引集合的并集和交集运算所满足的恒等式。其中许多性质与二元并集和二元交集的性质类似,后者已在子集定理、集合相等性定理和乘积集定理中给出。

3.74 Theorem on General Unions and Intersections. For all nonempty index sets I and J and all sets X and A i and B i :

3.74 关于一般并集和交集的定理。对于所有非空索引集 I 和 J,以及所有集合 X 和 A i 和 B i

(a) Monotonicity: If A i B i for all i I , then i I A i i I B i and i I A i i I B i .

(a)单调性:如果对于所有 i ∈ I,A i ⊆ B i ,则 ⋃i∈IAi⊆⋃i∈IBi 且 ∩i∈IAi⊆∩i∈IBi。

(b) de Morgan Laws: X ( i I A i ) = i I ( X A i ) and X ( i I A i ) = i I ( X A i ) .

(b)德摩根定律:X−(⋃i∈IAi)=∩i∈I(X−Ai)和X−(∩i∈IAi)=⋃i∈I(X−Ai)。

(c) Distributive Laws: X ( i I A i ) = i I ( X A i ) and X ( i I A i ) = i I ( X A i ) and X × ( i I B i ) = i I ( X × B i ) and X × ( i I B i ) = i I ( X × B i ) ; similarly if X appears as the second input to ∩, ∪, or ×.

#

(c)分配律:X∩(⋃i∈IAi)=⋃i∈I(X∩Ai) 且 X∪(∩i∈IAi)=∩i∈I(X∪Ai) 且 X×(⋃i∈IBi)=⋃i∈I(X×Bi) 且 X×(∩i∈IBi)=∩i∈I(X×Bi);类似地,如果 X 作为 ∩、∪ 或 × 的第二个输入出现。

(d) Lower/Upper Bound: For all k I , i I A i A k and A k i I A i .

(d)下界/上界:对于所有 k ∈ I,∩i∈IAi⊆Ak 和 Ak⊆⋃i∈IAi。

(e) Greatest Lower Bound: X i I A i iff ( i I , X A i ).

(e)最大下界:X⊆∩i∈IAi 当且仅当(∀i∈I,X⊆Ai)。

(f) Least Upper Bound: i I A i X iff ( i I , A i X ).

(f)最小上界:⋃i∈IAi⊆X 当且仅当(∀i∈I,Ai⊆X)。

(g) Enlarging the Index Set: If I J , then i I A i j J A j and j J A j i I A i .

(g)扩大索引集:如果 I ⊆ J,则 ⋃i∈IAi⊆⋃j∈JAj 且 ∩j∈JAj⊆∩i∈IAi。

(h) Combining Index Sets:

(h)合并索引集:

( i I A 一个 i ) ( j j J J A 一个 j j ) = k k I J J A 一个 k k and ( i I A 一个 i ) ( j j J J A 一个 j j ) = k k I J J A 一个 k k .

The proofs of these results provide excellent practice with the proof templates involving quantifiers. We prove some parts to illustrate, and leave the other parts as exercises. Fix arbitrary sets I , J , X , A i , and B i (for i I or i J , as appropriate).

这些结果的证明为使用量词的证明模板提供了极佳的练习机会。我们证明其中一部分以作说明,其余部分留作练习。固定任意集合 I≠∅、J≠∅、X、A i 和 B i (对于 i ∈ I 或 i ∈ J,视情况而定)。

3.75 Proof of Lower/Upper Bound Property. Fix an arbitrary index k I . We first prove ∩ i I A i A k . Fix an arbitrary z ; assume z ∈ ∩ i I A i ; prove z A k . We have assumed i I , z A i . Since k I , we can let i = k in this assumed universal statement to conclude (by the Inference Rule for ALL) that z A k , as needed.

3.75 下界/上界性质的证明。固定任意索引 k ∈ I。我们首先证明 ∩ i I A i ⊆ A k 。固定任意 z;假设 z ∈ ∩ i I A i ;证明 z ∈ A k 。我们假设 ∀i∈I,z∈Ai。由于 k ∈ I,我们可以令此假设的全称命题中的 i = k,从而(根据全集推理规则)得出 z ∈ A k ,满足要求。

Next we prove A k ⊆ ∪ i I A i . Fix w ; assume w A k ; prove w ∈ ∪ i I A i . We must prove i I , w A i . Choose i = k , which is a member of I . Our assumption gives w A i , as needed.

接下来,我们证明 A k ⊆ ∪ i I A i 。固定 w;假设 w ∈ A k ;证明 w ∈ ∪ i I A i 。我们必须证明存在 i∈I,w∈Ai。选择 i = k,它是 I 的一个元素。我们的假设使得 w ∈ A i ,满足要求。

3.76 Proof of First de Morgan Law.

3.76 第一个德摩根定律的证明。

Part 1. We prove X −(∪ i I A i ) ⊆ ∩ i I ( X A i ). Fix z ; assume z X −(∪ i I A i ); prove z ∈ ∩ i I ( X A i ). We have assumed that z X and z i I A i ; we must prove z ∈ ∩ i I ( X A i ). Continuing to expand definitions, we have assumed that “ i I , z A i ” is false, so that “ i I , z A i ” is true. We must prove i I , z ( X A i ) . Fix an index i 0 I . Prove z X A i 0 . We must prove z X and z A i 0 . On one hand, z X was already assumed. On the other hand, taking i = i 0 in the assumption that i I , z A i , we see that z A i 0 , as needed.

第一部分。我们证明 X −(∪ i I A i ) ⊆ ∩ i I (X −A i )。固定 z;假设 z ∈ X −(∪ i I A i );证明 z ∈ ∩ i I (X −A i )。我们假设 z ∈ X 且 z∉⋃i∈IAi;我们必须证明 z ∈ ∩ i I (X −A i )。继续扩展定义,我们假设“∃i∈I,z∈Ai”为假,因此“∀i∈I,z∉Ai”为真。我们必须证明 ∀i∈I,z∈(X−Ai)。固定索引 i 0 ∈ I。证明 z∈X−Ai0。我们必须证明 z ∈ X 且 z∉Ai0。一方面,z ∈ X 已被假设。另一方面,在 ∀i∈I,z∉Ai 的假设下,取 i = i 0 ,我们可以看到 z∉Ai0,符合要求。

Part 2. We prove ∩ i I ( X A i ) ⊆ X −(∪ i I A i ). The proof is similar to part 1; you are asked to supply the details in Exercise 8.

第二部分。我们证明 ∩ i I (X −A i ) ⊆ X −(∪ i I A i )。证明过程与第一部分类似;请在练习 8 中提供详细信息。

3.77 Proof of Enlarging the Index Set Assume I J . First, we prove ∪ i I A i ⊆ ∪ j J A j . Fix z ; assume z ∈ ∪ i I A i ; prove z ∈ ∪ j J A j . We have assumed i I , z A i ; we must prove j J , z A j . Let i 0 be a fixed index in I with z A i 0 . Choose j 0 = i 0 . We know i 0 I and I J , so j 0 = i 0 J . Also z A j 0 for this choice of j 0 .

3.77 扩展索引集的证明 假设 I ⊆ J。首先,我们证明 ∪ i I A i ⊆ ∪ j J A j 。固定 z;假设 z ∈ ∪ i I A i ;证明 z ∈ ∪ j J A j 。我们已经假设 ∃i∈I,z∈Ai;我们必须证明 ∃j∈J,z∈Aj。设 i 0 是 I 中的一个固定索引,且 z∈Ai0。选择 j 0 = i 0 。我们知道 i 0 ∈ I 且 I ⊆ J,所以 j 0 = i 0 ∈ J。此外,对于 j 0 的这种选择,z∈Aj0。

Next, we prove ∩ j J A j ⊆ ∩ i I A i (note the reversal of the inclusion here). Fix w ; assume w ∈ ∩ j J A j ; prove w ∈ ∩ i I A i . We know j J , w A j ; we must prove i I , w A i . Fix i 0 I ; prove w A i 0 . Since i 0 I and I J , we know i 0 J . So we can take j = i 0 in the known universal statement to see that w A i 0 , as needed.

接下来,我们证明 ∩ j J A j ⊆ ∩ i I A i (注意此处包含关系的反转)。固定 w;假设 w ∈ ∩ j J A j ;证明 w ∈ ∩ i I A i 。我们已知 ∀j∈J,w∈Aj;我们必须证明 ∀i∈I,w∈Ai。固定 i 0 ∈ I;证明 w∈Ai0。因为 i 0 ∈ I 且 I ⊆ J,所以我们知道 i 0 ∈ J。因此,我们可以在已知的全称语句中取 j = i 0 ,以证明 w∈Ai0,这是必要的。

3.78 Proof of ( i I A i ) ( j J A j ) = k I J A k . We prove this set equality using a two-part proof.

3.78 证明 (∩i∈IAi)∩(∩j∈JAj)=∩k∈I∪JAk。我们使用两部分证明来证明这个集合等式。

Part 1. [proving ⊆ ] Fix z ; assume z ( i I A i ) ( j J A j ) ; prove z k I J A k . We have assumed z ∈ ∩ i I A i and z ∈ ∩ j J A j , which means i I , z A i and j J , z A j . We must prove k I J , z A k . Fix k I J ; prove z A k . We know k I or k J , so consider two cases.

第一部分 [证明 ⊆ ] 固定 z;假设 z∈(∩i∈IAi)∩(∩j∈JAj);证明 z∈∩k∈I∪JAk。我们假设 z ∈ ∩ i I A i 且 z ∈ ∩ j J A j ,这意味着 ∀i∈I, z∈Ai 且 ∀j∈J, z∈Aj。我们必须证明 ∀k∈I∪J, z∈Ak。固定 k∈I∪J;证明 z ∈ A k 。我们知道 k ∈ I 或 k ∈ J,因此考虑两种情况。

Case 1. Assume k I ; prove z A k . This follows by taking i = k in the first universal statement in our earlier assumption.

情况 1. 假设 k ∈ I;证明 z ∈ A k 。这可以通过在我们之前的假设中的第一个全称语句中取 i = k 得出。

Case 2. Assume k J ; prove z A k . This follows by taking j = k in the second universal statement in our earlier assumption.

情况 2. 假设 k ∈ J;证明 z ∈ A k 。这可以通过在我们之前的假设中,在第二个全称语句中取 j = k 得出。

Part 2. [proving ] Fix w ; assume w k I J A k ; prove w ( i I A i ) ( j J A j ) . We have assumed k I J , w A k . We must prove w ∈ ∩ i I A i and w ∈ ∩ j J A j . First, we must prove i I , w A i . Fix i I ; prove w A i . Since I I J , we know i I J and hence w A i (by taking k = i in the earlier assumption). Next, we must prove j J , w A j . Fix j J ; prove w A j . Since J I J , we know j I J and hence w A j (by taking k = j in the earlier assumption).

第二部分 [证明 ⊇] 固定 w;假设 w∈∩k∈I∪JAk;证明 w∈(∩i∈IAi)∩(∩j∈JAj)。我们假设 ∀k∈I∪J,w∈Ak。我们必须证明 w ∈ ∩ i I A i 和 w ∈ ∩ j J A j 。首先,我们必须证明 ∀i∈I,w∈Ai。固定 i ∈ I;证明 w ∈ A i 。由于 I⊆I∪J,我们知道 i∈I∪J,因此 w ∈ A i (通过在之前的假设中取 k = i)。接下来,我们必须证明 ∀j∈J,w∈Aj。固定 j ∈ J;证明 w ∈ A j 。由于 J⊆I∪J,我们知道 j∈I∪J,因此 w ∈ A j (通过在前面的假设中取 k = j)。

Set-Builder Notation

集合构建器符号

Before concluding our study of set theory, we need to consider a notational device called set-builder notation that is frequently used to introduce new sets. Suppose P ( x ) is an open sentence with free variable x . Informally, the notation S = { x : P ( x )} means that S is the set of all objects x such that P ( x ) is true. Similarly, if B is a given set, the notation S = { x B : P ( x )} means that S is the set of all objects x B such that P ( x ) is true. Many variations of this notation are possible; for example, we could define product sets by writing A × B = { ( x , y ) : x A y B } . This formula could be read: “ A × B is the set of all ordered pairs ( x , y ) such that x is in A and y is in B .” Here is a formal definition of the two basic forms of set-builder notation.

在结束集合论的学习之前,我们需要了解一种称为集合构造符号的记号工具,它常用于引入新的集合。假设 P(x) 是一个包含自由变量 x 的开语句。通俗地说,记号 S = {x: P(x)} 表示 S 是所有满足 P(x) 为真的对象 x 的集合。类似地,如果 B 是一个给定的集合,记号 S = {x ∈ B: P(x)} 表示 S 是所有满足 P(x) 为真的对象 x ∈ B 的集合。这种记号有很多变体;例如,我们可以定义乘积集为 A×B={(x,y):x∈A∧y∈B}。这个公式可以理解为:“A × B 是所有满足 x ∈ A 且 y ∈ B 的有序对 (x, y) 的集合。” 以下是集合构造符号两种基本形式的正式定义。

3.79. Definition: Set-Builder Notation. For all open sentences P ( x ), all sets B , and all objects z , z { x : P ( x ) } iff P ( z ) is true ; and z { x B : P ( x ) } iff z B and P ( z ) is true . The letter x can be replaced by any unused letter. For instance,

3.79. 定义:集合构造记号。对于所有开语句 P(x)、所有集合 B 和所有对象 z,z∈{x:P(x)} 当且仅当 P(z) 为真;z∈{x∈B:P(x)} 当且仅当 z∈B 且 P(z) 为真。字母 x 可以替换为任何未使用的字母。例如,

{ x x : P P ( x x ) } = { y : P P ( y ) } = { t t : P P ( t t ) } .

Table 3.1 illustrates set-builder notation by redefining the main set operations of this chapter using this notation. You must become familiar with this notation, as it is used universally in mathematics texts. You might wonder why I have avoided this notation so far in this text (and why I use it only sparingly in later chapters). The reason is this: in many years of teaching proofs, I have observed that set-builder notation is one of the biggest stumbling blocks for beginning students trying to write proofs about sets. In a proof involving set unions, say, one must fluidly transition back and forth between the statement “ z A B ” and the equivalent statement “ z A or z B ” — so I use this equivalence as the very definition of set unions. The set-builder definition A B = { z : z A z B } means exactly the same thing, but the extra level of translation needed to use this definition in a proof seems to cause endless trouble.

表 3.1 通过使用集合构造符号重新定义本章的主要集合运算,展示了集合构造符号的应用。你必须熟悉这种符号,因为它在数学教材中被广泛使用。你可能会好奇为什么我在本书中一直避免使用这种符号(以及为什么在后面的章节中也只是偶尔使用)。原因如下:在多年的证明教学中,我发现集合构造符号是初学者在尝试编写集合证明时遇到的最大障碍之一。例如,在涉及集合并的证明中,必须在语句“z∈A∪B”和等价语句“z ∈ A 或 z ∈ B”之间流畅地转换——所以我使用这种等价关系作为集合并的定义。集合构造定义 A∪B={z:z∈A∨z∈B} 的含义完全相同,但在证明中使用这个定义所需的额外转换似乎会造成无穷无尽的麻烦。

Table 3.1
表 3.1 Definitions of Set Operations in Set-Builder Notation.
集合构造符号中的集合运算定义。

A 一个 B B = { x x : x x A 一个 x x B B } A 一个 B B = { x x : x x A 一个 x x B B } A 一个 B B = { x x : x x A 一个 x x B B } = { x x A 一个 : x x B B } P P ( A 一个 ) = { S S : S S A 一个 } A 一个 × × B B = { ( x x , y ) : x x A 一个 y B B } = { z z : x x A 一个 , y B B , z z = ( x x , y ) } ( a 一个 , b b ) = { x x ∈: ∈: a 一个 < x x < b b } [ a 一个 , b b ] = { x x ∈: ∈: a 一个 x x b b } i I A 一个 i = { x x : i I , x x A 一个 i } i I A 一个 i = { x x : i I , x x A 一个 i }

Section Summary

章节概要

1. Definitions. Suppose I is an index set and, for each i I , A i is a given set.

1.定义。假设 I≠∅ 是一个索引集,对于每个 i ∈ I,A i 是一个给定的集合。

General Union: For all z , z i I A i iff i I , z A i .

一般联盟:对于所有 z,z∈⋃i∈IAi 当且仅当 ∃i∈I,z∈Ai。

General Intersection: For all z , z i I A i iff i I , z A i .

一般交集:对于所有 z,z∈∩i∈IAi 当且仅当 ∀i∈I,z∈Ai。

2. Properties of General Unions and Intersections. The general union and intersection operations satisfy properties such as monotonicity, de Morgan laws, and the distributive laws. Also, ∩ i I A i is the largest set contained in every A i , whereas ∪ i I A i is the smallest set containing every A i . Replacing I by a larger index set J produces a larger union and a smaller intersection. We can simplify a union of unions or an intersection of intersections by combining the index sets.

2. 一般并集和交集的性质。一般并集和交集运算满足单调性、德摩根定律和分配律等性质。此外,∩ i I A i 是包含于所有 A i 中的最大集合,而 ∪ i I A i 是包含于所有 A i 中的最小集合。将 I 替换为更大的索引集 J 会得到更大的并集和更小的交集。我们可以通过合并索引集来简化并集的并集或交集的交集。

3. Set-Builder Notation. Informally, { x B : P ( x )} is the set of all objects x in the set B that make P ( x ) true. Formally, for all open sentences P ( x ), all sets B , and all objects z , z { x B : P ( x ) } iff z B P ( z ) . Notation such as { x : P ( x )} and {( x , y ): P ( x , y )} is defined similarly. The letter x can be replaced by any unused letter; e.g., { x : P ( x )} = { t : P ( t )}.

3. 集合构造器符号。非正式地,{x ∈ B:P(x)} 表示集合 B 中所有使 P(x) 为真的对象 x 的集合。形式上,对于所有开语句 P(x)、所有集合 B 和所有对象 z,z∈{x∈B:P(x)} 当且仅当 z∈B∧P(z)。诸如 {x:P(x)} 和 {(x, y):P(x, y)} 之类的符号定义类似。字母 x 可以被任何未使用的字母替换;例如,{x:P(x)} = {t:P(t)}。

Exercises

练习

1. Let I be the set of odd integers. For n I , define A n = ( n − 1, n + 1). Sketch and describe the set ℝ − ∪ n I A n .

1.设 I 为奇数集合。对于 n ∈ I,定义 A n = (n − 1, n + 1)。绘制并描述集合 ℝ − ∪ n I A n

2. For each index set I and sets A i below, describe i I A i and i I A i .

2.对于下面的每个索引集 I 和集合 A i ,描述 ⋃i∈IAi 和 ∩i∈IAi。

Do not prove your answers; it can help to draw pictures in ℝ and ℝ 2 .

不要证明你的答案;在ℝ和ℝ 2 中画图可能会有所帮助。

(a) I = ℤ >0 ; for each n I , A n is the half-open interval (1 + 1/ n , 5 − 1/ n ].

(a)I = ℤ >0 ;对于每个 n ∈ I,A n 是半开区间 (1 + 1/n, 5 − 1/n)。

(b) I = [0, 3]; for each r I , A r is the open interval ( r − 2, r + 2).

(b)I = [0, 3];对于每个 r ∈ I,A r 是开区间 (r − 2, r + 2)。

(c) I = ℝ >0 ; for each r I , A r is the rectangle [0, r ] × [0, 1/ r ].

(c)I = ℝ >0 ;对于每个 r ∈ I,A r 是矩形 [0, r] × [0, 1/r]。

(d) I = ℝ >0 ; for each r I , A r is the set of ( x , y ) with ( x r ) 2 + y 2 r 2 .

(d)I = ℝ >0 ;对于每个 r ∈ I,A r 是满足 (x − r) 2 + y 2 ≤ r 2 的 (x, y) 的集合。

3. Let a , b be fixed real numbers with a < b . (a) Express the closed interval [ a , b ] as an intersection of infinitely many open intervals. (b) Express the open interval ( a , b ) as a union of infinitely many closed intervals.

3.设a、b为固定的实数,且a < b。(a)将闭区间[a, b]表示为无穷多个开区间的交集。(b)将开区间(a, b)表示为无穷多个闭区间的并集。

4. In Example 3.73 , carefully prove the claimed set equalities n = 1 A n = Q and n = 1 A n = Z .

4.在例 3.73 中,仔细证明所声称的集合等式 ⋃n=1∞An=Q 和 ∩n=1∞An=Z。

5. For each n ∈ ℤ >0 , let

5.对于每个 n ∈ ℤ >0 ,令

A 一个 n n = { m Z Z : m divides 分割 n n } and B B n n = { k k Z Z : n n divides 分割 k k } .

(a) Compute n = 1 A n and n = 1 A n .

(a)计算⋃n=1∞An和∩n=1∞An。

(b) Compute n = 1 B n and n = 1 B n .

(b)计算⋃n=1∞Bn和∩n=1∞Bn。

(c) Compute n = 1 ( A n B n ) .

(c)计算⋃n=1∞(An∩Bn)。

6. (a) Suppose A 1 , A 2 , …, A n , … are sets such that A n A n +1 for all n ∈ ℤ >0 . For i j , what are the sets n = i j A n and n = i j A n ?

6.(a)假设 A 1 , A 2 , …, A n , … 是集合,使得对于所有 n ∈ ℤ >0 ,A n ⊆ A n +1 。对于 i ≤ j,集合 ⋃n=ijAn 和 ∩n=ijAn 是什么?

(b) Suppose B 1 , B 2 , …, B n , … are sets such that B n B n + 1 for all n ∈ ℤ >0 . For i j , what are the sets n = i j B n and n = i j B n ?

(b)假设B 1 , B 2 , …, B n , …是集合,使得对于所有n ∈ ℤ >0 ,都有Bn⊇Bn+1。对于i ≤ j,集合⋃n=ijBn和∩n=ijBn是什么?

7. Prove part (a) of the Theorem on General Unions and Intersections (monotonicity).

7.证明关于一般并集和交集的定理(a)部分(单调性)。

8. (a) Finish the proof of the de Morgan Law in Proof 3.76 .

8.(a)完成证明 3.76 中德摩根定律的证明。

(b) Prove the second de Morgan Law in part (b) of the Theorem on General Unions and Intersections.

(b)证明一般并集和交集定理(b)部分中的第二德摩根定律。

9. Prove part (e) of the Theorem on General Unions and Intersections (greatest lower bound property). (b) Prove part (f) of the Theorem on General Unions and Intersections (least upper bound property).

9.证明一般并集与交集定理的(e)部分(最大下界性质)。(b)证明一般并集与交集定理的(f)部分(最小上界性质)。

10. Prove the distributive laws for general unions and intersections (part (c) of the Theorem on General Unions and Intersections).

10.证明一般并集和交集的分配律(关于一般并集和交集的定理的(c)部分)。

11. Use set-builder notation to describe the following sets:

11. 使用集合构建符号描述下列集合:

(a) ( − ∞, 3] (b)[ a , b ) (c) the set of odd integers (d) ℚ.

(a)( − ∞, 3] (b)[a, b) (c)奇数集合 (d)ℚ。

12. Draw pictures of these sets described in set-builder notation:

12.用集合构建符号画出下列集合的图像:

(a) { x ∈ ℝ: x > 3∨ x < −3}

(a){x ∈ ℝ: x > 3∨x < −3}

(b) {( x , y ) ∈ ℝ × ℝ: 0 ≤ y ≤ 4 − x 2 }

(b){(x, y) ∈ ℝ × ℝ: 0 ≤ y ≤ 4 −x 2

(c) {( x , y , z ) ∈ ℝ 3 :max (| x |, | y |, | z |) = 1}

(c){(x, y, z) ∈ ℝ 3 :max (|x|, |y|, |z|) = 1}

(d) {( x , y ) ∈ [0, 3] × [0, 3]: x + y ∈ ℤ}

(d){(x, y) ∈ [0, 3] × [0, 3]: x + y ∈ ℤ}

13. Give specific examples of sets B n ⊆ ℝ (for each n ∈ ℤ >0 ) such that B n B n + 1 for all n and n = 1 B n is: (a) the empty set; (b) {7}; (c) [ − 1, 1]; (d) ℤ.

13.给出集合 B n ⊆ ℝ (对于每个 n ∈ ℤ >0 ) 的具体例子,使得对于所有 n,Bn⊋Bn+1,并且 ∩n=1∞Bn 是: (a) 空集; (b) {7}; (c) [ − 1, 1]; (d) ℤ。

14. The goal of this problem is to get practice using known theorems to help prove new theorems. Fill in all the blanks in the proof below; the reasons for each step should be appropriate parts of the Theorem on Subsets, the Theorem on Set Equality, and the Theorem on General Unions and Intersections (excluding the first identity in part (h), which we are proving).

14. 本题的目的是练习运用已知定理来证明新定理。请填写以下证明中的所有空白处;每一步的理由应属于子集定理、集合相等定理以及一般并集与交集定理的适当部分(不包括 (h) 部分中的第一个恒等式,因为我们正在证明它)。

Theorem. For all I , J , A i , A j , ( i I A i ) ( j J A j ) = k I J A k .

定理。对于所有 I,J≠∅,A i , A j , (⋃i∈IAi)∪(⋃j∈JAj)=⋃k∈I∪JAk。

Proof. Fix arbitrary sets I , J , A i , and A j .

证明。固定任意集合 I,J≠∅,A i 和 A j

Part 1. We must prove ( i I A i ) ( j J A j ) k I J A k .

第 1 部分。我们必须证明 (⋃i∈IAi)∪(⋃j∈JAj)⊆⋃k∈I∪JAk。

We know I I J (by (a) from the Theorem on Subsets), and so i I A i k I J A k (by (b) from the Theorem on General Unions and Intersections). Similarly, we know (c) (by (d) from the Theorem on Subsets), and so j J A j k I J A k (by (e) from the Theorem on General Unions and Intersections). Combining these two facts with (f) from the Theorem on Subsets, part 1 is proved.

我们知道 I⊆I∪J(由子集定理的 (a) 可知),因此 ⋃i∈IAi⊆⋃k∈I∪JAk(由一般并集和交集定理的 (b) 可知)。类似地,我们知道 (c)(由子集定理的 (d) 可知),因此 ⋃j∈JAj⊆⋃k∈I∪JAk(由一般并集和交集定理的 (e) 可知)。结合这两个事实以及子集定理的 (f),第一部分得证。

Part 2. We must prove k I J A k ( i I A i ) ( j J A j ) .

第 2 部分。我们必须证明 ⋃k∈I∪JAk⊆(⋃i∈IAi)∪(⋃j∈JAj)。

By (g) from the Theorem on General Unions and Intersections, we may instead prove the equivalent statement:

由一般并集与交集定理的 (g) 可知,我们可以证明以下等价命题:

k k I J J , A 一个 k k ( i I A 一个 i ) ( j j J J A 一个 j j ) .

Fix k I J , so we know k I or k J . To prove A k ( i I A i ) ( j J A j ) , use cases.

固定 k∈I∪J,因此我们知道 k ∈ I 或 k ∈ J。为了证明 Ak⊆(⋃i∈IAi)∪(⋃j∈JAj),使用案例。

Case 1. Assume k I . First, A k ⊆ ∪ i I A i by (h) .

情况 1. 假设 k ∈ I。首先,根据 (h),A k ⊆ ∪ i I A i

Second, i I A i ( i I A i ) ( j J A j ) by (i) .

其次,⋃i∈IAi⊆(⋃i∈IAi)∪(⋃j∈JAj) 由 (i) 。

Therefore, A k ( i I A i ) ( j J A j ) by (j) .

因此,由 (j) 可知 Ak⊆(⋃i∈IAi)∪(⋃j∈JAj)。

Case 2. Assume k J . First, A k ⊆ ∪ j J A j by (k) .

情况 2. 假设 k ∈ J。首先,由 (k) 可知 A k ⊆ ∪ j J A j

Second, j J A j ( i I A i ) ( j J A j ) by (l) .

其次,⋃j∈JAj⊆(⋃i∈IAi)∪(⋃j∈JAj) 由 (l) 。

Therefore, A k ( i I A i ) ( j J A j ) by (m) .

因此,由 (m) 可知 Ak⊆(⋃i∈IAi)∪(⋃j∈JAj)。

15. Recall that an indexed collection of sets { S i : i I } is called pairwise disjoint iff for all i , j I , if i j , then S i S j = . Suppose A 1 , A 2 , …, A n , … are sets such that A n A n +1 for all n ∈ ℤ >0 . Define B 1 = A 1 and B n = A n A n −1 for all n ∈ ℤ >1 . Prove that { B n : n ∈ ℤ >0 } is pairwise disjoint, and n = 1 B n = n = 1 A n . Illustrate with a Venn diagram. [ Hint: If x n = 1 A n , then there exists a least positive integer n 0 with x A n 0 . Show that x B n 0 .]

15. 回顾一下,集合 {S i :i ∈ I} 被称为两两不相交,当且仅当对于所有 i, j ∈ I,如果 i ≠ j,则 Si∩Sj=∅。假设 A 1 , A 2 , …, A n , … 是集合,使得对于所有 n ∈ ℤ >0 ,A n ⊆ A n +1 。定义 B 1 = A 1 和 B n = A n − A n −1 ,对于所有 n ∈ ℤ >1 。证明 {B n :n ∈ ℤ >0 两两不相交,且 ⋃n=1∞Bn=⋃n=1∞An。用维恩图表示。[提示:若 x∈⋃n=1∞An,则存在最小正整数 n 0 使得 x∈An0。证明 x∈Bn0。]

16. Let { A n : n ∈ ℤ >0 } be an indexed family of sets. Define B 1 = A 1 and, for each n > 1, define B n = A n j = 1 n 1 A j . Prove that { B n : n ∈ ℤ >0 } is pairwise disjoint, and n = 1 B n = n = 1 A n . Illustrate A n and B n for 1 ≤ n ≤ 4 in a Venn diagram.

16.设{A n :n ∈ ℤ >0 为一个带索引的集合族。定义B 1 = A 1 ,并且对于每个n > 1,定义Bn=An−⋃j=1n−1Aj。证明{B n :n ∈ ℤ >0 两两不相交,且⋃n=1∞Bn=⋃n=1∞An。用维恩图表示1 ≤ n ≤ 4的A n 和B n

17. Let { A n : n ∈ ℤ >0 } be an indexed family of sets. Define B n = j = 1 n A j for each n ∈ ℤ >0 . Prove that B n B n +1 for all n ∈ ℤ >0 , and n = 1 B n = n = 1 A n .

17.设{A n :n ∈ ℤ >0 为一个带索引的集合族。定义Bn=⋃j=1nAj,其中n∈ℤ >0 。证明对于所有n∈ℤ >0 ,B n ⊆B n +1 ,且⋃n=1∞Bn=⋃n=1∞An。

18. Let { C n : n ∈ ℤ >0 } be an indexed family of subsets of a given set X . Define D n = j = 1 n C j for each n ∈ ℤ >0 . Prove that D n D n + 1 for all n ∈ ℤ >0 , and n = 1 D n = n = 1 C n . Do this by using the result of the previous exercise and other known theorems. [ Hint: Consider A n = X C n .]

18.设{C n :n ∈ ℤ >0 为给定集合X的子集族。定义Dn=∩j=1nCj,其中n ∈ ℤ >0 。证明对于所有n ∈ ℤ >0 ,Dn⊇Dn+1,且∩n=1∞Dn=∩n=1∞Cn。利用上一题的结果和其他已知定理证明此结论。[提示:考虑A n = X − C n 。]

19. (a) Prove or disprove: for all I and all sets A i , B i ,

19.(a)证明或反驳:对于所有 I≠∅ 和所有集合 A i , B i ,

i I ( A 一个 i B B i ) ( j j I A 一个 j j ) ( k k I B B k k ) .

(b) Prove or disprove: for all I and all sets A i , B i ,

(b)证明或反驳:对于所有 I≠∅ 和所有集合 A i , B i ,

i I ( A 一个 i B B i ) ( j j I A 一个 j j ) ( k k I B B k k ) .

20. A subset U of ℝ is called an open set iff x U , r R > 0 , ( x r , x + r ) U .

20.ℝ 的子集 U 被称为开集,当且仅当对于 ∀x∈U,∃r∈R>0,(x−r,x+r)⊆U。

(a) Prove: ∅ and ℝ are open sets.

(a)证明:∅ 和 ℝ 是开集。

(b) Prove: every open interval ( a , b ) is an open set.

(b)证明:每个开区间(a,b)都是一个开集。

(c) Prove: for all U , V ⊆ ℝ, if U and V are open sets, then U V is an open set.

(c)证明:对于所有 U, V ⊆ ℝ,如果 U 和 V 是开集,则 U ∩ V 是开集。

(d) Prove: for any indexed collection { U i : i I } of subsets of ℝ, if U i is an open set for all i I , then ∪ i I U i is an open set.

(d)证明:对于 ℝ 的子集的任意索引集合 {U i :i ∈ I},如果 U i 是所有 i ∈ I 的开集,则 ∪ i I U i 是开集。

21. A subset C of ℝ is called a closed set iff ℝ − C is an open set (see Exercise 20).

21. ℝ 的子集 C 被称为闭集,当且仅当 ℝ −C 是开集(见练习 20)。

(a) Prove: ∅ and ℝ are closed sets.

(a)证明:∅ 和 ℝ 是闭集。

(b) Prove: every closed interval [ a , b ] is a closed set.

(b)证明:每个闭区间[a, b]都是一个闭集。

(c) Prove: ℤ is a closed set.

(c)证明:ℤ 是一个闭集。

(d) Find (with proof) a specific subset of ℝ that is neither an open set nor a closed set.

(d)找出(并证明)ℝ 的一个特定子集,该子集既不是开集也不是闭集。

(e) Prove: the union of any two closed sets is a closed set.

(e)证明:任意两个闭集的并集仍然是闭集。

(f) Prove: the intersection of any nonempty family of closed sets is a closed set.

(f)证明:任意非空闭集族的交集是闭集。

22. An open disk is a subset of ℝ 2 of the form {( x , y ):( x h ) 2 + ( y k ) 2 < r 2 }, where h , k , r ∈ ℝ are fixed real numbers with r > 0. Express each of the following subsets of ℝ 2 as a union of an indexed collection of open disks.

22. 一个开圆盘是 ℝ 2 的一个子集,其形式为 {(x, y):(x − h) 2 + (y − k) 2 < r 2 ,其中 h, k, r ∈ ℝ 是固定的实数,且 r > 0。将 ℝ 2 的每个子集表示为一组开圆盘的索引集合的并集。

(a) 2 (b) { ( x , y ) R 2 : x > 0 y < 0 } (c) (−1, 1) × ℝ (d) (0, 5) × (2, 4) (e) {( x , y ) ∈ ℝ 2 : y > | x |}.

(a)ℝ 2 (b) {(x,y)∈R2:x>0∧y<0} (c) (−1, 1) × ℝ (d) (0, 5) × (2, 4) (e) {(x, y) ∈ ℝ 2 : y > |x|}.

23. Show that the intersection of two open disks in ℝ 2 is either empty or is the union of an indexed collection of open disks.

23.证明ℝ 2 中两个开圆盘的交集要么为空,要么是开圆盘的索引集合的并集。

24. Let X be a fixed set, and let A be a collection of subsets of X with the following properties: (i) A ; (ii) for all S A , X S A ; (iii) given { S n : n ∈ ℤ >0 } with every S n A , n = 1 S n A . Prove: (a) X A ; (b) for any { S n : n ∈ ℤ >0 } with every S n A , n = 1 S n A ; (c) for all S , T A , S T A ; (d) for all S , T A , S T A .

24.设 X 为一个固定的集合,A 为 X 的子集族,具有以下性质:(i) ∅∈A;(ii) 对于所有 S ∈ A,X−S∈A;(iii) 给定 {S n :n ∈ ℤ >0 ,其中每个 Sn∈A,⋃n=1∞Sn∈A。证明:(a) X∈A;(b) 对于任意 {S n :n ∈ ℤ >0 ,其中每个 Sn∈A,∩n=1∞Sn∈A;(c) 对于所有 S,T∈A,S∪T∈A;(d) 对于所有 S,T∈A,S∩T∈A。

25. Let X = {1, 2, 3}. (a) Give three different examples of collections A of subsets of X satisfying conditions (i) through (iii) in the previous exercise. (b) How many such collections A are there for this set X ?

25.设 X = {1, 2, 3}。(a)给出三个不同的 X 的子集集合 A,满足上题中的条件 (i) 到 (iii)。(b)对于集合 X,有多少个这样的子集集合 A?

3.7 Axiomatic Set Theory (Optional)

3.7 公理化集合论(选修)

So far in this chapter, we have defined several operations on sets and learned how to prove theorems involving these set operations. However, there is a subtle logical omission in our presentation of set theory so far. In many instances, we have introduced a new set (such as the union A B , or the intersection A B , or the power set P ( A )) without giving any formal justification for why the new set actually exists . Such justifications may seem unnecessary, but (as we will see below) if we indiscriminately assume the existence of certain sets, logical paradoxes can result.

在本章到目前为止,我们已经定义了集合上的几种运算,并学习了如何证明涉及这些集合运算的定理。然而,我们目前对集合论的阐述存在一个微妙的逻辑疏漏。在很多情况下,我们引入了一个新的集合(例如并集 A ∪ B、交集 A ∩ B 或幂集 P(A)),却没有给出任何正式的证明来解释为什么这个新集合确实存在。这种证明看似没有必要,但(正如我们将在下文看到的)如果我们不加区分地假设某些集合的存在,就可能导致逻辑悖论。

A completely rigorous logical development of set theory (as opposed to the semi-formal treatment given in earlier sections) is called axiomatic set theory In axiomatic set theory, we must use explicitly stated axioms to justify the existence (and uniqueness) of each set under consideration. This optional section gives a sketch of the framework and axioms for one version of axiomatic set theory called the ZFC (Zermelo–Fraenkel–Ch system. More detailed introductions to axiomatic set theory can be found in the texts listed in the Suggestions for Further Reading at the end of the book.

对集合论进行完全严谨的逻辑推导(与前几节中半形式化的处理方式相对)被称为公理化集合论。在公理化集合论中,我们必须使用明确陈述的公理来证明所考虑的每个集合的存在性(以及唯一性)。本选读章节概述了公理化集合论的一个版本——ZFC(策梅洛-弗兰克尔-切尔系统)的框架和公理。关于公理化集合论的更详细介绍,请参阅本书末尾“延伸阅读建议”中列出的文献。

Undefined Concepts

未定义概念

Axiomatic set theory begins with two undefined notions: the idea of a set , and the concept of membership in a set. One aspect of the theory that may seem unusual at first is that all objects under consideration are sets . You might think that we would need to allow other types of mathematical objects such as numbers, functions, geometric figures, and so forth. Remarkably, all of these objects can be defined as certain kinds of sets. It follows that all variables appearing in the axioms and formulas below represent sets. As above, we use ∈ as a symbol for the undefined concept of set membership. Thus, for any two objects (sets) x and y , exactly one of the following alternatives holds: x y (“ x is a member of y ”), or x y (“ x is not a member of y ”). You must learn not to be distracted by the fact that a lowercase letter is used for the set y . We could have equally well used uppercase letters for both these sets, or a mixture of lowercase and uppercase letters.

公理化集合论始于两个未定义的概念:集合的概念和集合成员关系的概念。该理论的一个特点乍看之下可能有些不同寻常,那就是所有被考虑的对象都是集合。你可能会认为我们需要允许其他类型的数学对象,例如数字、函数、几何图形等等。但令人惊讶的是,所有这些对象都可以被定义为某种类型的集合。由此可见,下文公理和公式中出现的所有变量都代表集合。如上所述,我们使用 ∈ 作为集合成员关系这个未定义概念的符号。因此,对于任意两个对象(集合)x 和 y,以下两种情况中只有一种成立:x ∈ y(“x 是 y 的成员”),或者 x ∉ y(“x 不是 y 的成员”)。你必须学会不要被集合 y 的首字母小写形式所迷惑。我们完全可以使用大写字母来表示这两个集合,或者使用大小写混合的字母。

Russell’s Paradox

罗素悖论

Our intuition tells us that, for any logical property, there should exist a set whose members are precisely the objects that satisfy the given property. Formalizing this intuition, we are led to propose the following axiom.

我们的直觉告诉我们,对于任何逻辑性质,都应该存在一个集合,其成员恰好是满足该性质的对象。将这种直觉形式化,我们便提出了以下公理。

3.80 Axiom of Abstraction (Tentative) Let P ( z ) be a statement with a free variable z , where the variable x does not occur in P ( z ). Axiom: x , z , ( z x ) P ( z ) .

3.80 抽象公理(暂定)设 P(z) 是一个包含自由变量 z 的语句,其中变量 x 不出现在 P(z) 中。公理:∃x,∀z,(z∈x)⇔P(z)。

This axiom is really a family of axioms, one for each statement P ( z ). The members of the set x in the axiom are precisely those objects (sets) z for which P ( z ) is true. In set-builder notation, we could write x = { z : P ( z )} (although use of this notation suggests the set x is unique, which we have not proved yet).

这条公理实际上是一族公理,每个命题 P(z) 对应一条公理。公理中集合 x 的成员恰好是满足 P(z) 为真的所有对象(集合)z。用集合构造法表示,我们可以写成 x = {z:P(z)}(尽管使用这种表示法暗示集合 x 是唯一的,而这一点我们尚未证明)。

3.81 Example Let P ( z ) be the statement “ z z .” By the Axiom of Abstraction, there is a set x such that for all z , z x iff z z . By properties of logical equality, z z is false for any z . It follows that for all z , z x . So, we have used the axiom to prove the existence of an empty set x .

3.81 例 设 P(z) 为命题“z ≠ z”。根据抽象公理,存在一个集合 x,使得对于所有 z,z ∈ x 当且仅当 z ≠ z。根据逻辑等式的性质,对于任意 z,z ≠ z 都为假。由此可知,对于所有 z,z∉ x。因此,我们利用抽象公理证明了空集 x 的存在性。

3.82 Example On the other hand, letting P ( z ) be the statement “ z = z ,” which holds for every z , we obtain a set x such that z x for all sets z . This set x is a universal set that co all objects (sets) in the universe being studied. In particular, since x itself is a set, we have x x ( x is a member of itself).

3.82 例 另一方面,令 P(z) 为命题“z = z”,该命题对每个 z 都成立,我们得到一个集合 x,使得对于所有集合 z,都有 z ∈ x。这个集合 x 是一个全集,它包含了所研究的宇宙中的所有对象(集合)。特别地,由于 x 本身也是一个集合,因此我们有 x ∈ x(x 是它自身的成员)。

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3.83 Example Now let P ( z ) be the statement “ z z .” The Axiom of Abstraction asserts that there is a set x for which the following statement is true:

3.83 示例 现在设 P(z) 为命题“z∉z”。抽象公理断言存在一个集合 x,使得以下命题为真:

z z , ( z z x x ) ( z z z z ) .

In words, x is the set of all sets z such that z is not a member of itself. In the universal statement just written, we can choose z to be the particular set x , thereby obtaining the proposition

换句话说,x 是所有集合 z 的集合,其中 z 不属于自身。在刚才写的全称命题中,我们可以选择 z 为特指集合 x,从而得到该命题。

( x x x x ) ( x x x x ) .

But this proposition is false, since the statements on the two sides of the biconditional symbol must have opposite truth values. So, the Axiom of Abstraction has led us to a contradiction!

但这个命题是错误的,因为双条件符号两侧的语句必须具有相反的真值。因此,抽象公理导致了矛盾!

The phenomenon in the last example is called Russell’s Paradox (named after its discoverer, Bertrand Russell). The only way to avoid this paradox is to give up the Axiom of Abstraction, despite its initial intuitive appeal, and replace it with more delicate existence axioms for sets. We describe these axioms below.

最后一个例子中的现象被称为罗素悖论(以其发现者伯特兰·罗素的名字命名)。避免这个悖论的唯一方法是放弃抽象公理,尽管它最初看起来很直观,并用更精细的集合存在性公理来代替它。我们将在下文中描述这些公理。

The Axiom of Extension

外延公理

We now state the axioms of the ZFC theory. For each axiom, we give a brief discussion of the intuitive meaning of the axiom and how it is used.

现在我们陈述ZFC理论的公理。对于每一条公理,我们简要讨论其直观含义及其用途。

3.84 Axiom of Extension x , y , [ x = y ( z , z x z y ) ] .

3.84 扩展公理 ∀x,∀y,[x=y⇔(∀z,z∈x⇔z∈y)].

Intuitively, this axiom indicates the relationship between the logical equality of sets and the undefined concept of set membership. The axiom says that two sets x and y are equal iff they have exactly the same members. Recall that we defined x y to mean z , z x z y . We also saw earlier that an equivalent way to formulate the Axiom of Extension is to say that x = y iff x y and y x . The Axiom of Extension is often used to verify the uniqueness of a newly constructed set, as we will see in examples below.

直观地说,这条公理揭示了集合的逻辑相等性与集合成员关系这一未定义概念之间的联系。该公理指出,两个集合 x 和 y 相等当且仅当它们拥有完全相同的元素。回想一下,我们定义 x ⊆ y 为 ∀z, z∈x⇒z∈y。我们之前也看到,外延公理的等价表述方式是:x = y 当且仅当 x ⊆ y 且 y ⊆ x。外延公理常用于验证新构造集合的唯一性,我们将在下面的例子中看到这一点。

We should mention here that the axioms of set theory are overlaid on top of certain logical axioms, which we have not listed explicitly. Some of these logical axioms describe the properties of the equality symbol =. For example, we might have axioms asserting that equality is reflexive, symmetric, and transitive (although we have seen that these properties can be proved from the Axiom of Extension). Some other axioms are needed to ensure that equal objects can be substituted for one another wherever they appear in a statement. In the context of set theory, it suffices to adopt these two Substitution Axioms:

这里需要指出的是,集合论的公理建立在某些逻辑公理之上,这些逻辑公理我们并未明确列出。其中一些逻辑公理描述了等号“=”的性质。例如,我们可能有一些公理断言等式具有自反性、对称性和传递性(尽管我们已经看到这些性质可以从外延公理证明)。还需要一些其他的公理来确保相等的对象可以在语句中任意位置相互替换。在集合论的语境下,采用以下两个替换公理就足够了:

x x , y , [ x x = y ( z z , x x z z y z z ) ] ;
x x , y , [ x x = y ( z z , z z x x z z y ) ] .

With these logical axioms already in place, the new content provided by the Axiom of Extension is the converse of the second Substitution Axiom:

在这些逻辑公理已经存在的情况下,外延公理所提供的新内容是第二个替换公理的逆命题:

x x , y , [ ( z z , z z x x z z y ) x x = y ] .

Additional substitution properties involving defined concepts such as union and intersection can then be proved as theorems (see part (d) of the Theorem on Set Equality).

然后,可以证明涉及并集和交集等已定义概念的其他替换性质为定理(参见集合相等定理的(d)部分)。

The Axiom of Specification

规范公理

Our first set construction axiom lets us create subsets of given sets, if we can state a condition that determines which objects belong to the subset.

我们的第一个集合构造公理允许我们创建给定集合的子集,只要我们能陈述一个条件来确定哪些对象属于该子集。

3.85 Axiom of Specification Let P ( z ) be a statement with a free variable z , in which the variable x does not occur. Axiom: u , x , z , [ z x ( z u P ( z ) ) ] .

3.85 规范公理 设 P(z) 为一个包含自由变量 z 的语句,其中变量 x 不出现。公理:∀u,∃x,∀z,[z∈x⇔(z∈u∧P(z))]。

Intuitively, given any logical property P ( z ) and given a particular set u , we can form a new set x consisting of all objects (sets) belonging to u that have the property P ( z ). In set-builder notation, we could write x = { z U : P ( z )}. Note how closely this axiom resembles the forbidden Axiom of Abstraction. The key difference is that we must already have a set u given in advance, and then we use the property P ( z ) to select or “specify” which elements should belong to the new set x (which is automatically a subset of u ).

直观地说,给定任意逻辑属性 P(z) 和一个特定的集合 u,我们可以构造一个新的集合 x,其中包含所有属于 u 且具有属性 P(z) 的对象(集合)。用集合构造器表示法,我们可以写成 x = {z ∈ U:P(z)}。注意这条公理与被禁止的抽象公理非常相似。关键区别在于,我们必须预先给定一个集合 u,然后使用属性 P(z) 来选择或“指定”哪些元素应该属于新的集合 x(它自动成为 u 的子集)。

3.86 Example: Binary Intersections. Let us use this axiom to justify the existence of the intersection of two sets. Earlier, we “defined” A B by saying that for all z , z A B iff z A and z B . To obtain this set by invoking the Axiom of Specification, we let P ( z ) be the statement “ z B ” and we let u be the set A . The axiom tells us there is a set x such that for all z , z x iff z A and z B . We define this set to be the intersection of A and B , denoted A B . Before introducing this notation for the intersection, we should also check that the set x (whose existence is guaranteed by the Axiom of Specification) is uniquely determined by the given sets A and B . This follows from the Axiom of Extension. For suppose x ′ is another set such that for all z , z x ′ iff z A and z B . We deduce that for all z , z x iff z x ′, hence x = x ′.

3.86 示例:二元交集。让我们利用这条公理来证明两个集合交集的存在性。之前,我们“定义”了 A ∩ B,即对于所有 z,z∈A∩B 当且仅当 z ∈ A 且 z ∈ B。为了通过调用规范公理得到这个集合,我们令 P(z) 为命题“z ∈ B”,并令 u 为集合 A。该公理告诉我们,存在一个集合 x,使得对于所有 z,z ∈ x 当且仅当 z ∈ A 且 z ∈ B。我们将这个集合定义为 A 和 B 的交集,记为 A ∩ B。在引入交集的这种表示方法之前,我们还应该验证集合 x(其存在性由规范公理保证)是否由给定的集合 A 和 B 唯一确定。这由扩展公理得出。假设 x′ 是另一个集合,使得对于所有 z,z ∈ x′ 当且仅当 z ∈ A 且 z ∈ B。我们推断对于所有 z,z ∈ x 当且仅当 z ∈ x′,因此 x = x′。

Similarly, whenever we use the Axiom of Specification to create a new set x , this set is uniquely determined by P ( z ) and u . For if another set x ′ satisfies the conditions in the axiom, then for all z , z x ′ iff ( z u and P ( z )) iff z x . So x = x ′ by the Axiom of Extension. It is now safe to use the notation x = { z u : P ( z )} to denote the set of all z such that z u and P ( z ). On the other hand, the related notation { z : P ( z )} must not be used.

类似地,每当我们使用规范公理创建一个新集合 x 时,该集合都由 P(z) 和 u 唯一确定。因为如果另一个集合 x′ 满足该公理的条件,那么对于所有 z,z ∈ x′ 当且仅当 (z ∈ u 且 P(z)) 当且仅当 z ∈ x。因此,根据扩展公理,x = x′。现在可以安全地使用符号 x = {z ∈ u: P(z)} 来表示所有满足 z ∈ u 且 P(z) 的 z 的集合。另一方面,相关的符号 {z: P(z)} 则不能使用。

3.87 Example: Set Difference. Given any sets A and B , we previously “defined” A B by declaring that for all z , z A B iff z A and z B . The Axioms of Specification and Extension justify this definition by proving the existence and uniqueness of the set denoted A B . This follows by writing A B = { z A : z B } .

3.87 示例:集合差集。给定任意集合 A 和 B,我们之前“定义”了 A − B,即对于所有 z,z ∈ A − B 当且仅当 z ∈ A 且 z∉B。规范公理和扩展公理通过证明集合 A − B 的存在性和唯一性来证明这一定义。这可以通过记作 A−B={z∈A:z∉B} 来得出。

Let us see what happens if we try to recreate Russell’s Paradox using the Axiom of Specification. Let u be a fixed set, and take P ( z ) to be the statement “ z z .” Our axiom provides a set x satisfying

让我们看看如果用规范公理重现罗素悖论会发生什么。设 u 为一个固定的集合,并取 P(z) 为命题“z∉z”。我们的公理提供了一个满足以下条件的集合 x:

z z , [ z z x x ( z z u z z z z ) ] .

Taking z = x here leads to the statement

令 z = x,即可得出以下结论

x x x x ( x x u x x x x ) .

If x u , the statement just written must be false, as in the original discussion of Russell’s Paradox. But now we can escape the contradiction by concluding that x u . Since u was a perfectly arbitrary set, we have converted Russell’s Paradox into the following theorem.

如果 x ∈ u,那么刚才的陈述必然是假的,正如罗素悖论最初讨论时那样。但现在我们可以通过得出 x∉u 来消除矛盾。由于 u 是一个完全任意的集合,我们就将罗素悖论转化为以下定理。

3.88. No-Universe Theorem. For every set u , there exists a set x with x u .

3.88. 无宇宙定理。对于每个集合 u,存在一个集合 x 使得 x∉u。

This theorem tells us that, in the ZFC theory, there does not exist a “universal set” that contains every set. Note also that if such a set u did exist, the Axiom of Specification (using this u ) would reduce to the Axiom of Abstraction.

该定理告诉我们,在ZFC理论中,不存在包含所有集合的“全集”。还要注意的是,如果这样的集合u确实存在,那么规范公理(使用这个u)就会简化为抽象公理。

The Axioms of Pairs, Power Sets, and Unions

配对、幂集和并集的公理

Our next three axioms allow us to combine several sets to produce a larger set.

接下来的三个公理允许我们将多个集合合并成一个更大的集合。

3.89. Axiom of Pairs. a , b , x , z , [ z x ( z = a z = b ) ] .

3.89. 配对公理。∀a,∀b,∃x,∀z,[z∈x⇔(z=a∨z=b)].

Informally, given any two sets (objects) a and b , there exists a set x whose members are precisely a and b . The Axiom of Extension shows that this set x is uniquely determined by a and b . So we can introduce the notation x = { a , b }; as before, we call this set the unordered pair with members a and b . Observe that we can take a = b here, in which case we obtain the singleton set { a } whose only member is a . By using the Axiom of Pairs on the sets { a } and { a , b }, we obtain the set {{ a }, { a , b }}, which was our formal definition of the ordered pair ( a , b ) in Definition 3.66 . In § 3.5 , we used this definition to prove that for all objects (sets) a , b , c , d , ( a , b ) = ( c , d ) iff a = c and b = d .

通俗地说,给定任意两个集合(对象)a 和 b,存在一个集合 x,其元素恰好为 a 和 b。扩展公理表明,这个集合 x 由 a 和 b 唯一确定。因此,我们可以引入记号 x = {a, b};和之前一样,我们称这个集合为元素为 a 和 b 的无序对。注意,这里我们可以取 a = b,此时我们得到一个单元素集合 {a},其唯一元素为 a。通过对集合 {a} 和 {a, b} 应用对公理,我们得到集合 {{a}, {a, b}},这正是我们在定义 3.66 中对有序对 (a, b) 的正式定义。在 §3.5 中,我们利用这个定义证明,对于所有对象(集合)a、b、c、d,(a, b) = (c, d) 当且仅当 a = c 且 b = d。

3.90. Axiom of Power Sets. a , x , z , [ z x z a ] .

3.90. 幂集公理。∀a,∃x,∀z,[z∈x⇔z⊆a].

Informally, given any set a , there exists a set x such that the members of x are precisely the subsets of a . By the Axiom of Extension, the set x is uniquely determined by this condition. As before, we write x = P ( a ) and call x the power set of a .

通俗地说,对于任意集合 a,存在一个集合 x,使得 x 的所有元素恰好都是 a 的子集。根据外延公理,集合 x 由此条件唯一确定。和之前一样,我们记 x=P(a),并称 x 为 a 的幂集。

3.91. Axiom of Unions. a , x , z , [ z x s a , z s ] .

3.91. 并集公理。∀a,∃x,∀z,[z∈x⇔∃s∈a,z∈s].

Informally, given any set a , there exists a set x such that the members of x are precisely the members of the members of a . By the Axiom of Extension, the set x is uniquely determined by this condition. We write x = a and call x the union of the sets in a . The following examples may clarify what is going on he

通俗地说,对于任意集合 a,存在一个集合 x,使得 x 的元素恰好是 a 的元素。根据外延公理,集合 x 由此条件唯一确定。我们记 x = ⋃a,并称 x 为 a 中所有集合的并集。以下示例或许能更好地说明这一点。

3.92 Example. Suppose we have a set a = { b , c , d , e }. Applying the Axiom of Unions to this set, we get a set x = ∪ a such that for all z , z x iff z is a member of some member of a . As the members of a are b , c , d , and e , we see that z x iff z b or z c or z d or z e . Using informal notation from earlier in the chapter, this says that z x iff z b c d e .

3.92 例。假设我们有一个集合 a = {b, c, d, e}。对这个集合应用并集公理,我们得到一个集合 x = ∪ a,使得对于所有 z,z ∈ x 当且仅当 z 是集合 a 中某个元素的元素。由于集合 a 的元素是 b、c、d 和 e,我们看到 z ∈ x 当且仅当 z ∈ b 或 z ∈ c 或 z ∈ d 或 z ∈ e。用本章前面提到的非正式符号表示,这意味着 z ∈ x 当且仅当 z∈b∪c∪d∪e。

3.93. Example: Binary Unions. Given any two sets B and C , define B C = { B , C } . In more detail, we use the Axiom of Pairs to obtain the unordered pair a = { B , C }, and then use the Axiom of Unions to obtain the set ∪ a . As in the preceding example, it follows that for all z , z B C iff z B or z C . Iterating this construction, we can define the union of three sets by letting B C D = ( B C ) D . Similarly, we can define the union of four sets, five sets, or any specific finite number of sets by repeated use of the binary union operation.

3.93. 示例:二元并集。给定任意两个集合 B 和 C,定义 B∪C=⋃{B,C}。更详细地说,我们利用二元对公理得到无序对 a = {B, C},然后利用并集公理得到集合 ∪ a。与前面的例子类似,对于任意 z,z∈B∪C 当且仅当 z ∈ B 或 z ∈ C。重复此构造,我们可以通过令 B∪C∪D=(B∪C)∪D 来定义三个集合的并集。类似地,我们可以通过重复使用二元并集运算来定义四个集合、五个集合或任意特定有限个集合的并集。

3.94. Example. Suppose a = { B 0 , B 1 , B 2 , …, B k , …} = { B k : k ∈ ℤ ≥0 }. (This is an informal example, since we have not formally discussed ℤ ≥0 yet or defined what is meant by { B k : k ∈ ℤ ≥0 }.) Here, x = ∪ a consists of all z such that z B k for some k ∈ ℤ ≥0 . This is the set that we denoted earlier by k = 0 B k .

3.94. 示例。假设 a = {B 0 , B 1 , B 2 , …, B k , …} = {B k :k ∈ ℤ ≥0 。(这是一个非正式示例,因为我们尚未正式讨论 ℤ ≥0 ,也未定义 {B k :k ∈ ℤ ≥0 的含义。)这里,x = ∪ a 由所有满足 z ∈ B k 的 z 组成,其中 k ∈ ℤ ≥0 。这就是我们之前用 ⋃k=0∞Bk 表示的集合。

3.95. Example: Product Sets. Given any two sets A and B , we can now justify the definition of the product set A × B , which we defined earlier by the condition

3.95. 示例:乘积集。给定任意两个集合 A 和 B,我们现在可以证明乘积集 A × B 的定义,该定义之前已由以下条件给出:

z z , [ z z A 一个 × × B B a 一个 A 一个 , b b B B , z z = ( a 一个 , b b ) ] .

By the Axiom of Extension, there is at most one set satisfying this condition. But, how can we prove that A × B exists ? To begin, recall that ( a , b ) = {{ a }, { a , b }}. Given a A and b B , we know { a } and { a , b } are subsets of A B , hence { a } and { a , b } are members of P ( A B ) . Then ( a , b ), which is a set of two members of P ( A B ) , is a subset of P ( A B ) and hence a member of P ( A B ) . Finally, A × B is a set of ordered pairs, each of which is a member of P ( A B ) , so that A × B is a certain subset of P ( A B ) . To be precise,

根据外延公理,至多只有一个集合满足这个条件。但是,我们如何证明 A × B 存在呢?首先,回顾一下 (a, b) = {{a}, {a, b}}。给定 a ∈ A 和 b ∈ B,我们知道 {a} 和 {a, b} 是 A ∪ B 的子集,因此 {a} 和 {a, b} 是 P(A∪B) 的成员。那么,(a, b) 是 P(A∪B) 中两个元素的集合,也是 P(A∪B) 的子集,因此也是 P(A∪B) 的成员。最后,A × B 是有序对的集合,每个有序对都是 P(A∪B) 的成员,因此 A × B 是 P(A∪B) 的一个子集。更准确地说,

A 一个 × × B B = { z z P P ( A 一个 B B ) : a 一个 A 一个 , b b B B , z z = { { a 一个 } , { a 一个 , b b } } } .

We have seen that A B exists using the Axiom of Pairs and the Axiom of Unions. Using the Axiom of Power Sets twice, we see that P ( A B ) exists. Finally, the displayed formula and the Axiom of Specification shows that the set A × B exists. We can iterate this construction to define the product of three sets, of four sets, and so on; for instance, A × B × C = ( A × B ) × C .

我们已经利用二元组公理和并集公理证明了 A ∪ B 的存在。两次运用幂集公理,我们证明了 P(A∪B) 的存在。最后,利用所展示的公式和规范公理,我们证明了集合 A × B 的存在。我们可以重复这个构造过程来定义三个集合的乘积、四个集合的乘积等等;例如,A × B × C = (A × B) × C。

The Axioms of ∅ and Infinity

∅与无穷的公理

Based on the axioms stated so far, it is impossible to prove that there exists any set whatsoever! The reason is that each set-construction axiom can only be invoked if we already have at least one set that is already known to exist (such as the set u in the Axiom of Specification). The following axiom gets us started by postulating the existence of the empty set.

根据目前所述的公理,我们无法证明任何集合的存在!原因在于,每个集合构造公理都只能在已知至少一个集合存在的情况下才能被调用(例如规范公理中的集合 u)。以下公理通过假设空集的存在性来帮助我们进行论证。

3.96. Axiom of the Empty Set. x , z , z x .

3.96. 空集公理。∃x,∀z,z∉x。

Informally, there is a set x having no members. By the Axiom of Extension, if another set x ′ satisfies z , z x , then x = x ′. So we can denote the unique set x with no members by ∅, and we can call x the empty set.

通俗地说,存在一个不包含任何元素的集合 x。根据外延公理,如果另一个集合 x′ 满足 ∀z,z∉x′,则 x = x′。因此,我们可以用 ∅ 表示唯一不包含任何元素的集合 x,并将 x 称为空集。

Once we have the empty set, we can start applying the other set-construction axioms to build more and more new sets. Intuitively, we can get infinitely many distinct sets in this manner, but we claim that each individual set that we build is still finite. Reasoning informally (as we have not yet rigorously defined “finite”), the preceding claim holds since we can only obtain a given set by finitely many applications of the axioms, and each axiom stated so far produces a finite output set when applied to finite input sets. Thus, with the apparatus developed so far, there is no way to justify the existence of an infinite set such as ℤ or ℝ. The next axiom resolves this difficulty.

一旦我们有了空集,就可以开始应用其他集合构造公理来构造越来越多的新集合。直观上,我们可以用这种方法得到无穷多个不同的集合,但我们声称我们构造的每个集合仍然是有限的。非正式地推理(因为我们还没有严格定义“有限”),上述论断成立,因为我们只能通过有限次应用公理来获得给定的集合,而迄今为止陈述的每个公理在应用于有限的输入集合时都会产生一个有限的输出集合。因此,用目前建立的机制,无法证明像 ℤ 或 ℝ 这样的无限集合的存在。下一个公理解决了这个难题。

3.97 Axiom of Infinity x , [ x z , ( z x z { z } x ) ] .

3.97 无穷公理 ∃x,[∅∈x∧∀z,(z∈x⇒z∪{z}∈x)].

Although it may not be evident from an initial reading of the axiom, this axiom guarantees the existence of an infinite set x . In fact, the implication appearing in the axiom will assure that all the natural numbers {0, 1, 2, …} (to be defined later as certain sets) appear as members of x . To give a hint of what is going on here, we mention that the natural number 0 is defined to be the empty set ∅, and for a natural number z (which is a certain type of set), z + 1 is defined to be the set z { z } . Then the axiom is saying (among other things) that 0 ∈ x , and for any natural number z x , z + 1 also belongs to x . It should then be intuitively believable that Z 0 x (compare to the discussion of mathematical induction in the next chapter), so that the set x is infinite.

虽然乍看之下可能并不明显,但这条公理保证了无限集 x 的存在。事实上,公理中的蕴含关系保证了所有自然数 {0, 1, 2, …}(稍后会定义为特定类型的集合)都属于 x。为了更好地理解这一点,我们提到自然数 0 被定义为空集 ∅,而对于自然数 z(它本身也是某种类型的集合),z + 1 被定义为集合 z∪{z}。因此,这条公理(除其他含义外)表明 0 ∈ x,并且对于任意自然数 z ∈ x,z + 1 也属于 x。由此,我们可以直观地理解 Z≥0⊆x(参见下一章关于数学归纳法的讨论),从而得出集合 x 是无限集。

The Axiom of Replacement

替换公理

Although our discussion of the Axiom of Infinity and ℤ ≥0 is not yet complete, let us suppose for a moment that we have constructed the set ℤ ≥0 = {0, 1, 2, 3, …} within axiomatic set theory. Returning to a previous example, we could now hope to give meaning to the indexed union k = 0 B k , where B 0 , B 1 , B 2 , … are sets indexed by the nonnegative integers. We would like to use the Axiom of Unions to define k = 0 B k = { B k : k Z 0 } . In order to do this, we need to justify the existence of the indexed set { B k : k Z 0 } . We can use the next axiom to accomplish this task.

尽管我们对无穷公理和 ℤ ≥0 的讨论尚未完成,但让我们暂时假设我们已经在公理化集合论中构造了集合 ℤ ≥0 = {0, 1, 2, 3, …}。回到之前的例子,我们现在可以赋予索引并集 ⋃k=0∞Bk 以意义,其中 B 0 , B 1 , B 2 , … 是由非负整数索引的集合。我们希望使用并集公理来定义 ⋃k=0∞Bk=⋃{Bk:k∈Z≥0}。为了做到这一点,我们需要证明索引集 {Bk:k∈Z≥0} 的存在性。我们可以使用下一个公理来完成这项任务。

3.98 Axiom of Replacement Let X be a given set, and let P ( x , z ) be a given statement with free variables x and z . Suppose that for each x X , there exists a set y (depending on x ) such that for all z , z y iff P ( x , z ). By the Axiom of Extension, y is uniquely determined by x , so we denote y as y ( x ). Then there exists a set Y such that for all w , w Y iff there exists x X with w = y ( x ).

3.98 替换公理 设 X 为给定的集合,P(x, z) 为给定的命题,其中 x 和 z 为自由变量。假设对于每个 x ∈ X,存在一个集合 y(依赖于 x),使得对于所有 z,z ∈ y 当且仅当 P(x, z) 成立。根据外延公理,y 由 x 唯一确定,因此我们将 y 记为 y(x)。那么存在一个集合 Y,使得对于所有 w,w ∈ Y 当且仅当存在 x ∈ X 使得 w = y(x)。

Intuitively, the purpose of this axiom is to let us apply a “function” to all members of a given set X to create a new set Y . In this case, a “function” is not a set of ordered pairs (as discussed in later chapters), but rather a logical statement that tells us how each input set x X is related to the corresponding output set y Y . In more detail, our “function” F with domain X is defined by letting F ( x ) = y ( x ) = { z : P ( x , z ) is true } for each x X . Part of the hypothesis of the Axiom of Replacement is that each output y ( x ), as x varies over X , is already known to be a set. Given that this hypothesis holds, the axiom asserts the existence of a new set Y = { F ( x ): x X } = { y ( x ): x X }, which is obtained by “applying F to each member of X ” and collecting together all the resulting outputs into a new set. In other words, we obtain a new set Y from the set X by replacing each x X by the set F ( x ).

直观地说,这条公理的目的是让我们能够将一个“函数”应用于给定集合 X 的所有元素,从而创建一个新集合 Y。这里的“函数”并非有序对的集合(如后续章节所述),而是一个逻辑语句,它告诉我们每个输入集合 x ∈ X 与对应的输出集合 y ∈ Y 之间的关系。更详细地说,定义域为 X 的“函数”F 的定义是:对于每个 x ∈ X,都有 F(x) = y(x) = {z: P(x,z) 为真}。替换公理的部分假设是:当 x 在 X 上变化时,每个输出 y(x) 已知都是一个集合。基于此假设,该公理断言存在一个新集合 Y = {F(x): x ∈ X} = {y(x): x ∈ X},它是通过将“F 应用于 X 的每个元素”并将所有结果输出收集到一个新集合中得到的。换句话说,我们通过将集合 X 中的每个 x ∈ X 替换为集合 F(x),从集合 X 得到一个新的集合 Y。

3.99 Example. This example assumes we have already constructed the sets ℤ ≥0 and ℝ within axiomatic set theory. Let P ( x , z ) be the statement ( x Z 0 ) ( z R ) ( 2 x z x ) . By the Axiom of Specification, for each x ∈ ℤ ≥0 , the set y ( x ) = { z : P ( x , z )} = { z ∈ ℝ: −2 x z x } exists. So, the Axiom of Replacement transforms the known set Z 0 into the set of closed intervals Y = {[ −2 x , x ]: x ∈ ℤ ≥0 }. We should point out that, for this particular set Y , we can construct Y from other axioms without invoking the Axiom of Replacement.

3.99 示例。本示例假设我们已经在公理化集合论中构造了集合 ℤ ≥0 和 ℝ。令 P(x, z) 表示命题 (x∈Z≥0)∧(z∈R)∧(−2x≤z≤x)。根据规范公理,对于每个 x ∈ ℤ ≥0 ,集合 y(x) = {z:P(x, z)} = {z ∈ ℝ: −2x ≤ z ≤ x} 存在。因此,替换公理将已知的集合 Z≥0 转换为闭区间集合 Y = {[ −2x, x]:x ∈ ℤ ≥0 。需要指出的是,对于这个特定的集合 Y,我们可以利用其他公理构造 Y,而无需调用替换公理。

Generalizing the last example, if we have an index set I and a formula that characterizes a unique set B i for each i I , then the Axiom of Replacement converts the index set I into the indexed set { B i : i I }. Then, for instance, we could use the Axiom of Unions to obtain the set ∪ i I B i . Also, provided I is nonempty, we could use the Axiom of Specification to obtain ∩ i I B i

推广上一个例子,如果我们有一个索引集 I 和一个公式,该公式为每个 i ∈ I 都刻画了一个唯一的集合 B i ,那么替换公理会将索引集 I 转换为索引集 {B i :i ∈ I}。然后,例如,我们可以使用并集公理来获得集合 ∪ i I B i 。此外,如果 I 非空,我们可以使用指定公理来获得 ∩ i I B i

The Axiom of Choice

选择公理

Suppose X is a set of nonempty sets. Then each x X is a nonempty set, so there exists at least one object (set) y with y x . The y appearing here depends on x , and y is usually not unique. In many parts of mathematics, it is technically helpful to have access to a set G of ordered pairs with the property that for each x X , there is exactly one ordered pair ( x , y ) ∈ G with first component x , and for all ( x , y ) ∈ G we have y x . In order to prove the existence of such a set G , we need a new axiom, called the Axiom of Choice . The name arises because the set G represents a simultaneous choice of one element y from each set x in the given collection of sets X .

假设 X 是一个非空集合的集合。那么每个 x ∈ X 都是一个非空集合,因此至少存在一个对象(集合)y,使得 y ∈ x。这里出现的 y 取决于 x,并且 y 通常不唯一。在数学的许多领域,我们需要一个有序对集合 G,其性质为:对于每个 x ∈ X,恰好存在一个有序对 (x, y) ∈ G,其第一个元素为 x,并且对于所有 (x, y) ∈ G,都有 y ∈ x。为了证明这样的集合 G 的存在性,我们需要一个新的公理,称为选择公理。之所以称之为选择公理,是因为集合 G 表示从给定集合 X 中的每个集合 x 中同时选择一个元素 y。

3.100 Axiom of Choice. X , [ X F , x X , ! y x , ( x , y ) F ] .

3.100 选择公理。∀X,[∅∉X⇒∃F,∀x∈X,∃!y∈x,(x,y)∈F].

The set F provided by the axiom may have “too many” elements. For example, F could have members that are not ordered pairs, or members of the form ( a , b ) with a X , or members ( x , z ) with x X and z x . However, using the Axiom of Specification, we can shrink F to a set G having the properties stated above (see Exercise 15).

由公理提供的集合 F 可能包含“过多”的元素。例如,F 可能包含非有序对的元素,或者形如 (a, b) 且 a∉X 的元素,或者形如 (x, z) 且 x ∈ X 且 z∉x 的元素。然而,利用规范公理,我们可以将 F 缩小为一个具有上述性质的集合 G(参见练习 15)。

The Axiom of Choice may seem intuitively obvious, since we can readily imagine forming a collection of ordered pairs ( x , y ) by considering each nonempty set x in turn and arbitrarily choosing some member y x . However, being intuitively convinced that a set ought to exist is not the same as being able to prove that such a set exists based on explicitly stated axioms. In some special situations, we can avoid the Axiom of Choice, by using a specific choice rule (in the form of a logical statement) together with the Axiom of Specification to build the set of ordered pairs G . For example, if every x X is a nonempty subset of ℤ ≥0 , we could define

选择公理似乎直观易懂,因为我们可以很容易地想象,通过依次考虑每个非空集合 x 并任意选择 x 中的某个元素 y,来构建有序对 (x, y) 的集合。然而,直觉上确信某个集合应该存在,并不等同于能够基于明确陈述的公理来证明该集合的存在。在某些特殊情况下,我们可以通过使用特定的选择规则(以逻辑语句的形式)以及规范公理来构建有序对集合 G,从而避免使用选择公理。例如,如果每个 x ∈ X 都是 ℤ ≥0 的非空子集,我们可以定义

G G = { z z X X × × Z Z 0 : x x X X , y Z Z 0 , z z = ( x x , y ) y x x w 西 x x , y w 西 } .

This definition works because of the fact (discussed in more detail later) that every nonempty subset of ℤ ≥0 has a unique least element.

这个定义之所以有效,是因为(稍后会更详细地讨论)ℤ ≥0 的每个非空子集都有一个唯一的最小元素。

It is a famous and difficult result of axiomatic set theory that the Axiom of Choice is independent of the other axioms in the ZFC theory, assuming that those axioms are themselves consistent. This means that we could add either the Axiom of Choice, or the negation of this axiom, to the other axioms without introducing new contradictions. In particular, it is impossible to prove the Axiom of Choice (or its negation) as a consequence of the other axioms, unless the other axioms are inconsistent (in which case every statement could be proved from those axioms, and the entire theory is worthless).

公理集合论中一个著名且难以证明的结论是:选择公理独立于ZFC理论中的其他公理,前提是这些公理本身是自洽的。这意味着我们可以将选择公理或其否定添加到其他公理中,而不会引入新的矛盾。特别地,除非其他公理不自洽(在这种情况下,任何命题都可以从这些公理中证明,整个理论也就失去了意义),否则不可能将选择公理(或其否定)证明为其他公理的推论。

There are many logically equivalent versions of the Axiom of Choice, some of which are discussed later in this text. Two such versions are: the product of any family of nonempty sets is nonempty; and for any set X , there exists a well-ordering ≤ on X . Many key theorems in various areas of mathematics require the Axiom of Choice in their proofs. Some of these theorems include: the union of countably many countable sets if countable (proved in §7.3 when we discuss cardinality); every vector space has a basis (proved in linear algebra); the product of any collection of compact topological spaces is compact (proved in topology); every commutative ring with identity has a maximal ideal (proved in abstract algebra); every consistent set of sentences has a model (proved in formal logic); and not every subset of ℝ is a Lebesgue-measurable set (proved in real analysis). Most mathematicians today accept the Axiom of Choice in order to have access to these and many other theorems that depend on this axiom.

选择公理有许多逻辑等价的版本,本文稍后将讨论其中一些。例如:任意非空集合族的乘积非空;以及对于任意集合 X,存在 X 上的良序 ≤。数学各个领域的许多关键定理的证明都需要用到选择公理。这些定理包括:可数个可数集的并集是可数的(在讨论基数时于第 7.3 节中证明);每个向量空间都有一组基(在线性代数中证明);任意紧拓扑空间族的乘积是紧的(在拓扑学中证明);每个带恒等元的交换环都有极大理想(在抽象代数中证明);每个相容语句集都有模型(在形式逻辑中证明);以及并非 ℝ 的每个子集都是勒贝格可测集(在实分析中证明)。为了能够运用这些定理以及许多其他依赖于该公理的定理,当今大多数数学家都接受选择公理。

The Axiom of Foundation

基础公理

The final axiom of the ZFC theory is the following odd-looking statement.

ZFC理论的最后一个公理是以下这个看似奇怪的陈述。

3.101. Axiom of Foundation. a , [ a ( b , b a a b = ) ] .

3.101. 基础公理。∀a,[a≠∅⇒(∃b,b∈a∧a∩b=∅)].

In words, for any nonempty set a , a has a member b that has empty intersection with a . The hypothesis a is not surprising, since the empty set does not have any members. Next we give some intuition for what the conclusion means.

换句话说,对于任意非空集合 a,a 中存在一个元素 b,使得 b 与 a 的交集为空。假设 a≠∅ 并不令人意外,因为空集本身不包含任何元素。接下来,我们将对这一结论给出一些直观的解释。

We begin with an informal example involving the set ℤ ≥0 = {0, 1, 2, 3, …}. It is intuitively evident that there does not exist an infinite strictly decreasing sequence of nonnegative integers

我们首先来看一个涉及集合ℤ ≥0 = {0, 1, 2, 3, …}的非正式例子。直观上显然,不存在无限严格递减的非负整数序列。

a 一个 0 > a 一个 1 > a 一个 2 > > a 一个 k k > .

(We discuss this point in more detail in later chapters.) Roughly speaking, the goal of the Axiom of Foundation is to rule out analogous strictly decreasing sequences in the world of sets. In this setting, the concept b < c for integers is replaced by the analogous concept x y for sets. So, we are trying to rule out the existence of an infinite sequence of sets x 0 , x 1 , x 2 , x 3 , … such that ·· · ∈ x 3 x 2 x 1 x 0 , i.e., such that x k +1 x k for all k Z 0 . As special cases of this, we want to forbid a set x such that x x (which would lead to the infinite sequence ·· · x x x x x ), as well as a pair of sets x , y such that x y and y x (which would lead to the sequence ·· · x y x y x y ), and so on.

(我们将在后续章节中更详细地讨论这一点。)简而言之,基础公理的目标是排除集合世界中类似的严格递减序列。在这种情况下,整数中的 b < c 概念被集合中的类似概念 x ∈ y 所取代。因此,我们试图排除是否存在一个无限集合序列 x 0 , x 1 , x 2 , x 3 , … 使得 ·· · ∈x 3 ∈ x 2 ∈ x 1 ∈ x 0 ,即对于所有 k∈Z≥0,都有 x k +1 ∈ x k 。作为此问题的特殊情况,我们想要禁止存在一个集合 x,使得 x ∈ x(这将导致无限序列 ·· · x ∈ x ∈ x ∈ x ∈ x),以及一对集合 x、y,使得 x ∈ y 且 y ∈ x(这将导致序列 ·· · x ∈ y ∈ x ∈ y ∈ x ∈ y),依此类推。

Thus, although the Axiom of Foundation is phrased as an existence statement, its true goal is to force the non-existence of certain pathological combinations of set membership relationships. We now sketch how this goal is achieved. Suppose (to get a contradiction) that there did exist specific sets x k with x k +1 x k for all k ∈ ℤ ≥0 . By the Axiom of Replacement, we can form the set a = { x k : k ∈ ℤ ≥0 }. The set a is nonempty, since x 0 a , so the Axiom of Foundation provides a set b a with a b = . Now, b = x k for some k ∈ ℤ ≥0 . We deduce x k +1 b and x k +1 a , so that x k + 1 a b = , which is a contradiction.

因此,尽管基础公理被表述为存在性陈述,但其真正的目的是强制某些病态的集合成员关系组合不存在。现在我们简要概述一下如何实现这一目标。假设(为了得出矛盾)确实存在特定的集合 x k ,使得对于所有 k ∈ ℤ ≥0 ,都有 x k +1 ∈ x k 。根据替换公理,我们可以构造集合 a = {x k :k ∈ ℤ ≥0 。由于 x 0 ∈ a,集合 a 非空,因此基础公理提供了一个集合 b ∈ a,使得 a∩b=∅。现在,对于某个 k ∈ ℤ ≥0 ,有 b = x k 。我们推断 x k +1 ∈ b 且 x k +1 ∈ a,因此 xk+1∈a∩b=∅,这与假设矛盾。

The Axiom of Foundation is also called the Axiom of Regularity .

基础公理又称为正则性公理。

Section Summary

章节概要

All of mathematics can be formalized within set theory. One development of set theory is based on the following axioms.

数学的方方面面都可以用集合论来形式化。集合论的一个发展方向是基于以下公理。

3.102. Axioms of ZFC Set Theory.

3.102. ZFC集合论的公理。

(a) Axiom of Extension: x , y , [ x = y ( z , z x z y ) ] .

(a)扩展公理:∀x,∀y,[x=y⇔(∀z,z∈x⇔z∈y)].

(b) Axiom of Specification: For all open sentences P ( z ), u , x , z , [ z x ( z u P ( z ) ) ] .

(b)规范公理:对于所有开语句 P(z),∀u,∃x,∀z,[z∈x⇔(z∈u∧P(z))].

(c) Axiom of Pairs: a , b , x , z , [ z x ( z = a z = b ) ] .

(c)配对公理:∀a,∀b,∃x,∀z,[z∈x⇔(z=a∨z=b)].

(d) Axiom of Power Sets: a , x , z , [ z x z a ] .

(d)幂集公理:∀a,∃x,∀z,[z∈x⇔z⊆a].

(e) Axiom of Unions: a , x , z , [ z x s a , z s ] .

(e)并集公理:∀a,∃x,∀z,[z∈x⇔∃s∈a,z∈s].

(f) Axiom of the Empty Set: x , z , z x .

(f)空集公理:∃x,∀z,z∉x。

(g) Axiom of Infinity: x , [ x z , ( z x z { z } x ) ] .

(g)无穷公理:∃x,[∅∈x∧∀z,(z∈x⇒z∪{z}∈x)].

(h) Axiom of Replacement: For all open sentences P ( x , z ) and all sets X , [ x X , y , z , ( z y P ( x , z ) ) ] Y , w , [ w Y x X , z , ( z w P ( x , z ) ) ] .

(h)替换公理:对于所有开语句 P(x, z) 和所有集合 X,[∀x∈X,∃y,∀z,(z∈y⇔P(x,z))]⇒∃Y,∀w,[w∈Y⇔∃x∈X,∀z,(z∈w⇔P(x,z))]。

(i) Axiom of Choice: X , [ X F , x X , ! y x , ( x , y ) F ] .

(i)选择公理:∀X,[∅∉X⇒∃F,∀x∈X,∃!y∈x,(x,y)∈F].

(j) Axiom of Foundation: a , [ a ( b , b a a b = ) ] .

(j)基础公理:∀a,[a≠∅⇒(∃b,b∈a∧a∩b=∅)].

These axioms can be used to prove the existence and uniqueness of sets such as ∅, ℤ ≥0 , A B , A B , A B , P ( A ) , A × B , ∪ i I A i , ∩ i I A i , and so on. The No-Universe Theorem states that for every set u , there exists some set x u . The Axiom of Foundation rules out infinite descending chains of sets belonging to other sets: i.e., there do not exist sets x k (for k ∈ ℤ ≥0 ) with x k +1 x k for all k ≥ 0.

这些公理可用于证明集合的存在性和唯一性,例如 ∅、ℤ ≥0 、A ∪ B、A ∩ B、A − B、P(A)、A × B、∪ i I A i 、∩ i I A i 等等。无宇宙定理指出,对于每个集合 u,都存在某个集合 x∉u。基础公理排除了属于其他集合的无限递减链:也就是说,不存在集合 x k (对于 k ∈ ℤ ≥0 )使得对于所有 k ≥ 0,x k +1 ∈ x k

Exercises

练习

1. Prove or disprove: x , y , [ z , x z y z ] x = y .

1.证明或反证:∀x,∀y,[∀z,x∈z⇔y∈z]⇒x=y。

2. Use the axioms to prove: for all sets a , b , c , there exists a unique set x such that for all z , z x iff z = a or z = b or z = c . (The set x is the unordered triple { a , b , c }.)

2. 利用公理证明:对于任意集合 a、b、c,存在唯一的集合 x,使得对于任意 z,z ∈ x 当且仅当 z = a 或 z = b 或 z = c。(集合 x 是无序三元组 {a, b, c}。)

3. Use the axioms to prove: for all sets a , b , c , d , e , there exists a unique set x such that for all z , z x iff z = a or z = b or z = c or z = d or z = e .

3.利用公理证明:对于所有集合a、b、c、d、e,存在唯一的集合x,使得对于所有z,z∈x当且仅当z=a或z=b或z=c或z=d或z=e。

4. Assuming that ℝ and the order relation < on ℝ have already been constructed, use the axioms to prove the existence and uniqueness of the intervals ( a , b ), [ a , b ], ( a , b ], and [ a , b ) for all real numbers a < b .

4.假设 ℝ 和 ℝ 上的序关系 < 已经构造,利用公理证明对于所有实数 a < b,区间 (a, b)、[a, b]、(a, b] 和 [a, b] 的存在性和唯一性。

5. Assuming that ℤ and the arithmetic operations and order relation on ℤ have already been constructed, use the axioms to prove the existence and uniqueness of the following subsets of ℤ: (a) ℤ <0 ; (b) the set of odd integers; (c) the set of prime integers (these are integers p > 1 such that the only positive integers dividing p are 1 and p ).

5.假设 ℤ 以及 ℤ 上的算术运算和序关系已经构造,利用公理证明 ℤ 的下列子集的存在性和唯一性:(a)ℤ <0 ;(b)奇数集合;(c)素数集合(这些是大于 1 的整数,使得 p 只能被 1 和 p 整除)。

6. Assuming that ℝ and ℤ ≥0 have already been constructed, prove the existence of the set Y = {[ −2 x , x ]: x ∈ ℤ ≥0 } without using the Axiom of Replacement.

6.假设 ℝ 和 ℤ ≥0 已经构造,证明集合 Y = {[ −2x, x]:x ∈ ℤ ≥0 的存在性,不使用替换公理。

7. Show that the Axiom of the Empty Set follows from the alternate axiom x , x = x and the other axioms of ZFC set theory.

7.证明空集公理可由替代公理∃x,x=x和ZFC集合论的其他公理得出。

8. (a) Show that the Axiom of Pairs (as stated in this section) follows from the alternate version a , b , x , z , [ ( z = a z = b ) z x ] along with the other axioms of ZFC set theory. (b) State and prove similar results for the Axiom of Power Sets and the Axiom of Unions.

8.(a)证明本节所述的二元组公理可由以下替代形式 ∀a,∀b,∃x,∀z,[(z=a∨z=b)⇒z∈x] 以及 ZFC 集合论的其他公理推出。(b)陈述并证明幂集公理和并集公理的类似结果。

9. Compute: (a) (b) { } (c) { , { } , { { } } } (d) P ( S ) (e) ∪ ( A × B ).

9.计算:(a) ⋃∅ (b) ⋃{∅} (c) ⋃{∅,{∅},{{∅}}} (d) ⋃P(S) (e) ∪ (A × B)。

10. For A , informally define the set ∩ A by the condition: for all objects z , z ∈ ∩ A iff for all y A , z y . (a) Use the axioms to justify the existence and uniqueness of the set ∩ A . (b) In the case A = { B , C }, prove that A = B C . (c) Explain carefully why our informal definition cannot be used to define the set . In contrast, what is { } ?

10. 对于 A≠∅,非正式地定义集合 ∩ A 满足以下条件:对于所有对象 z,z ∈ ∩ A 当且仅当对于所有 y ∈ A,z ∈ y。(a)利用公理证明集合 ∩ A 的存在性和唯一性。(b)当 A = {B, C} 时,证明 ∩A=B∩C。(c)仔细解释为什么我们的非正式定义不能用于定义集合 ∩∅。相反,∩{∅} 是什么?

11. Prove or disprove each variation of the Axiom of Foundation.

11.证明或反驳基础公理的每一个变体。

(a) a , b , b a a b = .

(a)∀a,∃b,b∈a∧a∩b=∅。

(b) a , [ a b , a b = ] .

(b)∀a,[a≠∅⇒∃b,a∩b=∅].

(c) a , [ a b , b a a b = ] .

(c)∀a,[a≠∅⇒∃b,b⊆a∧a∩b=∅].

12. (a) Use the Axiom of Foundation to prove x , x x without referencing ℤ ≥0 .

12.(a)利用基础公理证明∀x,x∉x,而不引用ℤ ≥0

(b) Similarly, prove x , y , ( x y y x ) .

(b)类似地,证明∼∃x,∃y,(x∈y∧y∈x)。

13. Suppose we try to define ordered pairs by letting ( a , b ) = { a , { a , b }} for all objects a and b . Prove or disprove the Ordered Pair Axiom using this definition of ( a , b ).

13.假设我们尝试定义有序对,令 (a, b) = {a, {a, b}},其中 a 和 b 为所有对象。使用 (a, b) 的这个定义证明或反驳有序对公理。

14. Suppose we define the ordered pair ( a , b ) to be the set { { , a } , { { } , b } } (compare to Exercise 21 in § 3.5 ). (a) Use the axioms to justify the existence and uniqueness of ( a , b ) for all sets a and b . (b) Use the axioms to justify the existence and uniqueness of A × B for all sets A and B .

14. 假设我们定义有序对 (a, b) 为集合 {{∅,a},{{∅},b}}(参见 §3.5 中的练习 21)。(a)利用公理证明对于所有集合 a 和 b,有序对 (a, b) 的存在性和唯一性。(b)利用公理证明对于所有集合 A 和 B,有序对 A × B 的存在性和唯一性。

15. In the discussion following the statement of the Axiom of Choice, show how to use the Axiom of Specification to convert the set F in the axiom to the set G described prior to the axiom.

15.在选择公理的陈述之后的讨论中,说明如何使用规范公理将公理中的集合 F 转换为公理之前描述的集合 G。

16. Rewrite the Axiom of Infinity without using the defined symbols ∅, ∪, and { z }; use only logical operators, variables, =, and ∈.

16. 重写无穷公理,不要使用已定义的符号 ∅、∪ 和 {z};只使用逻辑运算符、变量、= 和 ∈。

17. Show how the Axiom of the Empty Set can be deduced from the Axiom of Infinity and other axioms.

17.说明如何从无穷公理和其他公理推导出空集公理。

18. Define 0 = , 1 = {0}, and 2 = {0, 1}. Prove the existence of the set P ( { 0 , 1 , 2 } ) without using the Axiom of Power Sets.

18.定义 0=∅,1={0},2={0,1}。证明集合 P({0,1,2}) 的存在性,不使用幂集公理。

19. (a) Without using the Axiom of Pairs, prove the existence of a set S such that there exist exactly two objects a with a S . (b) Use (a) to show that the Axiom of Pairs can be proved from the Axiom of Replacement.

19.(a)不使用配对公理,证明存在一个集合S,使得恰好存在两个对象a,其中a∈S。(b)利用(a)证明配对公理可以从替换公理证明。

20. Consider a universe in which there exists only one object (denoted 0), and suppose 0 ∈ 0 is false. Assuming all quantified variables range over only the objects in this universe, which of the axioms of ZFC set theory are true?

20.考虑一个只存在一个对象(记为0)的宇宙,并假设0∈0为假。假设所有量化变量都只取值于该宇宙中的对象,那么ZFC集合论的哪些公理是正确的?

1 Actually, in an axiomatic development of mathematics from set theory, 7 (and every other mathematical object) is defined to be a certain set. Without knowing the definition of 7 (and 8 and 9), we cannot decide whether 7⊆{7, 8, 9}. We ignore this technical point in this section.

1 实际上,在数学从集合论出发的公理化发展过程中,7(以及任何其他数学对象)都被定义为某个特定的集合。在不知道7(以及8和9)的定义的情况下,我们无法判断7是否⊆{7, 8, 9}。本节暂且忽略这一技术细节。

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4

Integers

整数

4.1  Recursive Definitions and Proofs by Induction

4.1 递归定义和归纳证明

We turn now from set theory to a closer examination of facts involving integers, including proofs by induction, integer division with remainder, prime numbers, and the theorem that every integer has a unique factorization into a product of primes. Proofs by induction are often needed when dealing with concepts defined recursively. So we begin by discussing recursive definitions.

现在我们从集合论转向更深入地探讨与整数相关的事实,包括归纳法证明、带余数的整数除法、素数以及每个整数都可以唯一分解为素数乘积的定理。当处理递归定义的概念时,通常需要用到归纳法证明。因此,我们首先讨论递归定义。

Recursive Definitions

递归定义

Let n be a fixed positive integer, and let x , a 1 , a 2 , …, a n be fixed real numbers. Consider the following informal definitions.

设 n 为一个固定的正整数,x, a 1 , a 2 , …, a n 为固定的实数。考虑以下非正式定义。

(a) Factorials: Define n ! = n · ( n − 1) · ( n − 2) · … · 3 · 2 · 1.

(a)阶乘:定义 n! = n · (n − 1) · (n − 2) · … · 3 · 2 · 1。

(b) Summation Notation: Define k = 1 n a k = a 1 + a 2 + + a n .

(b)求和符号:定义 ∑k=1nak=a1+a2+⋯+an。

(c) Product Notation: Define k = 1 n a k = a 1 a 2 a n .

(c)乘积符号:定义 ∏k=1nak=a1⋅a2⋅…⋅an。

(d) Exponent Notation: Define x n = x · x · … · x , where the right side has n copies of x .

(d)指数表示法:定义 x n = x · x · … · x,其中右边有 n 个 x。

These definitions may seem perfectly legitimate, but they all contain a subtle flaw. Each definition uses the ellipsis symbol “…” to represent missing terms in the middle of each formula. The reader is supposed to know what terms are intended here by following the pattern of the terms that are present. But, in a formal development of these concepts, we cannot omit information in this way. One solution is to rephrase these definitions recursively . We introduce the idea of a recursive definition with the following formal definition of n !.

这些定义看似完全合理,但都存在一个微妙的缺陷。每个定义都使用省略号“…”来表示公式中间缺失的项。读者应该能够通过观察现有项的模式来推断此处所指的项。但是,在对这些概念进行正式阐述时,我们不能以这种方式省略信息。一种解决方法是递归地重述这些定义。我们引入递归定义的概念,并给出 n! 的如下正式定义。

4.1. Recursive Definition of Factorials. Define 0 ! = 1 .

4.1. 阶乘的递归定义。定义 0!=1。

For each n Z 0 , recursively define ( n + 1 ) ! = ( n + 1 ) n ! .

对于每个 n∈Z≥0,递归定义 (n+1)!=(n+1)⋅n!。

This recursive definition begins with an initial condition , which provides an ordinary (non-recursive) definition for 0!. We are told that 0! = 1. Next, the recursive part of the definition gives us a formula for computing ( n + 1)! that assumes we already know the value of n !. For instance, taking n = 0 in the recursive formula, we see that 1! = (0 + 1)! = (0 + 1) · 0! = 1 · 1 = 1. Next, taking n = 1 in the recursive formula, we get 2! = (1 + 1)! = (1 + 1) · 1! = 2 · 1 = 2. Continuing similarly, we see that 3! = 3 · 2! = 3 · 2 = 6, 4! = 4 · 3! = 4 · 6 = 24, 5! = 5 · 4! = 5 · 24 = 120, and so on. We can also iterate the recursive definition to recover the original informal definition; for instance,

这个递归定义从一个初始条件开始,该初始条件提供了一个 0! 的普通(非递归)定义。我们已知 0! = 1。接下来,定义的递归部分给出了一个计算 (n + 1)! 的公式,该公式假设我们已经知道 n! 的值。例如,在递归公式中取 n = 0,我们看到 1! = (0 + 1)! = (0 + 1) · 0! = 1 · 1 = 1。接下来,在递归公式中取 n = 1,我们得到 2! = (1 + 1)! = (1 + 1) · 1! = 2 · 1 = 2。以此类推,我们看到 3! = 3 · 2! = 3 · 2 = 6,4! = 4 · 3! = 4 · 6 = 24,5! = 5 · 4! = 24。 = 5 × 24 = 120,依此类推。我们还可以迭代递归定义来恢复原始的非正式定义;例如,

4 ! = 4 3 ! = 4 3 2 ! = 4 3 2 1 ! = 4 3 2 1 0 ! = 4 3 2 1 1.

A basic recursive definition introduces a sequence of quantities f (0), f (1), …, f ( n ), … by specifying what f (0) is, and then telling us (for each n ≥ 0) how to compute f ( n + 1) if we already know the value of f ( n ). In a more elaborate recursive definition, f ( n + 1) might depend on any of the values f (0), f (1), …, f ( n ) preceding it in the sequence, not just f ( n ). The sequence might start at n = 1 instead of n = 0, or at any other fixed integer. We illustrate with a few more recursive definitions. As always, memorize all of these definitions !

一个基本的递归定义引入了一系列量 f(0), f(1), …, f(n), …,首先指定 f(0) 的值,然后告诉我们(对于每个 n ≥ 0),如果已知 f(n) 的值,如何计算 f(n + 1)。在一个更复杂的递归定义中,f(n + 1) 可能取决于序列中它前面的任意值 f(0), f(1), …, f(n),而不仅仅是 f(n)。序列可以从 n = 1 而不是 n = 0 开始,或者从任何其他固定的整数开始。我们用几个递归定义来举例说明。和往常一样,记住所有这些定义!

4.2. Definition: Exponent Notation. Let x be a fixed real number.

4.2 定义:指数记号。设 x 为一个固定的实数。

Define x 0 = 1 . For all n Z 0 , define x n + 1 = x n x .

定义 x0=1。对于所有 n∈Z≥0,定义 xn+1=xn⋅x。

For example, x 3 = x 2 · x = x 1 · x · x = x 0 · x · x · x = 1 · x · x · x = x · x · x , as expected.

例如,x 3 = x 2 · x = x 1 · x · x = x 0 · x · x · x = 1 · x · x · x = x · x · x,正如预期的那样。

4.3. Definition: Summation Notation. Suppose we are given a real number a n for each n Z 1 . We define k = 1 1 a k = a 1 . For all n Z 1 , define k = 1 n + 1 a k = ( k = 1 n a k ) + a n + 1 .

4.3. 定义:求和符号。假设对于每个 n∈Z≥1,我们给定一个实数 a n 。我们定义 ∑k=11ak=a1。对于所有 n∈Z≥1,定义 ∑k=1n+1ak=(∑k=1nak)+an+1。

Intuitively, the sum of the first n + 1 terms is found by taking the sum of the first n terms and adding a n +1 . We can introduce variations of this definition for sums starting at index zero, such as k = 0 n a k , or sums starting at any other integer. By convention, for all integers j < i , we let k = i j a k = 0 . The summation index k can be replaced by any other letter that does not already have a meaning. For instance, k = 1 n a k = j = 1 n a j = s = 1 n a s . But, we could not write n = 1 n a n , since n is already being used as the upper limit of the sum.

直观地说,前 n + 1 项之和可以通过计算前 n 项之和并加上 a n +1 得到。我们可以引入此定义的变体,例如从索引零开始的和 ∑k=0nak,或者从任何其他整数开始的和。按照惯例,对于所有整数 j < i,我们令 ∑k=ijak=0。求和索引 k 可以替换为任何其他没有特定含义的字母。例如,∑k=1nak=∑j=1naj=∑s=1nas。但是,我们不能写成 ∑n=1nan,因为 n 已经用作和的上限。

4.4. Definition: Product Notation. Suppose we are given a real number a n for each n Z 1 . We define k = 1 1 a k = a 1 . For all n Z 1 , define k = 1 n + 1 a k = ( k = 1 n a k ) a n + 1 .

4.4. 定义:乘积符号。假设对于每个 n∈Z≥1,我们给定一个实数 a n 。我们定义 ∏k=11ak=a1。对于所有 n∈Z≥1,定义 ∏k=1n+1ak=(∏k=1nak)⋅an+1。

We could have defined powers and factorials using product notation. For any real x and positive integer n , you can check that x n = k = 1 n x and n ! = j = 1 n j .

我们可以用乘法符号来定义幂和阶乘。对于任意实数 x 和正整数 n,你可以验证 xn=∏k=1nx 且 n!=∏j=1nj。

The Induction Axiom

归纳公理

In § 2.1 , we listed axioms describing some algebraic properties of Z (the set of integers). To prove facts about recursively defined concepts, we need a new axiom giving a fundamental property of the set Z 1 = { 1 , 2 , 3 , } of positive integers.

在第 2.1 节中,我们列出了描述整数集 Z 的一些代数性质的公理。为了证明关于递归定义概念的事实,我们需要一个新的公理来给出正整数集 Z≥1={1,2,3,…} 的一个基本性质。

4.5 Induction Axiom for Z 1 .

4.5 Z≥1 的归纳公理。

For all open sentences P ( n ), the following two statements are equivalent:

对于所有开放语句 P(n),以下两个陈述是等价的:

(a) n Z 1 , P ( n ) .

(a)∀n∈Z≥1,P(n)。

(b) P ( 1 ) n Z 1 , ( P ( n ) P ( n + 1 ) ) .

(b)P(1)∧∀n∈Z≥1,(P(n)⇒P(n+1))。

Recall that the axioms of a mathematical theory do not need to be proved. Nevertheless, in the case of the Induction Axiom, we can provide an intuitive justification for the axiom that helps us understand what the axiom is saying. Let P ( n ) be a fixed open sentence. On one hand, it is a routine exercise using the proof templates to prove that statement (a) in the Induction Axiom implies statement (b). The real content of the axiom is the assertion that statement (b) implies statement (a). Suppose we know that statement (b) is true; why should we believe that statement (a) must follow? By using the Inference Rule for ALL (taking n = 1, 2, 3, … in statement (b)), we see that statement (b) contains the following information:

回想一下,数学理论的公理不需要证明。然而,对于归纳公理,我们可以提供一个直观的解释,帮助我们理解公理的含义。设 P(n) 是一个固定的开语句。一方面,使用证明模板证明归纳公理中的语句 (a) 蕴含语句 (b) 是一个常规练习。该公理的真正内容是断言语句 (b) 蕴含语句 (a)。假设我们知道语句 (b) 为真;为什么我们应该相信语句 (a) 必然成立呢?通过使用全集推理规则(在语句 (b) 中取 n = 1, 2, 3, …),我们发现语句 (b) 包含以下信息:

  1. P (1) is true.
    P(1)为真。
  2. If P (1) is true, then P (2) is true.
    如果 P(1) 为真,则 P(2) 为真。
  3. If P (2) is true, then P (3) is true.
    如果 P(2) 为真,则 P(3) 为真。
  4. If P (3) is true, then P (4) is true.
    如果 P(3) 为真,则 P(4) 为真。
  5. If P (4) is true, then P (5) is true.
    如果 P(4) 为真,则 P(5) 为真。

… and so on …

… 等等 …

Now, by (1) and (2) and the Inference Rule for IF, we deduce that P (2) is true. It then follows from (3) and the Inference Rule for IF that P (3) is true. It then follows from (4) and the Inference Rule for IF that P (4) is true. It then follows from (5) and the Inference Rule for IF that P (5) is true. Continuing this chain of reasoning forever, we see that P ( n ) is indeed true for every positive integer n , as asserted in statement (a).

现在,根据(1)和(2)以及“如果”的推理规则,我们推断P(2)为真。然后,根据(3)以及“如果”的推理规则,我们得出P(3)为真。接着,根据(4)以及“如果”的推理规则,我们得出P(4)为真。最后,根据(5)以及“如果”的推理规则,我们得出P(5)为真。以此类推,我们最终会发现,对于每个正整数n,P(n)确实为真,正如语句(a)中所述。

This argument can be informally visualized as follows. We imagine setting up an infinite row of dominos standing on edge, with one domino for each positive integer n .

这个论证可以用以下非正式的方式形象化表示。我们想象摆放一排无限长的立式多米诺骨牌,每张骨牌代表一个正整数n。

ufig4_1.jpg

Let P ( n ) be the statement “domino n falls down.” Statement (b) in the Induction Axiom says that domino 1 falls down; and for every positive integer n , if domino n falls down then domino n + 1 falls down. The axiom itself tells us that this information is enough to conclude that all the dominos fall down. The argument in the previous paragraph spells out in more detail how the cascade of falling dominos occurs. First domino 1 falls down, which causes domino 2 to fall, which causes domino 3 to fall, and so on.

设 P(n) 为命题“第 n 个多米诺骨牌倒下”。归纳公理中的命题 (b) 指出,第 1 个多米诺骨牌倒下;并且对于每个正整数 n,如果第 n 个多米诺骨牌倒下,则第 n + 1 个多米诺骨牌也倒下。该公理本身告诉我们,这些信息足以得出所有多米诺骨牌都倒下的结论。上一段的论证更详细地阐述了多米诺骨牌级联倒下的经过。首先,第 1 个多米诺骨牌倒下,导致第 2 个多米诺骨牌倒下,第 3 个多米诺骨牌倒下,依此类推。

We hope that this intuitive argument convinces the reader of the truth of the Induction Axiom. The argument is not a proof of the Induction Axiom, however. This is because we would need to use the Inference Rule for IF infinitely many times to finish proving the universal statement n Z 1 , P ( n ) , and proofs are required to be finite in length. In a similar vein, we can think of the Induction Axiom as indirectly characterizing what the positive integers are : they are precisely the objects we get by starting with 1 and then adding 1 again and again, forever.

我们希望这个直观的论证能够使读者信服归纳公理的正确性。然而,这个论证并非归纳公理的证明。这是因为我们需要无限次地使用IF的推理规则才能完成对全称命题∀n∈Z≥1,P(n)的证明,而证明的长度必须是有限的。类似地,我们可以将归纳公理理解为间接地刻画了正整数的本质:它们正是我们从1开始,然后无限地加1而得到的。

Now that we have intuition for why the Induction Axiom is true, let us see how to use this axiom in proofs. The key idea is that if we need to prove statement (a) in the axiom (a universal statement about all positive integers), it is sufficient to instead prove the equivalent statement (b). We can use the proof templates for AND, , and IF to generate the outline of a proof of statement (b). This leads to the following template for ordinary induction proofs , which should be carefully memorized.

既然我们已经对归纳公理的正确性有了直观的理解,接下来让我们看看如何在证明中使用这条公理。关键在于,如果我们需要证明公理中的命题 (a)(关于所有正整数的全称命题),只需证明等价的命题 (b) 即可。我们可以利用 AND、∀ 和 IF 的证明模板来构建命题 (b) 的证明框架。由此可以得到以下用于普通归纳证明的模板,请务必牢记。

4.6. Template for Ordinary Induction Proofs. To prove n Z 1 , P ( n ) :

4.6. 普通归纳证明的模板。证明对于所有 n∈Z≥1,P(n):

Say “We use induction on n .”

说“我们对 n 使用归纳法。”

Part 1. Prove P (1).

第一部分。证明 P(1)。

Part 2. Fix an arbitrary integer n 0 Z 1 . Assume P ( n 0 ) is true. Prove P ( n 0 + 1) is true.

第二部分。固定任意整数 n0∈Z≥1。假设 P(n 0 ) 为真。证明 P(n 0 + 1) 为真。

The two parts of this template arise since statement (b) in the Induction Axiom is an AND-statement. In part 2, we prove n Z 1 , ( P ( n ) P ( n + 1 ) ) by combining the generic-element proof template with the direct proof template for IF-statements. Part 1 of an induction proof is called the base case . Part 2 is called the induction step . The assumption P ( n 0 ) in part 2 is called the induction hypothesis .

该模板分为两部分,因为归纳公理中的语句 (b) 是一个 AND 语句。在第二部分中,我们通过将通用元素证明模板与 IF 语句的直接证明模板相结合,证明了 ∀n∈Z≥1,(P(n)⇒P(n+1))。归纳证明的第一部分称为基本情况。第二部分称为归纳步骤。第二部分中的假设 P(n 0 ) 称为归纳假设。

Returning to statement (a) and statement (b) in the Induction Axiom, we can now see that statement (b) will likely be easier to prove than statement (a), even though statement (b) appears to be more complex. The reason is that we are allowed to assume certain information in the course of proving statement (b) (namely, the induction hypothesis P ( n 0 )) to help us prove the target statement P ( n 0 + 1). We write n 0 here to emphasize that n 0 is a fixed integer (though in many proofs below, we drop the subscript to reduce clutter). We are not using circular reasoning when using P ( n 0 ) to prove P ( n 0 + 1), since n 0 and n 0 + 1 are different integers, and hence P ( n 0 ) and P ( n 0 + 1) are different statements.

回到归纳公理中的陈述 (a) 和陈述 (b),我们现在可以看出,尽管陈述 (b) 看起来更复杂,但它可能比陈述 (a) 更容易证明。原因是,在证明陈述 (b) 的过程中,我们可以假设某些信息(即归纳假设 P(n 0 )),以帮助我们证明目标陈述 P(n 0 + 1)。这里我们写 n 0 是为了强调 n 0 是一个固定的整数(尽管在下面的许多证明中,为了简化表达式,我们省略了下标)。当我们使用 P(n 0 ) 来证明 P(n 0 + 1) 时,我们并没有使用循环论证,因为 n 0 和 n 0 + 1 是不同的整数,因此 P(n 0 ) 和 P(n 0 + 1) 是不同的陈述。

Examples of Induction Proofs

We now give some examples illustrating the template for proofs by induction. Our first result states that the sum of the first n odd positive integers is n 2 .

4.7. Theorem on Sum of Odd Integers. For all n Z 1 , k = 1 n ( 2 k 1 ) = n 2 .

Proof. We use induction on n . Here, P ( n ) is the summation formula k = 1 n ( 2 k 1 ) = n 2 .

Base Case. We prove P (1), which says k = 1 1 ( 2 k 1 ) = 1 2 . [To prove this, we use the initial condition in the definition of summation notation, together with a chain proof.] We know k = 1 1 ( 2 k 1 ) = 2 1 1 = 2 1 = 1 = 1 2 .

Induction Step. Let n be a fixed positive integer. Assume P ( n ), i.e., assume k = 1 n ( 2 k 1 ) = n 2 . Prove P ( n + 1), i.e., prove k = 1 n + 1 ( 2 k 1 ) = ( n + 1 ) 2 . [We now use the recursive part of the definition of summation notation, and another chain proof.] We know

k = 1 n + 1 ( 2 k 1 ) = [ k = 1 n ( 2 k 1 ) ] + 2 ( n + 1 ) 1 (by definition of sums) = n 2 + 2 ( n + 1 ) 1 (by induction hypothesis) = n 2 + 2 n + 1 = ( n + 1 ) 2 (by algebra).

This chain of known equations proves the target equation P ( n + 1).

The following diagram offers a visual justification of the formula k = 1 n ( 2 k 1 ) = n 2 . A square of area 5 2 has been decomposed into L-shaped pieces of areas 1, 3, 5, 7, and 9; hence, 1 + 3 + 5 + 7 + 9 = 25 = 5 2 .

ufig4_2.jpg

Although the diagram provides compelling visual intuition for why the summation formula is true, it does not prove the formula for all positive integers n . At best, by considering the full diagram and the smaller squares inside it, the diagram proves the first five instances of the summation formula.

4.8. Theorem on Sum of Squares. For all n Z 1 , k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 .

Proof. We use induction on n . Here, P ( n ) is the statement k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) / 6 .

Base Case. We prove P (1), which says k = 1 1 k 2 = 1 ( 1 + 1 ) ( 2 1 + 1 ) / 6 . [To prove this, we use the initial condition in the definition of summation notation, together with a chain proof.] We know k = 1 1 k 2 = 1 2 = 1 = 1 ( 2 ) ( 3 ) / 6 = 1 ( 1 + 1 ) ( 2 1 + 1 ) / 6 .

Induction Step. Let n be a fixed positive integer. Assume P ( n ), i.e., assume

k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 .
Prove P ( n + 1), i.e., prove k = 1 n + 1 k 2 = ( n + 1 ) ( [ n + 1 ] + 1 ) ( 2 [ n + 1 ] + 1 ) / 6 . [We now use the recursive part of the definition of summation notation and another chain proof.] We know
k = 1 n + 1 k 2 = [ k = 1 n k 2 ] + ( n + 1 ) 2 (by definition of sums) = n ( n + 1 ) ( 2 n + 1 ) 6 + 6 ( n + 1 ) 2 6 (by induction hypothesis) = ( n + 1 ) ( n ( 2 n + 1 ) + 6 ( n + 1 ) ) 6 (by factoring n + 1 out) = ( n + 1 ) ( 2 n 2 + 7 n + 6 ) 6 (by algebra) = ( n + 1 ) ( n + 2 ) ( 2 n + 3 ) 6 (by factoring) = ( n + 1 ) ( [ n + 1 ] + 1 ) ( 2 [ n + 1 ] + 1 ) 6 (by algebra).

This chain of known equalities proves the target formula P ( n + 1).

归纳证明示例 我们现在给出一些示例来说明归纳证明的模板。我们的第一个结果表明,前 n 个奇正整数之和为 n 2 。4.7. 奇整数之和定理。对于所有 n∈Z≥1,∑k=1n(2k−1)=n2。证明。我们对 n 进行归纳。这里,P(n) 是求和公式 ∑k=1n(2k−1)=n2。基本情况。我们证明 P(1),即 ∑k=11(2k−1)=12。[为了证明这一点,我们使用求和符号定义中的初始条件,并结合链式证明。] 我们知道 ∑k=11(2k−1)=2⋅1−1=2−1=1=12。归纳步骤。设 n 为一个固定的正整数。假设 P(n) 成立,即假设 ∑k=1n(2k−1)=n²。证明 P(n + 1),即证明 ∑k=1n+1(2k−1)=(n+1)²。[我们现在使用求和符号定义的递归部分,以及另一个链式证明。] 我们知道 ∑k=1n+1(2k−1)=[∑k=1n(2k−1)]+2(n+1)−1(根据求和的定义)=n²+2(n+1)−1(根据归纳假设)=n²+2n+1=(n+1)²(根据代数)。这一系列已知方程证明了目标方程 P(n + 1)。下图直观地展示了公式 ∑k=1n(2k−1)=n²。面积为 5 的正方形被分解成面积分别为 1、3、5、7 和 9 的 L 形区域;因此,1 + 3 + 5 + 7 + 9 = 25 = 5。虽然该图提供了令人信服的视觉直观解释,说明求和公式为何成立,但它并不能证明该公式对所有正整数 n 都成立。充其量,通过考虑整个图及其内部的小正方形,该图证明了求和公式的前五个实例。4.8. 平方和定理。对于所有 n∈Z≥1,∑k=1nk2=n(n+1)(2n+1)6。证明。我们对 n 使用归纳法。这里,P(n) 是命题 ∑k=1nk2=n(n+1)(2n+1)/6。基本情况。我们证明 P(1),即 ∑k=11k2=1(1+1)(2⋅1+1)/6。[为了证明这一点,我们使用求和符号定义中的初始条件,并结合链式证明。] 我们知道 ∑k=11k2=12=1=1(2)(3)/6=1(1+1)(2⋅1+1)/6。归纳步骤。设 n 为一个固定的正整数。假设 P(n) 成立,即假设 ∑k=1nk2=n(n+1)(2n+1)6。证明 P(n+1),即证明 ∑k=1n+1k2=(n+1)([n+1]+1)(2[n+1]+1)/6。 [我们现在使用求和符号定义的递归部分和另一个链式证明。] 我们知道 ∑k=1n+1k2=[∑k=1nk2]+(n+1)2(根据求和的定义)=n(n+1)(2n+1)6+6(n+1)26(根据归纳假设)=(n+1)(n(2n+1)+6(n+1))6(通过因式分解n+1)=(n+1)(2n2+7n+6)6(通过代数运算)=(n+1)(n+2)(2n+3)6(通过因式分解)=(n+1)([n+1]+1)(2[n+1]+1)6(通过代数运算)。这一系列已知的等式证明了目标公式 P(n + 1)。

Section Summary

章节概要

  1. Recursive Definitions. A recursive definition introduces a sequence of quantities f (0), f (1), …, f ( n ), … by defining f (0) explicitly and giving a formula for f ( n + 1) involving the previous terms f (0), f (1), …, f ( n ).
    递归定义。递归定义通过显式定义 f(0) 并给出包含前项 f(0)、f(1)、…、f(n) 的 f(n + 1) 的公式,引入量序列 f(0)、f(1)、…、f(n)。
  2. Sums, Products, Powers, and Factorials. Given a k R and n Z 1 , recursively define
    k = 1 1 a k = a 1 , k = 1 n + 1 a k = ( k = 1 n a k ) + a n + 1 ,
    k = 1 1 a k = a 1 , k = 1 n + 1 a k = ( k = 1 n a k ) a n + 1 .
    For x R and n Z 0 , define x 0 = 1, x n +1 = x n · x , 0! = 1, and ( n + 1)! = ( n + 1) · n !.
    和、积、幂和阶乘。给定 ak∈R 和 n∈Z≥1,递归定义 ∑k=11ak=a1,∑k=1n+1ak=(∑k=1nak)+an+1,∏k=11ak=a1,∏k=1n+1ak=(∏k=1nak)⋅an+1。对于 x∈R 和 n∈Z≥0,定义 x 0 = 1, x n +1 = x n · x, 0! = 1, 和 (n + 1)! = (n + 1) · n!。
  3. Induction Axiom. For any open sentence P ( n ), the statement “ n Z 1 , P ( n ) ” is equivalent to the statement “ P ( 1 ) n Z 1 , ( P ( n ) P ( n + 1 ) ) .”
    归纳公理。对于任何开语句 P(n),命题“∀n∈Z≥1,P(n)”等价于命题“P(1)∧∀n∈Z≥1,(P(n)⇒P(n+1))”。
  4. Ordinary Induction Proof Template. To prove “ n Z 1 , P ( n ) ” by induction: Say “We use induction on n .”

    普通归纳证明模板。要用归纳法证明“∀n∈Z≥1,P(n)”,可以这样写:“我们对n进行归纳。”

    Step 1. Prove P (1) [the base case].

    步骤 1. 证明 P(1) [基本情况]。

    Step 2a. Fix n Z 1 . Assume P ( n ) is true [the induction hypothesis].

    步骤 2a. 固定 n∈Z≥1。假设 P(n) 为真[归纳假设]。

    Step 2b. Aided by this assumption, prove P ( n + 1) is true [the induction step].

    步骤 2b. 借助此假设,证明 P(n + 1) 为真 [归纳步骤]。

Exercises

练习

  1. Compute each quantity by expanding a recursive definition. (a) 6! (b) 2 5 (c) 5 2 (d) 0 0 (e) k = 1 4 k 2 (f) k = 1 4 k 2 (g) j = 1 5 1 j (h) i = 0 4 ( 2 i + 1 ) .
    通过展开递归定义来计算每个量。 (a) 6! (b) 2 5 (c) 5 2 (d) 0 0 (e) ∑k=14k2 (f) ∏k=14k2 (g) ∑j=151j (h) ∏i=04(2i+1).
  2. (a) Prove by induction: for all c R and all n Z 1 , k = 1 n c = c n . (b) Find the value of k = 0 n c .
    (a)用数学归纳法证明:对于所有 c∈R 和所有 n∈Z≥1,∑k=1nc=cn。(b)求 ∑k=0nc 的值。
  3. (a) Prove by induction: for all n Z 1 , k = 1 n k = n ( n + 1 ) / 2 . (b) Prove by induction: for all n Z 1 , k = 1 n k 3 = n 2 ( n + 1 ) 2 / 4 . [Thus the sum of the first n cubes is the square of the sum of the first n positive integers.]
    (a) 用数学归纳法证明:对于所有 n∈Z≥1,∑k=1nk=n(n+1)/2。 (b) 用数学归纳法证明:对于所有 n∈Z≥1,∑k=1nk3=n2(n+1)2/4。 [因此,前 n 个立方数的和等于前 n 个正整数之和的平方。]
  4. Prove by induction: for all n Z 1 , k = 1 n 2 3 k = 3 n + 1 3 .
    用归纳法证明:对于所有 n∈Z≥1,∑k=1n2⋅3k=3n+1−3。
  5. Prove by induction: for all n Z 1 , k = 1 n k 2 k = ( n 1 ) 2 n + 1 + 2 .
    用归纳法证明:对于所有 n∈Z≥1,∑k=1nk2k=(n−1)2n+1+2。
  6. Prove: for all n Z 1 , k = 1 n k k ! = ( n + 1 ) ! 1 .
    证明:对于所有 n∈Z≥1,∑k=1nk⋅k!=(n+1)!−1。
  7. Prove: for all n Z 1 , k = 1 n ( 1 + ( 3 / k ) ) = ( n + 1 ) ( n + 2 ) ( n + 3 ) / 6 .
    证明:对于所有 n∈Z≥1,∏k=1n(1+(3/k))=(n+1)(n+2)(n+3)/6。
  8. (a) Find k = 1 n 1 k ( k + 1 ) for 1 ≤ n ≤ 5. (b) Based on your answers to (a), guess the value of this sum for general n . Then prove your guess by induction.
    (a) 求 ∑k=1n1k(k+1),其中 1 ≤ n ≤ 5。(b) 根据你对 (a) 的回答,猜测该和对于一般 n 的值。然后用数学归纳法证明你的猜测。
  9. Use induction and the recursive definitions to prove: (a) for all a R and all n Z 1 , a n = k = 1 n a . (b) for all n Z 1 , n ! = j = 1 n j .
    使用归纳法和递归定义证明: (a) 对于所有 a∈R 和所有 n∈Z≥1,an=∏k=1na. (b) 对于所有 n∈Z≥1,n!=∏j=1nj.
  10. Find the error in the following proposed proof that for all positive integers n , n = n + 3. Proof by induction. Fix an arbitrary positive integer n ; assume n = n + 3; prove n + 1 = ( n + 1) + 3. Add 1 to both sides of the induction hypothesis to get n + 1 = ( n + 3) + 1. By algebra, this becomes n + 1 = ( n + 1) + 3, completing the induction step.
    找出以下证明中的错误,该证明表明对于所有正整数 n,n = n + 3。证明方法为数学归纳法。固定一个任意正整数 n;假设 n = n + 3;证明 n + 1 = (n + 1) + 3。在归纳假设两边同时加 1,得到 n + 1 = (n + 3) + 1。通过代数运算,这可以简化为 n + 1 = (n + 1) + 3,从而完成归纳步骤。
  11. Identify the logical error in the following proposed proof template for an induction proof of n Z 1 , P ( n ) . Explain exactly how the proof template in the section differs from the one here. Part 1. Prove P (1) is true. Part 2. Assume P ( n ) is true for all positive integers n . Prove P ( n + 1) is true for all positive integers n .
    找出以下归纳证明模板中的逻辑错误,该模板用于证明对于所有 n∈Z≥1,P(n) 成立。并详细解释该部分中的证明模板与此处的证明模板有何不同。 第一部分:证明 P(1) 为真。 第二部分:假设 P(n) 对所有正整数 n 都成立。证明 P(n + 1) 对所有正整数 n 都成立。
  12. (a) Prove: for all real a , b , c , d , ( a + b ) + ( c + d ) = ( a + c ) + ( b + d ). Justify every step using one of the axioms from § 2.1 . (b) Additivity of Sums. Suppose a k and b k are given real numbers for each positive integer k . Prove by induction: for all n Z 1 , k = 1 n ( a k + b k ) = k = 1 n a k + k = 1 n b k . Explain how part (a) is used in the proof.
    (a) 证明:对于任意实数 a、b、c、d,(a + b) + (c + d) = (a + c) + (b + d)。请用 §2.1 中的公理之一来证明每一步。(b) 和的可加性。假设对于每个正整数 k,a 和 b 是给定的实数。用数学归纳法证明:对于任意 n∈Z≥1,∑k=1n(ak+bk)=∑k=1nak+∑k=1nbk。解释 (a) 部分是如何在证明中应用的。
  13. Let c and a k (for each positive integer k ) be given real numbers. Prove by induction: for all n Z 1 , k = 1 n ( c a k ) = c k = 1 n a k .
    设 c 和 a k (对于每个正整数 k)为给定的实数。用归纳法证明:对于所有 n∈Z≥1,∑k=1n(cak)=c⋅∑k=1nak。
  14. Without invoking the Induction Axiom, prove that statement (a) in the Induction Axiom implies statement (b).
    不使用归纳公理,证明归纳公理中的陈述(a)蕴含陈述(b)。
  15. (a) Use the Induction Axiom to prove that the statement “ n Z 1 , P ( n ) ” is equivalent to “ P ( 1 ) m Z > 1 , P ( m 1 ) P ( m ) .” (b) Create a variation of the ordinary induction proof template based on the equivalence in part (a).
    (a)利用归纳公理证明语句“∀n∈Z≥1,P(n)”等价于“P(1)∧∀m∈Z>1,P(m−1)⇒P(m)”。(b)基于(a)部分的等价性,创建普通归纳证明模板的变体。
  16. Use intuitive reasoning (such as a picture of dominos) to create a variation of the Induction Axiom leading to a method for proving a statement of the form n Z < 0 , P ( n ) . Make a proof template for your proof method.
    运用直觉推理(例如想象多米诺骨牌),创建归纳公理的变体,从而得出证明形如 ∀n∈Z<0,P(n) 的命题的方法。为你的证明方法创建一个证明模板。
  17. Let A k be a fixed set for each k Z > 0 . (a) Formulate recursive definitions for k = 1 n A k and k = 1 n A k similar to the recursive definitions of sums and products. (b) Let I n = { j Z : 1 j n } for each positive integer n . Prove by induction: for all n Z 1 , k = 1 n A k = j I n A j and k = 1 n A k = j I n A j .
    设 A k 为任意 k∈Z>0 的固定集合。(a) 类似于和与积的递归定义,为 ⋃k=1nAk 和 ⋂k=1nAk 构建递归定义。(b) 设 In={j∈Z:1≤j≤n},其中 n 为正整数。用数学归纳法证明:对于所有 n∈Z≥1,⋃k=1nAk=⋃j∈InAj 且 ⋂k=1nAk=⋂j∈InAj。
  18. We say that a set S R has a least element iff z S , x S , z x . (a) Prove by induction on n : for all S Z 1 , if S { 1 , 2 , , n } , then S has a least element. [You can use the fact that for all n Z 1 , there is no integer k with n < k < n + 1.] (b) Prove that every nonempty subset of Z 1 has a least element.
    我们称集合 S⊆R 有最小元素当且仅当存在 z∈S,∀x∈S,z≤x。 (a) 用数学归纳法证明:对于所有 S⊆Z≥1,如果 S∩{1,2,…,n}≠∅,则 S 有最小元素。[你可以利用以下事实:对于所有 n∈Z≥1,不存在整数 k 使得 n < k < n + 1。] (b) 证明 Z≥1 的每个非空子集都有最小元素。

4.2 Induction Starting Anywhere and Backwards Induction

4.2 任意起始归纳法和逆归纳法

This section introduces some variations of the ordinary induction proof template that allow us to apply induction to more general situations. In particular, induction starting anywhere gives us a way to prove statements of the form n Z b , P ( n ) . Backwards induction lets us prove statements of the form n Z b , P ( n ) . Finally, we consider various methods for proving a universal statement n Z , P ( n ) where n ranges over all integers (positive, negative, and zero).

本节介绍一些常规归纳证明模板的变体,这些变体允许我们将归纳法应用于更一般的情况。特别地,从任意位置开始的归纳法使我们能够证明形如 ∀n∈Z≥b,P(n) 的命题。反向归纳法使我们能够证明形如 ∀n∈Z≤b,P(n) 的命题。最后,我们考虑证明全称命题 ∀n∈Z,P(n) 的各种方法,其中 n 取遍所有整数(正数、负数和零)。

Induction Starting Anywhere

感应加热可以从任何地方开始

Sometimes, we need to prove universal statements of the form n Z 0 , P ( n ) , which start at n = 0 rather than n = 1. The Induction Axiom and Induction Proof Template are readily adapted to this situation. The only change is that in the base case, we prove P (0) instead of P (1); and in the induction step, we fix n in the set Z 0 rather than fixing n in the set Z 1 . More generally, we have the following proof template for induction proofs starting anywhere.

有时,我们需要证明形如 ∀n∈Z≥0,P(n) 的全称命题,其中 n 从 0 开始,而不是从 1 开始。归纳公理和归纳证明模板可以很容易地应用于这种情况。唯一的区别在于,在基本情况下,我们证明的是 P(0) 而不是 P(1);并且在归纳步骤中,我们将 n 固定在集合 Z≥0 中,而不是固定在集合 Z≥1 中。更一般地,对于从任意位置开始的归纳证明,我们有以下证明模板。

4.9. Template for Induction Proofs Starting Anywhere. Let b be a fixed integer.

4.9. 从任意位置开始的归纳证明模板。设 b 为一个固定的整数。

To prove n Z b , P ( n ) :

证明 ∀n∈Z≥b,P(n):

Say “We prove the statement by induction on n .”

说“我们用归纳法证明这个命题”。

Part 1 (Base Case). Prove P ( b ).

第 1 部分(基本情况)。证明 P(b)。

Part 2 (Induction Step). Fix an arbitrary integer n 0 Z b . Assume P ( n 0 ) is true.

第二部分(归纳步骤)。固定任意整数 n0∈Z≥b。假设 P(n 0 ) 为真。

Prove P ( n 0 + 1) is true.

证明 P(n 0 + 1) 为真。

Exercise 16 outlines how this proof template may be justified using the Induction Axiom.

练习 16 概述了如何使用归纳公理来证明此证明模板的合理性。

4.10. Example. Prove: for all n Z 0 , i = 0 n 2 i = 2 n + 1 1 .

4.10. 例。证明:对于所有 n∈Z≥0,∑i=0n2i=2n+1−1。

Proof. We use induction on n .

证明。我们对 n 使用归纳法。

Base Case. We prove i = 0 0 2 i = 2 0 + 1 1 . We know i = 0 0 2 i = 2 0 = 1 = 2 1 = 2 0 + 1 1 .

基本情况。我们证明 ∑i=002i=20+1−1。我们知道 ∑i=002i=20=1=2−1=20+1−1。

Induction Step. Fix n Z 0 . Assume i = 0 n 2 i = 2 n + 1 1 . Prove i = 0 n + 1 2 i = 2 ( n + 1 ) + 1 1 . [Use a chain proof.] We know

归纳步骤。固定 n∈Z≥0。假设 ∑i=0n2i=2n+1−1。证明 ∑i=0n+12i=2(n+1)+1−1。[使用链式证明。] 我们知道

i = 0 n n + 1 2 i = ( i = 0 n n 2 i ) + 2 n n + 1 (by definition of sums) (根据求和的定义) = 2 n n + 1 1 + 2 n n + 1 (by induction hypothesis) (根据归纳假设) = 2 n n + 1 2 1 (by algebra) (通过代数方法) = 2 ( n n + 1 ) + 1 1 (by definition of exponents). (根据指数的定义)。

All of our examples so far have used induction to prove summation formulas. The next examples show how induction can be used to prove other kinds of statements.

到目前为止,我们所有的例子都使用了归纳法来证明求和公式。接下来的例子将展示如何用归纳法来证明其他类型的命题。

4.11. Example. Prove: n Z 3 , k = 3 n ( 1 2 k ) = 2 n ( n 1 ) .

4.11. 例。证明:∀n∈Z≥3,∏k=3n(1−2k)=2n(n−1)。

Proof. We use induction on n .

证明。我们对 n 使用归纳法。

Base Case. We prove k = 3 3 ( 1 2 k ) = 2 3 ( 3 1 ) . By the initial condition in the recursive definition of products and by arithmetic, we know

基本情况。我们证明 ∏k=33(1−2k)=23(3−1)。根据乘积递归定义的初始条件和算术运算,我们知道

k k = 3 3 ( 1 2 k k ) = 1 2 3 = 1 3 = 2 6 = 2 3 ( 3 1 ) .

Induction Step. Fix an arbitrary integer n Z 3 . Assume k = 3 n ( 1 2 k ) = 2 n ( n 1 ) .

归纳步骤。固定任意整数 n∈Z≥3。假设 ∏k=3n(1−2k)=2n(n−1)。

Prove k = 3 n + 1 ( 1 2 k ) = 2 ( n + 1 ) ( n + 1 1 ) . We know

证明 ∏k=3n+1(1−2k)=2(n+1)(n+1−1)。我们知道

k k = 3 n n + 1 ( 1 2 k k ) = [ k k = 3 n n ( 1 2 k k ) ] ( 1 2 n n + 1 ) (by definition of products) (根据产品的定义) = 2 n n ( n n 1 ) ( 1 2 n n + 1 ) (by induction hypothesis) (根据归纳假设) = 2 n n ( n n 1 ) n n 1 n n + 1 (by algebra) (通过代数方法) = 2 ( n n + 1 ) n n = 2 ( n n + 1 ) ( n n + 1 1 ) (by algebra). (通过代数方法)

4.12. Example. Prove: for all integers n ≥ 4, n ! > 2 n .

4.12. 例。证明:对于所有整数 n ≥ 4,n! > 2 n

Proof. We use induction on n . [Here, P ( n ) is the statement “ n ! > 2 n .”]

证明。我们对 n 使用归纳法。[这里,P(n) 表示命题“n! > 2 n .”]

Base Case. We must prove 4! > 2 4 . By the recursive definitions, we know 4! = 24 > 16 = 2 4 .

基本情况。我们必须证明 4! > 2 4 。根据递归定义,我们知道 4! = 24 > 16 = 2 4

Induction Step. Fix an arbitrary integer n ≥ 4; assume n ! > 2 n ; prove ( n + 1)! > 2 n +1 . [We use a chain proof for inequalities.] We know

归纳步骤。固定任意整数 n ≥ 4;假设 n! > 2;证明 (n + 1)! > 2。[我们对不等式使用链式证明。] 我们知道

( n n + 1 ) ! = ( n n + 1 ) n n ! (by definition of factorials) (根据阶乘的定义) > ( n n + 1 ) 2 n n (by induction hypothesis; note (根据归纳假设;注) n n + 1 > 0 ) > 2 2 n n (since (自从 n n + 1 > 2 and 2 n n > 0 ) = 2 n n + 1 (by definition of exponents). (根据指数的定义)。

[To explain the two inequalities in more detail, recall this theorem: a R , b R , c R , if a > b and c > 0, then ca > cb and ac > bc . The first inequality in the chain follows from this theorem by taking a = n !, b = 2 n , and c = n + 1; note that the hypothesis a > b is our assumption in the induction proof, and c > 0 since n + 1 must be at least 5. Similarly, the second inequality follows from this theorem by taking a = n + 1, b = 2 and c = 2 n , which is allowed since we know n + 1 > 2 and 2 n > 0. Finally, by transitivity of >, the displayed chain of steps lets us deduce that ( n + 1)! > 2 n +1 , which is the goal of the induction step.]

为了更详细地解释这两个不等式,回顾一下这个定理:∀a∈R,∀b∈R,∀c∈R,如果 a > b 且 c > 0,则 ca > cb 且 ac > bc。链中的第一个不等式由该定理得出,只需令 a = n!,b = 2 n ,c = n + 1;注意,假设 a > b 是我们在归纳证明中的假设,而 c > 0 是因为 n + 1 必须至少为 5。类似地,第二个不等式由该定理得出,只需令 a = n + 1,b = 2,c = 2 n ,这是允许的,因为我们知道 n + 1 > 2 且 2 n > 0。最后,根据 > 的传递性,上述步骤链使我们能够推断出 (n + 1)! > 2 n +1 ,这是归纳步骤的目标。

The next example shows why we started the induction at n = 4 in the last proof.

下一个例子说明了为什么我们在上一个证明中从 n = 4 开始进行归纳推理。

4.13. Example. Disprove: for all integers n ≥ 1, n ! > 2 n .

4.13. 示例。反证:对于所有整数 n ≥ 1,n! > 2 n

Disproof. We must prove n Z 1 , n ! 2 n . [Induction is not required here; we prove the existence statement by giving a single example.] Choose n = 3, which is in Z 1 . We compute 3! = 6 ≤ 8 = 2 3 .

反证。我们必须证明存在 n∈Z≥1,n!≤2n。[这里不需要归纳法;我们通过给出一个例子来证明存在性命题。] 选择 n = 3,它在 Z≥1 中。我们计算 3! = 6 ≤ 8 = 2 3

More Examples of Induction Proofs

更多归纳证明示例

4.14. Example. Prove that for all n Z 0 , 8 divides 5 2 n − 1.

4.14. 例。证明对于所有 n∈Z≥0,8 能整除 5 2 n − 1。

Proof. We use induction on n . [Here, P ( n ) is the statement “8 divides 5 2 n − 1.”]

证明。我们对 n 使用归纳法。[这里,P(n) 表示命题“8 能整除 5 2 n − 1。”]

Base Case. We must prove 8 divides 5 2·0 − 1, i.e., that 8 divides 0. We must prove k Z , 0 = 8 k . Choose k = 0; we know 0 Z and 0 = 8 · 0.

基本情况。我们必须证明 8 能整除 5 2·0 − 1,即 8 能整除 0。我们必须证明存在 k∈Z,0=8k。选择 k = 0;我们知道 0∈Z 且 0 = 8 · 0。

Induction Step. Fix n Z 0 ; assume 8 divides 5 2 n − 1; prove 8 divides 5 2( n +1) − 1. We know a Z , 5 2 n 1 = 8 a ; we must prove b Z , 5 2 n + 2 1 = 8 b . [How can we choose b ? The key is to rewrite 5 2 n +2 in a way that lets us use the induction hypothesis.] On one hand, we know 5 2 n +2 − 1 = 5 2 n · 5 · 5 − 1 = 25 · 5 2 n − 1. Now, from the induction hypothesis, we know 5 2 n = 8 a + 1. Substituting, we get 5 2 n +2 − 1 = 25(8 a + 1) − 1 = 25(8 a ) + 25 − 1 = 8(25 a ) + 24 = 8(25 a + 3). Choose b = 25 a + 3, which is an integer by closure. By the above algebra, we have 5 2 n +2 − 1 = 8 b , as needed.

归纳步骤。固定 n∈Z≥0;假设 8 能整除 5 2 n − 1;证明 8 能整除 5 2( n +1) − 1。我们知道 ∃a∈Z,52n−1=8a;我们必须证明 ∃b∈Z,52n+2−1=8b。[我们如何选择 b?]关键在于将 5 2 n +2 改写成能够运用归纳假设的形式。一方面,我们知道 5 2 n +2 − 1 = 5 2 n · 5 · 5 − 1 = 25 · 5 2 n − 1。另一方面,根据归纳假设,我们知道 5 2 n = 8a + 1。代入后,我们得到 5 2 n +2 − 1 = 25(8a + 1) − 1 = 25(8a) + 25 − 1 = 8(25a) + 24 = 8(25a + 3)。根据封闭性,选择 b = 25a + 3,这是一个整数。根据上述代数运算,我们有 5 2 n +2 − 1 = 8b,符合要求。

The next example introduces the idea of a recursively defined sequence . The example also illustrates how we can gather data to guess a theorem that might be true, and then use induction to prove that theorem.

下一个例子引入了递归定义序列的概念。这个例子还说明了我们如何收集数据来猜测一个可能为真的定理,然后用归纳法来证明这个定理。

4.15. Example. Define numbers a 1 , a 2 , …, a n , … recursively by setting a 1 = 5 and a n +1 = 3 a n − 4 for all integers n ≥ 1. (a) Calculate a 1 , a 2 , a 3 , a 4 , a 5 . (b) Use the data in (a) to guess a non-recursive formula for a n . (c) Prove the formula in (b) by induction.

4.15. 示例。递归地定义数 a 1 , a 2 , …, a n , …,令 a 1 = 5 且 a n +1 = 3a n − 4,其中 n 为所有整数 n ≥ 1。(a) 计算 a 1 , a 2 , a 3 , a 4 , a 5 。(b) 利用 (a) 中的数据,猜测 a n 的一个非递归公式。(c) 用数学归纳法证明 (b) 中的公式。

Solution. (a) We know a 1 = 5. Taking n = 1 in the recursive formula, we find a 2 = 3 a 1 − 4 = 3(5) − 4 = 11. Taking n = 2 in the recursive formula, we get a 3 = 3 a 2 − 4 = 3(11) − 4 = 29. Continuing similarly, a 4 = 3 a 3 − 4 = 3(29) − 4 = 83, and a 5 = 3 a 4 − 4 = 3(83) − 4 = 245. (b) What is the pattern in the numbers 5, 11, 29, 83, 245? We might eventually notice that these numbers are close to powers of 3, which are 3, 9, 27, 81, 243. In fact, we see that a n = 3 n + 2 for 1 ≤ n ≤ 5. We can verify that this formula continues to work for n = 6, 7, 8, … by calculating more values from the recursion. So we guess that a n = 3 n + 2 for all integers n ≥ 1. (Notice that we cannot fully confirm this guess by calculating finitely many a n from the recursion.)

解:(a) 已知 a 1 = 5。在递归公式中取 n = 1,得 a 2 = 3a 1 − 4 = 3(5) − 4 = 11。在递归公式中取 n = 2,得 a 3 = 3a 2 − 4 = 3(11) − 4 = 29。以此类推,得 a 4 = 3a 3 − 4 = 3(29) − 4 = 83,a 5 = 3a 4 − 4 = 3(83) − 4 = 245。(b) 求数字 5、 11、29、83、245?我们最终可能会注意到这些数字接近 3 的幂,即 3、9、27、81、243。事实上,我们看到,对于 1 ≤ n ≤ 5,a n = 3 n + 2。我们可以通过递归计算更多值来验证该公式对于 n = 6、7、8、… 仍然成立。因此,我们猜测对于所有整数 n ≥ 1,a n = 3 n + 2。(请注意,我们无法通过递归计算有限个 a n 来完全验证此猜测。)

(c) We now prove n Z 1 , a n = 3 n + 2 by induction on n . For the base case, we must prove a 1 = 3 1 + 2. We know a 1 = 5 and 3 1 + 2 = 5, so the base case is true. For the induction step, fix an arbitrary integer n Z 1 , assume a n = 3 n + 2, and prove a n +1 = 3 n +1 + 2. We have the chain of known equalities:

(c) 现在我们用数学归纳法证明对于任意 n∈Z≥1,a<sub>n</sub>=3n+2。对于基本情况,我们必须证明 a<sub>n</sub> = 3<sub>n</sub> + 2。我们知道 a<sub>n</sub> = 5 且 3<sub>n</sub> + 2 = 5,所以基本情况成立。对于归纳步骤,固定任意整数 n∈Z≥1,假设 a<sub>n</sub> = 3<sub>n</sub> + 2,并证明 a<sub>n</sub> = 3<sub>n</sub> + 2。我们有如下已知的等式链:

a 一个 n n + 1 = 3 a 一个 n n 4 ( by recursive definition of 通过递归定义 a 一个 n n + 1 ) = 3 ( 3 n n + 2 ) 4 (by induction hypothesis) (根据归纳假设) = 3 ( 3 n n ) + 3 2 4 (by the distributive law) (根据分配法) = 3 n n + 1 + 2 (by definition of exponents). (根据指数的定义)。

The goal of the next example is to prove one of the laws of exponents from algebra. This statement contains multiple universal quantifiers; we prove it by combining a generic-element proof with an induction proof starting at 0.

下一个例子的目标是证明代数中的一条指数定律。该命题包含多个全称量词;我们通过结合一般元素证明和从 0 开始的归纳证明来证明它。

4.16. Example. Prove: for all real numbers x and all integers m , n Z 0 , x m + n = x m x n .

4.16. 例。证明:对于所有实数 x 和所有整数 m,n∈Z≥0,x m + n = x m x n

Proof. We must prove x R , m Z 0 , n Z 0 , x m + n = x m x n . [We begin by using the generic-element proof template twice.] Let x be a fixed, but arbitrary real number. Let m be a fixed, but arbitrary nonnegative integer. [For the universal statement involving n , we use induction.] We prove by induction on n that x m + n = x m x n .

证明。我们必须证明 ∀x∈R,∀m∈Z≥0,∀n∈Z≥0,xm+n=xmxn。[我们首先使用两次通用元素证明模板。] 设 x 为一个固定的任意实数。设 m 为一个固定的任意非负整数。[对于涉及 n 的全称命题,我们使用归纳法。] 我们对 n 进行归纳证明 x m + n = x m x n

Base Case. For n = 0, we must prove x m +0 = x m x 0 . On one hand, x m +0 = x m since m + 0 = m . On the other hand, x m x 0 = x m · 1 = x m by the initial condition in the definition of exponents and the multiplicative identity axiom. So the base case holds.

基本情况。对于 n = 0,我们必须证明 x m +0 = x m x 0 。一方面,由于 m + 0 = m,x m +0 = x m 。另一方面,根据指数定义中的初始条件和乘法恒等式公理,x m x 0 = x m · 1 = x m 。因此,基本情况成立。

Induction Step. Fix an arbitrary integer n ≥ 0. Assume x m + n = x m x n . Prove x m +( n +1) = x m x n +1 . We know

归纳步骤。固定任意整数 n ≥ 0。假设 x m + n = x m x n 。证明 x m +( n +1) = x m x n +1 。我们知道

x x m + ( n n + 1 ) = x x ( m + n n ) + 1 (by associativity of addition) (根据加法结合律) = x x m + n n x x (by recursive definition of exponents) (根据指数的递归定义) = ( x x m x x n n ) x x (by induction hypothesis) (根据归纳假设) = x x m ( x x n n x x ) (by associativity of multiplication) (根据乘法结合律) = x x m x x n n + 1 (by recursive definition of exponents). (根据指数的递归定义)。

Backwards Induction and Induction on Z (Optional)

反向感应和 Z 轴感应(可选)

So far, we have talked about induction proof templates for proving statements of the form n Z b , P ( n ) . What if we need to prove that P ( n ) holds for all integers n ? We discuss two ways to do this in a moment. First, we introduce backwards induction , which can be used to prove statements of the form n Z b , P ( n ) .

到目前为止,我们已经讨论了用于证明形如 ∀n∈Z≥b,P(n) 的命题的归纳证明模板。如果我们需要证明 P(n) 对所有整数 n 都成立,该怎么办呢?我们稍后会讨论两种方法。首先,我们介绍逆归纳法,它可以用来证明形如 ∀n∈Z≤b,P(n) 的命题。

4.17. Template for Backwards Induction Proofs Let b be a fixed integer.

4.17. 逆归纳证明的模板 设 b 为一个固定的整数。

To prove n Z b , P ( n ) :

证明 ∀n∈Z≤b,P(n):

Say “We prove the statement by backwards induction on n .”

说“我们用对 n 进行逆归纳法来证明这个命题。”

Part 1 (Base Case). Prove P ( b ).

第 1 部分(基本情况)。证明 P(b)。

Part 2 (Induction Step). Fix an arbitrary integer n 0 Z b . Assume P ( n 0 ) is true.

#

第二部分(归纳步骤)。固定任意整数 n0∈Z≤b。假设 P(n 0 ) 为真。

Prove P ( n 0 − 1) is true.

证明 P(n 0 − 1) 为真。

See the exercises for intuitive and formal justifications of why this proof template works. With this template in hand, we can now describe the first method for proving a universal statement about all integers.

请参阅练习,了解此证明模板为何有效的直观和正式解释。有了这个模板,我们现在可以描述证明关于所有整数的普遍命题的第一种方法。

4.18. First Template for Proving n Z , P ( n )

4.18. 证明 ∀n∈Z,P(n) 的第一个模板

Part 1 (Base Case). Prove P (0).

第一部分(基本情况)。证明 P(0)。

Part 2 (Forward Induction Step). Fix an arbitrary integer n 0 Z 0 . Assume P ( n 0 ) is true. Prove P ( n 0 + 1) is true.

第二部分(前向归纳步骤)。固定任意整数 n0∈Z≥0。假设 P(n 0 ) 为真。证明 P(n 0 + 1) 为真。

Part 3 (Backward Induction Step). Fix an arbitrary integer m 0 Z 0 . Assume P ( m 0 ) is true. Prove P ( m 0 − 1) is true.

第三部分(逆向归纳步骤)。固定任意整数 m0∈Z≤0。假设 P(m 0 ) 为真。证明 P(m 0 − 1) 为真。

This template works because “ n Z , P ( n ) ” is logically equivalent to “ n Z 0 , P ( n ) m Z 0 , P ( m ) .” We prove the latter statement by combining forwards induction (starting at 0) with backwards induction (also starting at 0). We could replace each occurrence of 0, here and in the proof template above, by any fixed integer b .

这个模板之所以有效,是因为“∀n∈Z,P(n)”在逻辑上等价于“∀n∈Z≥0,P(n)∧∀m∈Z≤0,P(m)”。我们通过结合正向归纳法(从0开始)和反向归纳法(也从0开始)来证明后一个命题。我们可以将这里以及上面证明模板中出现的每个0替换为任意固定的整数b。

Our second method for proving n Z , P ( n ) first handles nonnegative integers by ordinary induction, and then proves the statement for each negative integer by reducing to the positive case dealt with earlier. This technique is useful when the statement P ( − n ) is closely related to the statement P ( n ).

我们证明 ∀n∈Z,P(n) 的第二种方法首先用普通归纳法处理非负整数,然后通过简化到前面讨论过的正整数情况来证明每个负整数的命题。当命题 P(−n) 与命题 P(n) 密切相关时,这种方法非常有用。

4.19. Second Template for Proving n Z , P ( n )

4.19. 证明 ∀n∈Z,P(n) 的第二个模板

Part 1 (Base Case). Prove P (0).

第一部分(基本情况)。证明 P(0)。

Part 2 (Forward Induction Step). Fix an integer n 0 Z 0 . Assume P ( n 0 ) is true. Prove P ( n 0 + 1) is true.

第二部分(前向归纳步骤)。固定一个整数 n0∈Z≥0。假设 P(n 0 ) 为真。证明 P(n 0 + 1) 为真。

Part 3 (Proof for Negative Integers). Fix a positive integer n 0 Z > 0 . We have proved P ( n 0 ) is true. Prove P (− n 0 ) is true.

第三部分(负整数的证明)。固定一个正整数 n0∈Z>0。我们已经证明 P(n 0 ) 为真。证明 P(−n 0 ) 为真。

Refer to the exercises for a justification of this proof template.

请参考练习题来理解此证明模板的合理性。

4.20. Example. We outline a proof that for all integers n , n is even or n is odd , asking the reader to supply the missing details.

4.20. 例。我们概述一个证明,对于所有整数 n,n 要么是偶数,要么是奇数,请读者补充缺失的细节。

Part 1. Prove 0 is even or 0 is odd. Since 0 = 2 · 0, 0 is even, so the OR-statement is true.

第一部分:证明 0 是偶数或 0 是奇数。因为 0 = 2 × 0,所以 0 是偶数,因此“或”命题成立。

Part 2. Fix an integer n ≥ 0. Assume n is even or n is odd. Prove n + 1 is even or n + 1 is odd. This part is completed by considering two cases (see the exercises).

第二部分:固定一个整数 n ≥ 0。假设 n 为偶数或 n 为奇数。证明 n + 1 为偶数或 n + 1 为奇数。本部分通过考虑两种情况来完成(参见练习)。

Part 3. Now fix an integer n > 0. By part 2, we already know that n is even or n is odd. We must prove − n is even or − n is odd. Again this part can be completed by considering two cases. [The fastest way to finish is to use the previously proved results that n is even iff − n is even, and n is odd iff − n is odd.]

第三部分。现在固定一个大于 0 的整数 n。根据第二部分,我们已经知道 n 要么是偶数,要么是奇数。我们必须证明 −n 要么是偶数,要么是奇数。同样,这部分可以通过考虑两种情况来完成。[最快的完成方法是利用之前证明的结果:n 为偶数当且仅当 −n 为偶数,n 为奇数当且仅当 −n 为奇数。]

Combining this example with Example 2.35 (where we showed that no integer is both even and odd), we have finally proved the theorem “every integer is even or odd, but not both.” This theorem was announced without proof in § 2.3 . We give another proof of this theorem a little later, using the Integer Division Theorem.

结合例 2.35(其中我们证明了不存在既是偶数又是奇数的整数),我们最终证明了定理“每个整数要么是偶数,要么是奇数,但不可能既是偶数又是奇数”。该定理在 §2.3 中未给出证明。稍后我们将利用整数除法定理给出该定理的另一个证明。

Section Summary

章节概要

  1. Induction Starting Anywhere. To prove n Z b , P ( n ) by ordinary induction:

    从任意位置开始的归纳法。要用普通归纳法证明 ∀n∈Z≥b,P(n):

    Step 1. Prove P ( b ) [the base case].

    步骤 1. 证明 P(b) [基本情况]。

    Step 2a. Fix n Z b , and assume P ( n ) is true [the induction hypothesis].

    步骤 2a. 固定 n∈Z≥b,并假设 P(n) 为真[归纳假设]。

    Step 2b. Aided by this assumption, prove P ( n + 1) is true [the induction step].

    步骤 2b. 借助此假设,证明 P(n + 1) 为真 [归纳步骤]。

  2. Backwards Induction. To prove n Z b , P ( n ) by backwards induction:

    逆向归纳法。用逆向归纳法证明∀n∈Z≤b,P(n):

    Step 1. Prove P ( b ) [the base case].

    步骤 1. 证明 P(b) [基本情况]。

    Step 2a. Fix n Z b , and assume P ( n ) is true [the induction hypothesis].

    步骤 2a. 固定 n∈Z≤b,并假设 P(n) 为真[归纳假设]。

    Step 2b. Aided by this assumption, prove P ( n − 1) is true [the induction step].

    步骤 2b. 借助此假设,证明 P(n − 1) 为真 [归纳步骤]。

  3. First Method to Prove n Z , P ( n ) .

    第一种方法证明 ∀n∈Z,P(n)。

    Step 1. [Base case] Prove P (0).

    步骤 1. [基本情况] 证明 P(0)。

    Step 2. [Forward induction step] Fix n Z 0 ; assume P ( n ); prove P ( n + 1).

    步骤 2. [前向归纳步骤] 固定 n∈Z≥0;假设 P(n);证明 P(n + 1)。

    Step 3. [Backward induction step] Fix m Z 0 ; assume P ( m ); prove P ( m − 1).

    步骤 3. [逆归纳步骤] 固定 m∈Z≤0;假设 P(m);证明 P(m − 1)。

  4. Second Method to Prove n Z , P ( n ) .

    证明 ∀n∈Z,P(n) 的第二种方法。

    Step 1. [Base case] Prove P (0).

    步骤 1. [基本情况] 证明 P(0)。

    Step 2. [Forward induction step] Fix n Z 0 ; assume P ( n ); prove P ( n + 1).

    步骤 2. [前向归纳步骤] 固定 n∈Z≥0;假设 P(n);证明 P(n + 1)。

    Step 3. [Negative case] Fix n Z > 0 ; note P ( n ) has been proved; prove P ( − n ).

    步骤 3. [否定情况] 固定 n∈Z>0;注意 P(n) 已被证明;证明 P( − n)。

Exercises

练习

  1. Prove by induction: for all n Z 2 , k = 2 n ( 1 1 k 2 ) = n + 1 2 n .
    用归纳法证明:对于所有 n∈Z≥2,∏k=2n(1−1k2)=n+12n。
  2. (a) Compute the product k = 4 n ( 1 3 k ) for 4 ≤ n ≤ 7. (b) Based on the data in (a), guess the value of this product for general n ≥ 4, and prove your guess by induction.
    (a)计算 4 ≤ n ≤ 7 时 ∏k=4n(1−3k) 的乘积。(b)根据(a)中的数据,猜测一般 n ≥ 4 时该乘积的值,并用归纳法证明你的猜测。
    1. (a) Prove by induction: for all real r ≠ 1 and all n Z 0 ,
      k = 0 n r k = 1 r n + 1 1 r .

      (a)用归纳法证明:对于所有实数 r ≠ 1 和所有 n∈Z≥0,∑k=0nrk=1−rn+11−r。
    2. (b) Compute k = 0 n r k when r = 1.
      (b)当 r = 1 时,计算 ∑k=0nrk。
  3. Prove: x R 0 , n Z 0 , ( 1 + x ) n 1 + n x .
    证明:∀x∈R≥0,∀n∈Z≥0,(1+x)n≥1+nx。
  4. (a) Prove: for all n Z 5 , n 2 < 2 n . [To finish, you may need to prove a similar inequality involving a smaller power of n .] (b) Disprove: for all n Z 1 , n 2 < 2 n .
    (a) 证明:对于所有 n∈Z≥5,n 2 < 2 n 。[为了完成证明,你可能需要证明一个涉及 n 的较小幂的类似不等式。] (b) 反证:对于所有 n∈Z≥1,n 2 < 2 n
  5. Prove by induction: for all n Z 0 , 6 divides 7 n − 1.
    用归纳法证明:对于所有 n∈Z≥0,6 整除 7 n − 1。
  6. Telescoping Sums. For each k Z , let a k be a fixed real number. Prove: for all i , j Z , if i j , then k = i j ( a k + 1 a k ) = a j + 1 a i .
    伸缩和。对于每个 k∈Z,令 a k 为一个固定的实数。证明:对于所有 i,j∈Z,如果 i ≤ j,则 ∑k=ij(ak+1−ak)=aj+1−ai。
  7. Laws of Exponents. Prove: x R , m Z 0 , n Z 0 , ( x m ) n = x m n . [Fix x and m and use induction on n . You will need the law of exponents proved in Example 4.16.]
    指数定律。证明:∀x∈R,∀m∈Z≥0,∀n∈Z≥0,(xm)n=xmn。[固定x和m,对n使用数学归纳法。你需要用到例4.16中证明的指数定律。]
  8. Prove: x R , y R , n Z 0 , ( x y ) n = x n y n .
    证明:∀x∈R,∀y∈R,∀n∈Z≥0,(xy)n=xnyn。
  9. Define a 0 = 1 and a n +1 = 2 a n − 1 for all n Z 0 . (a) Compute a n for 0 ≤ n ≤ 5. (b) Based on the data in (a), guess a general formula for a n and prove it by induction.
    定义 a 0 = 1 和 a n +1 = 2a n − 1,其中 n∈Z≥0。(a)计算 0 ≤ n ≤ 5 时的 a n 。(b)基于(a)中的数据,猜测 a n 的一般公式,并用数学归纳法证明它。
  10. Define a 0 = 4 and a n +1 = 2 a n + 3 · 2 n for all n Z 0 . Prove: n Z 0 , a n = (3 n + 4)2 n .
    定义 a 0 = 4 和 a n +1 = 2a n + 3 · 2 n ,对于所有 n∈Z≥0。证明:∀n∈Z≥0,a n = (3n + 4)2 n
  11. Let x and y be fixed, unequal real numbers. (a) Simplify x n y n x y for 1 ≤ n ≤ 4. (b) For general n Z 1 , guess a summation formula that evaluates to x n y n x y , and prove it by induction on n .
    设 x 和 y 为固定的不相等实数。(a) 化简 xn−ynx−y,其中 1 ≤ n ≤ 4。(b) 对于一般的 n∈Z≥1,猜测一个求和公式,使其结果为 xn−ynx−y,并用数学归纳法证明该公式。
  12. (a) Prove by induction: for all n Z 1 , j = 1 n j 4 ( n + 1 ) 5 5 . (b) Prove the inequality in (a) using integral calculus.
    (a)用归纳法证明:对于所有 n∈Z≥1,∑j=1nj4≤(n+1)55。(b)用积分学证明(a)中的不等式。
  13. Fill in the missing details of the proof in Example 4.20.
    填写例 4.20 中证明的缺失细节。
  14. Negative Exponents. The goal of this problem is to prove the laws of exponents where the powers involved can be any integers (possibly negative). Given any nonzero real number x , the axioms for R tell us that there exists a unique real number x ′ (the multiplicative inverse of x ) such that x · x ′ = x ′ · x = 1. We have already defined x n when n is a nonnegative integer. For a positive integer n , we now define the negative power x n to be ( x ′) n . In particular, notice that x −1 = ( x ′) 1 = x ′ is the multiplicative inverse of x . (a) Prove: x R 0 , n Z > 0 , ( x ) n = ( x n ) . [So ( x −1 ) n = ( x n ) −1 .] (b) Prove: x R 0 , m Z , n Z , x m + n = x m x n . (c) Prove: x R 0 , m Z , n Z , ( x m ) n = x m n . (d) Prove: x R 0 , y R 0 , n Z , ( x y ) n = x n y n .
    负指数。本题的目标是证明指数定律,其中涉及的幂可以是任意整数(可能为负数)。给定任意非零实数 x,R 的公理告诉我们,存在唯一的实数 x′(x 的乘法逆元),使得 x · x′ = x′ · x = 1。我们已经定义了当 n 为非负整数时 x n 。对于正整数 n,我们现在定义负幂 x n 为 (x′) n 。特别地,注意到 x −1 = (x′) 1 = x′ 是 x 的乘法逆元。 (a) 证明:∀x∈R≠0,∀n∈Z>0,(x′)n=(xn)′。 [所以 (x −1 ) n = (x n ) −1 .] (b) 证明:∀x∈R≠0,∀m∈Z,∀n∈Z,xm+n=xmxn. (c) 证明:∀x∈R≠0,∀m∈Z,∀n∈Z,(xm)n=xmn. (d) 证明:∀x∈R≠0,∀y∈R≠0,∀n∈Z,(xy)n=xnyn.
  15. Proof Template 4.9 gives a method to prove statements of the form n Z b , P ( n ) by induction. Use the Induction Axiom 4.5 to justify this proof template. [ Hint: Given P ( n ) and b , let Q ( n ) be the statement P ( b + n − 1) for each positive integer n .]
    证明模板 4.9 提供了一种用归纳法证明形如 ∀n∈Z≥b,P(n) 的命题的方法。利用归纳公理 4.5 来证明该证明模板的合理性。[提示:给定 P(n) 和 b,令 Q(n) 为对于每个正整数 n,命题 P(b + n − 1)。]
  16. Give an intuitive justification for the Backwards Induction Proof Template 4.17 by imitating our informal explanation of the Induction Axiom, based on repeated use of the Inference Rules for IF and ALL. Illustrate your answer by drawing a row of dominos.
    请通过模仿我们对归纳公理的非正式解释,并基于对“如果”和“所有”推理规则的反复使用,对逆向归纳证明模板 4.17 给出直观的证明。请用一排多米诺骨牌来说明你的答案。
  17. Use induction starting anywhere (see Proof Template 4.9) to give a formal justification of backwards induction (see Proof Template 4.17).
    使用从任意位置开始的归纳法(参见证明模板 4.9)来给出反向归纳法的正式证明(参见证明模板 4.17)。
  18. It is known that for all m Z , either m ≥ 0 or − m > 0. Use this theorem to give a formal justification of Proof Template 4.19 for proving statements of the form n Z , P ( n ) .
    已知对于所有 m∈Z,要么 m ≥ 0,要么 −m > 0。利用此定理,对证明模板 4.19 给出形式化的证明,以证明形如 ∀n∈Z,P(n) 的陈述。
  19. Let P ( n ) be a fixed open sentence. Suppose we have proved “ n Z , P ( n ) ↠ P ( n + 1)” and “ n Z , P ( n ) ↠ P ( n − 1).” (a) Give a specific example to show that n Z , P ( n ) can be false. (b) Prove, however, that “ n Z , P ( n ) or m Z , ¬ P ( m ) ” must be true.
    设 P(n) 为一个固定的开语句。假设我们已经证明了“∀n∈Z, P(n) ↠ P(n + 1)”和“∀n∈Z, P(n) ↠ P(n − 1)”。 (a) 请给出一个具体的例子来证明 ∀n∈Z, P(n) 可能为假。 (b) 然而,证明“∀n∈Z, P(n) 或 ∀m∈Z, ¬P(m)” 必定为真。
  20. Induction Stepping by 2. Here is a new template for proving n Z 0 , P ( n ) : Step 1. Prove P (0). Prove P (1). Step 2. Fix n 0 Z 0 . Assume P ( n 0 ) is true. Prove P ( n 0 + 2) is true. Show that this template is a correct method for proving the given statement. [ Hint: Let Q ( n ) be the statement “ P ( n ) and P ( n + 1).” Prove n Z 0 , Q ( n ) by ordinary induction.]
    归纳法步骤 2. 以下是证明 ∀n∈Z≥0,P(n) 的新模板: 步骤 1. 证明 P(0)。证明 P(1)。 步骤 2. 固定 n0∈Z≥0。假设 P(n 0 ) 为真。证明 P(n 0 + 2) 为真。 证明此模板是证明给定命题的正确方法。 [提示:令 Q(n) 为命题“P(n) 且 P(n + 1)”。用普通归纳法证明 ∀n∈Z≥0,Q(n)。]
  21. Define a 0 = 1, a 1 = 4, and a n = 9 a n −2 − 8 for all n Z 2 . Guess a non-recursive formula for a n , and then prove it using the template in Exercise 21.
    定义 a 0 = 1,a 1 = 4,以及 a n = 9a n −2 − 8,其中 n∈Z≥2。猜测 a n 的一个非递归公式,然后使用练习 21 中的模板证明它。
  22. Prove: for every odd integer n , 8 divides n 2 − 1. [ Suggestion: Write n = 2 k + 1 and use induction on k .]
    证明:对于每个奇数 n,8 能整除 n 2 − 1。[提示:令 n = 2k + 1,并对 k 使用数学归纳法。]
  23. Let s ≥ 1 be a fixed integer (the step size ). State and prove a generalization of Exercise 21 where part 2 of the template fixes n 0 Z 0 , assumes P ( n 0 ) is true, and proves P ( n 0 + s ) is true.
    设 s ≥ 1 为固定整数(步长)。陈述并证明练习 21 的一个推广,其中模板的第 2 部分固定 n0∈Z≥0,假设 P(n 0 ) 为真,并证明 P(n 0 + s) 为真。

4.3  Strong Induction

4.3 强感应

In an ordinary induction proof of n Z 1 , P ( n ) , we are allowed to assume P ( n ) in the induction step (where n is fixed) to help us prove P ( n + 1). This method works well in many situations, but sometimes the assumption is not powerful enough to enable us to finish proving P ( n + 1). Strong induction lets us assume even more information as part of the induction hypothesis. Instead of assuming only P ( n ) when proving P ( n + 1), we are now allowed to assume that all of the preceding statements P (1), P (2), …, P ( n ) are true. This additional information gives us more to work with as we try to prove P ( n + 1). Strong induction is often needed to handle recursive definitions where the recursively defined quantity f ( n + 1) depends on several of the preceding objects f (1), f (2), …, f ( n ).

在对 ∀n∈Z≥1,P(n) 进行普通归纳证明时,我们可以在归纳步骤中假设 P(n) 为真(其中 n 为固定值),以帮助我们证明 P(n + 1)。这种方法在许多情况下都有效,但有时这种假设不足以帮助我们完成 P(n + 1) 的证明。强归纳法允许我们在归纳假设中包含更多信息。在证明 P(n + 1) 时,我们不再仅仅假设 P(n) 为真,而是可以假设所有前面的语句 P(1)、P(2)、…、P(n) 都为真。这些额外的信息为我们证明 P(n + 1) 提供了更多可用的信息。强归纳法通常用于处理递归定义,其中递归定义的量 f(n + 1) 依赖于多个前面的对象 f(1)、f(2)、…、f(n)。

The Strong Induction Proof Template

强归纳证明模板

We begin with an example of an attempted proof by ordinary induction where the induction hypothesis does not give us enough information to finish the proof.

我们首先举一个用普通归纳法进行证明的例子,其中归纳假设没有给我们足够的信息来完成证明。

4.21. Example. Define a sequence recursively by setting F 0 = 0, F 1 = 1, and F n = F n −1 + F n −2 for all integers n ≥ 2. (This is the famous Fibonacci sequence .)

4.21. 示例。递归地定义一个数列,令 F 0 = 0,F 1 = 1,且对于所有整数 n ≥ 2,F n = F n −1 + F n −2 。(这就是著名的斐波那契数列。)

Prove: n Z 0 , F n ≤ 2 n .

证明:∀n∈Z≥0,F n ≤ 2 n

Proof. We try a proof by induction on n . For the base case, we must prove F 0 ≤ 2 0 . Since F 0 = 0 and 2 0 = 1 and 0 ≤ 1, the base case is proved. Now fix an arbitrary integer n ≥ 0, assume F n ≤ 2 n , and try to prove F n +1 ≤ 2 n +1 . On one hand, if n = 0, we are proving F 1 ≤ 2 1 , which holds since F 1 = 1 and 2 1 = 2. On the other hand, if n > 0, then n + 1 ≥ 2, so the recursive definition of F n +1 tells us that F n +1 = F n + F n −1 . Here is where the ordinary induction proof gets stuck. We can use the induction hypothesis to bound the first term on the right side by 2 n . Similarly, we would like to bound the second term on the right side by writing F n −1 ≤ 2 n −1 . Unfortunately, our induction proof (as we have currently set it up) has not provided us with any assumptions about F n −1 . So the proof by induction cannot be completed.

证明。我们尝试对 n 进行归纳证明。对于基本情况,我们必须证明 F 0 ≤ 2 0 。由于 F 0 = 0 且 2 0 = 1 且 0 ≤ 1,基本情况得证。现在固定一个任意整数 n ≥ 0,假设 F n ≤ 2 n ,并尝试证明 F n +1 ≤ 2 n +1 。一方面,如果 n = 0,我们证明 F 1 ≤ 2 1 ,这成立,因为 F 1 = 1 且 2 1 = 2。另一方面,如果 n > 0,则 n + 1 ≥ 2,因此 F n +1 的递归定义告诉我们 F n +1 = F n + F n −1 。这就是普通归纳证明卡住的地方。我们可以利用归纳假设将右侧第一项限制在 2 n 以内。类似地,我们希望通过写出 F n −1 ≤ 2 n −1 来限制右侧第二项。遗憾的是,我们目前的归纳证明方法没有提供关于 F n −1 的任何假设。因此,归纳证明无法完成。

In the previous example, we could have finished the proof if we had been allowed to assume that P ( n − 1) and P ( n ) were both true, before trying to prove P ( n + 1). We now present a variation of induction, called strong induction or complete induction , that allows us to assume even more information as part of the induction hypothesis. These extra assumptions can make it easier to complete the induction step.

在前一个例子中,如果我们允许在证明 P(n + 1) 之前先假设 P(n − 1) 和 P(n) 都为真,我们就能完成证明。现在我们介绍一种归纳法的变体,称为强归纳法或完全归纳法,它允许我们在归纳假设中包含更多信息。这些额外的假设可以使归纳步骤更容易完成。

4.22. Strong Induction Theorem Let P ( n ) be any open sentence, and let b be a fixed integer. To prove n Z b , P ( n ) , we may instead prove this statement:

4.22. 强归纳定理 设 P(n) 为任意开语句,b 为固定整数。为了证明 ∀n∈Z≥b,P(n),我们可以改为证明以下命题:

For all integers n b , if (for all integers m such that b m < n , P ( m ) is true), then P ( n ) is true.

对于所有整数 n ≥ b,如果(对于所有整数 m 使得 b ≤ m < n,P(m) 为真),则 P(n) 为真。

We can recast this theorem as the following proof template for a strong induction proof.

我们可以将这个定理重新表述为以下强归纳证明的证明模板。

4.23. Strong Induction Proof Template To prove n Z b , P ( n ) by strong induction:

4.23. 强归纳法证明模板 要用强归纳法证明 ∀n∈Z≥b,P(n):

Fix an arbitrary integer n 0 b .

固定任意整数 n 0 ≥ b。

Assume: for all integers m in the range b m < n 0 , P ( m ) is true.

假设:对于所有在 b ≤ m < n 0 范围内的整数 m,P(m) 为真。

Prove P ( n 0 ) is true.

证明 P(n 0 ) 为真。

Five Remarks on Strong Induction

关于强归纳法的五点思考

(1) The Strong Induction Theorem can be proved from the (ordinary) Induction Axiom; we give a proof in an optional section later.

(1)强归纳定理可以从(普通的)归纳公理证明;我们将在后面的可选章节中给出证明。

(2) Here is intuition for why strong induction works. Intuitively, any induction proof starting at b proves the statements P ( b ), P ( b + 1), P ( b + 2), …, in this order. Let n 0 b be a fixed integer. When the proof reaches P ( n 0 ) and tries to prove it, we will have already successfully proved all preceding statements P ( b ), P ( b + 1), …, P ( n 0 − 1). So we might as well assume all those statements are true when attempting to prove P ( n 0 ). This is exactly the assumption in the proof template above; we call this assumption the (strong) induction hypothesis . The word “strong” (or “complete”) refers to the fact that we are allowed to assume a stronger statement in this proof template compared to the ordinary induction proof template.

(2) 以下是强归纳法有效的原因。直观地说,任何从 b 开始的归纳证明都会按顺序证明命题 P(b)、P(b + 1)、P(b + 2)、…。设 n 0 ≥ b 为一个固定的整数。当证明进行到 P(n 0 ) 并试图证明它时,我们已经成功证明了所有前面的命题 P(b)、P(b + 1)、…、P(n 0 − 1)。因此,在尝试证明 P(n 0 ) 时,我们可以假设所有这些命题都为真。这正是上述证明模板中的假设;我们将此假设称为(强)归纳假设。“强”(或“完备”)一词指的是,与普通归纳证明模板相比,我们可以在此证明模板中假设一个更强的命题。

This remark can be informally visualized using the domino analogy.

可以用多米诺骨牌的比喻来形象地说明这一点。

ufig4_3.jpg

Here P ( n ) is the statement “domino n falls down.” Suppose we are proving that all dominos fall down by induction. By the time we reach domino 6, we already know that the first 5 dominos have fallen down. So it is allowable to assume P ( 1 ) P ( 2 ) P ( 3 ) P ( 4 ) P ( 5 ) , not just P (5), to help us prove P (6).

这里 P(n) 表示“第 n 个多米诺骨牌倒下”。假设我们用归纳法证明所有多米诺骨牌都会倒下。当我们推导出第 6 个多米诺骨牌时,我们已经知道前 5 个多米诺骨牌已经倒下。因此,为了帮助我们证明 P(6),我们可以假设 P(1)∧P(2)∧P(3)∧P(4)∧P(5),而不仅仅是 P(5)。

(3) Now that strong induction is available, we could always use this proof technique instead of ordinary induction. However, ordinary induction proofs are conceptually simpler than strong induction proofs, so many people prefer ordinary induction for problems where this method succeeds. As a general rule, ordinary induction is applicable to recursive definitions where f ( n + 1) depends only on f ( n ); whereas strong induction is needed for more complicated recursive definitions.

(3) 既然有了强归纳法,我们当然可以用这种证明技巧来代替普通归纳法。然而,普通归纳法的证明在概念上比强归纳法更简单,因此许多人更倾向于在普通归纳法适用的情况下使用它。一般来说,普通归纳法适用于 f(n + 1) 仅依赖于 f(n) 的递归定义;而对于更复杂的递归定义,则需要使用强归纳法。

(4) The strong induction proof template does not specifically mention a base case, unlike the ordinary induction proof template. However, there is a hidden base case in strong induction. Consider the situation where the fixed integer n 0 happens to equal b . Then there are no integers m satisfying b m < n 0 , and so our strong induction hypothesis is not actually assuming anything for this value of n 0 . It follows that we need to prove P ( b ) without any prior knowledge, which is exactly what we must do in the base case of an ordinary induction proof.

(4) 与普通归纳证明模板不同,强归纳证明模板没有明确提及基本情况。然而,强归纳中却隐藏着一个基本情况。考虑固定整数 n 0 恰好等于 b 的情况。那么不存在满足 b ≤ m < n 0 的整数 m,因此,我们的强归纳假设实际上并未对 n 0 的值做任何假设。由此可知,我们需要在没有任何先验知识的情况下证明 P(b),这正是我们在普通归纳证明的基本情况下必须做的事情。

(5) In strong induction proofs, it often happens that several small values of n 0 (not just n 0 = b ) need to be treated in separate cases, which could be called the “base cases” of the strong induction proof. These special cases often correspond to the initial conditions in a recursive definition, as illustrated in the following examples.

(5) 在强归纳证明中,经常需要分别处理几个较小的 n 0 值(不仅仅是 n 0 = b),这些特殊情况可以称为强归纳证明的“基本情况”。这些特殊情况通常对应于递归定义中的初始条件,如下例所示。

Examples of Strong Induction Proofs

强归纳证明的例子

We begin by revisiting the failed proof above.

我们首先回顾一下上面失败的证明。

4.24. Example. Define F 0 = 0, F 1 = 1, and F n = F n −1 + F n −2 for all integers n ≥ 2.

4.24. 示例。定义 F 0 = 0,F 1 = 1,且对于所有整数 n ≥ 2,F n = F n −1 + F n −2

Prove: for all integers n ≥ 0, F n ≤ 2 n .

证明:对于所有整数 n ≥ 0,F n ≤ 2 n

Proof. We use strong induction on n . Fix an integer n ≥ 0. Assume that for all integers m in the range 0 ≤ m < n , F m ≤ 2 m . Prove F n ≤ 2 n . We know n = 0 or n = 1 or n ≥ 2, so use cases. [The previous statement was motivated by the recursive definition of F n , which has different cases for n = 0, n = 1, and n ≥ 2.]

证明。我们对 n 使用强归纳法。固定一个整数 n ≥ 0。假设对于所有 0 ≤ m < n 范围内的整数 m,F m ≤ 2 m 。证明 F n ≤ 2 n 。我们已知 n = 0 或 n = 1 或 n ≥ 2,因此可以使用这些情况。[前文的动机源于 F n 的递归定义,该定义对于 n = 0、n = 1 和 n ≥ 2 有不同的情况。]

Case 0. Assume n = 0; prove F 0 ≤ 2 0 . We know F 0 = 0 and 2 0 = 1 and 0 ≤ 1, so F 0 ≤ 2 0 .

情况 0. 假设 n = 0;证明 F 0 ≤ 2 0 。我们知道 F 0 = 0 且 2 0 = 1 且 0 ≤ 1,所以 F 0 ≤ 2 0

Case 1. Assume n = 1; prove F 1 ≤ 2 1 . We know F 1 = 1 and 2 1 = 2 and 1 ≤ 2, so F 1 ≤ 2 1 .

情况 1. 假设 n = 1;证明 F 1 ≤ 2 1 。我们知道 F 1 = 1 且 2 1 = 2 且 1 ≤ 2,所以 F 1 ≤ 2 1

Case 2. Assume n ≥ 2; prove F n ≤ 2 n . To start, we use the recursive definition to write F n = F n −1 + F n −2 . Now, 0 ≤ n − 1 < n , so the strong induction hypothesis (taking m = n − 1) tells us F n −1 ≤ 2 n −1 . Furthermore 0 ≤ n − 2 < n , so we also know F n −2 ≤ 2 n −2 by taking m = n − 2 in the induction hypothesis. Hence,

情况 2. 假设 n ≥ 2;证明 F n ≤ 2 n 。首先,我们使用递归定义写出 F n = F n −1 + F n −2 。现在,0 ≤ n − 1 < n,因此强归纳假设(取 m = n − 1)告诉我们 F n −1 ≤ 2 n −1 。此外,0 ≤ n − 2 < n,因此通过在归纳假设中取 m = n − 2,我们也知道 F n −2 ≤ 2 n −2 。因此,

F F n n = F F n n 1 + F F n n 2 2 n n 1 + 2 n n 2 = 2 n n ( 2 1 + 2 2 ) = 2 n n ( 3 / 4 ) < 2 n n ,

completing the proof of this case. The induction proof is now finished.

本案证明完毕。归纳证明到此结束。

4.25. Example. Define a 0 = 3, a 1 = 8, and a n = 4( a n −1 a n −2 ) for all integers n ≥ 2.

4.25. 示例。定义 a 0 = 3,a 1 = 8,以及 a n = 4(a n −1 − a n −2 ),其中 n 为所有整数 n ≥ 2。

Prove: for all integers n ≥ 0, a n = ( n + 3)2 n .

证明:对于所有整数 n ≥ 0,a n = (n + 3)2 n

Proof. We use strong induction on n . Fix an arbitrary integer n ≥ 0. Assume that for all m in the range 0 ≤ m < n , a m = ( m + 3)2 m . Prove a n = ( n + 3)2 n . We know n = 0 or n = 1 or n ≥ 2, so use cases. [This step was suggested by the cases in the recursive definition of a n .]

证明。我们对 n 使用强归纳法。固定任意整数 n ≥ 0。假设对于所有 0 ≤ m < n 范围内的 m,a m = (m + 3)2 m 。证明 a n = (n + 3)2 n 。我们知道 n = 0 或 n = 1 或 n ≥ 2,因此利用这些情况。[这一步是由 a n 的递归定义中的情况启发的。]

Case 0. Assume n = 0; prove a 0 = (0 + 3)2 0 . We know a 0 = 3 = (0 + 3)2 0 , so this case is proved.

情况 0. 假设 n = 0;证明 a 0 = (0 + 3)2 0 。我们知道 a 0 = 3 = (0 + 3)2 0 ,因此这种情况得到证明。

Case 1. Assume n = 1; prove a 1 = (1 + 3)2 1 . We know a 1 = 8 = 4 · 2 = (1 + 3)2 1 , so this case is proved.

情况 1. 假设 n = 1;证明 a 1 = (1 + 3)2 1 。我们知道 a 1 = 8 = 4 · 2 = (1 + 3)2 1 ,因此这种情况得证。

Case 2. Assume n ≥ 2; prove a n = ( n + 3)2 n . In this case, we have 0 ≤ n − 1 < n and 0 ≤ n − 2 < n , so the induction hypothesis tells us that a n −1 = ( n − 1 + 3)2 n −1 and a n −2 = ( n − 2 + 3)2 n −2 . Using the recursive definition and this information, we deduce

情况 2. 假设 n ≥ 2;证明 a n = (n + 3)2 n 。在这种情况下,我们有 0 ≤ n − 1 < n 且 0 ≤ n − 2 < n,因此归纳假设告诉我们 a n −1 = (n − 1 + 3)2 n −1 且 a n −2 = (n − 2 + 3)2 n −2 。利用递归定义和这些信息,我们可以推导出

a 一个 n n = 4 ( a 一个 n n 1 a 一个 n n 2 ) = 4 ( ( n n + 2 ) 2 n n 1 ( n n + 1 ) 2 n n 2 ) = 2 n n ( 2 ( n n + 2 ) ( n n + 1 ) ) = 2 n n ( n n + 3 ) .

So the result holds in all cases, completing the induction.

因此,该结论在所有情况下都成立,归纳推理完成。

4.26. Example. Define a 0 = 1 and a n = ( k = 0 n 1 a k ) + 1 for all n ≥ 1. Guess an explicit (non-recursive) formula for a n and then prove it.

4.26. 示例。定义 a 0 = 1 和 an=(∑k=0n−1ak)+1 对所有 n ≥ 1。猜出 a n 的一个显式(非递归)公式,然后证明它。

Solution. We use the recursive definition to compute a 0 = 1, a 1 = a 0 + 1 = 1 + 1 = 2, a 2 = a 0 + a 1 + 1 = 1 + 2 + 1 = 4, a 3 = a 0 + a 1 + a 2 + 1 = 1 + 2 + 4 + 1 = 8, a 4 = a 0 + a 1 + a 2 + a 3 + 1 = 1 + 2 + 4 + 8 + 1 = 16, and so on. It looks like a n = 2 n for all integers n ≥ 0. Since the recursive formula for a n involves all preceding values a k , we prove our guess by strong induction on n . Fix an arbitrary integer n ≥ 0. Assume: for all integers k in the range 0 ≤ k < n , a k = 2 k . Prove a n = 2 n . [The definition of a n breaks into two parts, suggesting the following two cases.] We know n = 0 or n > 0, so use cases.

解决方案。 我们使用递归定义来计算 a 0 = 1,a 1 = a 0 + 1 = 1 + 1 = 2,a 2 = a 0 + a 1 + 1 = 1 + 2 + 1 = 4,a 3 = a 0 + a 1 + a 2 + 1 = 1 + 2 + 4 + 1 = 8,a 4 = a 0 + a 1 + a 2 + a 3 + 1 = 1 + 2 + 4 + 8 + 1 = 16,以此类推。对于所有整数 n ≥ 0,a n = 2 n 似乎成立。由于 a n 的递归公式涉及所有前面的值 a k ,我们可以用强归纳法证明我们的猜测。固定一个任意整数 n ≥ 0。假设:对于所有 0 ≤ k < n 范围内的整数 k,a k = 2 k 。证明 a n = 2 n 。[a n 的定义分为两部分,暗示了以下两种情况。] 我们知道 n = 0 或 n > 0,因此使用这两种情况。

Case 1. Assume n = 0; prove a 0 = 2 0 . We know a 0 = 1 = 2 0 .

情况 1. 假设 n = 0;证明 a 0 = 2 0 。我们知道 a 0 = 1 = 2 0

Case 2. Assume n > 0; prove a n = 2 n . In this case, we know a n = ( k = 0 n 1 a k ) + 1 . Each term a k appearing in the sum has an index k in the range 0 ≤ k < n . So the strong induction hypothesis allows us to replace every term a k by 2 k in the sum defining a n . We now know a n = ( k = 0 n 1 2 k ) + 1 . By a previous theorem (proved in Example 4.10 by ordinary induction), the sum appearing here evaluates to 2 n −1+1 − 1 = 2 n − 1. So a n = (2 n − 1) + 1 = 2 n , completing the induction proof.

情况 2. 假设 n > 0;证明 a n = 2 n 。在这种情况下,我们知道 an=(∑k=0n−1ak)+1。求和式中出现的每一项 a k 的索引 k 都在 0 ≤ k < n 的范围内。因此,强归纳假设允许我们将定义 a n 的求和式中的每一项 a k 替换为 2 k 。现在我们知道 an=(∑k=0n−12k)+1。根据先前的定理(在例 4.10 中用普通归纳法证明),这里出现的和计算结果为 2 n −1+1 − 1 = 2 n − 1。因此 a n = (2 n − 1) + 1 = 2 n ,完成了归纳证明。

4.27. Example. Prove: n Z 18 , a Z 0 , b Z 0 , n = 7 a + 4 b . [This statement can be concretely interpreted as follows: for all n ≥ 18, we can pay n cents postage using a combination of 7-cent stamps and 4-cent stamps.]

4.27. 例。证明:∀n∈Z≥18,∃a∈Z≥0,∃b∈Z≥0,n=7a+4b。[此命题可具体解释如下:对于所有n≥18,我们可以使用7美分邮票和4美分邮票的组合支付n美分的邮资。]

Proof. We use strong induction on n . Fix an integer n ≥ 18. Assume: for all integers m in the range 18 ≤ m < n , c Z 0 , d Z 0 , m = 7 c + 4 d . Prove: a Z 0 , b Z 0 , n = 7 a + 4 b . [Note the use of non-conflicting letters for the existentially quantified variables in the induction hypothesis and the goal.] We know n = 18 or n = 19 or n = 20 or n = 21 or n ≥ 22, so use cases. [This is the tricky part — how did we know to introduce these particular cases? The reason will retroactively emerge at the end of the proof.]

证明。我们对 n 使用强归纳法。固定一个整数 n ≥ 18。假设:对于所有 18 ≤ m < n 范围内的整数 m,有 ∃c∈Z≥0,∃d∈Z≥0,m=7c+4d。证明:有 ∃a∈Z≥0,∃b∈Z≥0,n=7a+4b。[注意归纳假设和目标中存在量化变量使用了不冲突的字母。] 我们知道 n = 18 或 n = 19 或 n = 20 或 n = 21 或 n ≥ 22,因此使用特定情况。[这是棘手的部分——我们是如何知道要引入这些特定情况的?原因将在证明的最后部分追溯给出。]

Case 1. Assume n = 18. Choose a = 2 and b = 1, which are in Z 0 . By arithmetic, 7 a + 4 b = 7(2) + 4(1) = 14 + 4 = 18 = n .

情况 1. 假设 n = 18。选择 a = 2 和 b = 1,它们在 Z≥0 中。通过算术,7a + 4b = 7(2) + 4(1) = 14 + 4 = 18 = n。

Case 2. Assume n = 19. Choose a = 1 and b = 3, which are in Z 0 . By arithmetic, 7 a + 4 b = 7(1) + 4(3) = 7 + 12 = 19 = n .

情况 2. 假设 n = 19。选择 a = 1 和 b = 3,它们在 Z≥0 中。通过算术,7a + 4b = 7(1) + 4(3) = 7 + 12 = 19 = n。

Case 3. Assume n = 20. Choose a = 0 and b = 5, which are in Z 0 . By arithmetic, 7 a + 4 b = 7(0) + 4(5) = 0 + 20 = 20 = n .

情况 3. 假设 n = 20。选择 a = 0 和 b = 5,它们在 Z≥0 中。通过算术,7a + 4b = 7(0) + 4(5) = 0 + 20 = 20 = n。

Case 4. Assume n = 21. Choose a = 3 and b = 0, which are in Z 0 . By arithmetic, 7 a + 4 b = 7(3) + 4(0) = 21 + 0 = 21 = n . [Note that the induction hypothesis was not needed to complete the first four cases.]

情况 4. 假设 n = 21。选择 a = 3 和 b = 0,它们都在 Z≥0 中。通过算术运算,7a + 4b = 7(3) + 4(0) = 21 + 0 = 21 = n。[注意,前四个情况的证明并不需要归纳假设。]

Case 5. Assume n ≥ 22. The key observation is that for such an n , the integer m = n − 4 satisfies 18 ≤ m < n . So we can take m in the induction hypothesis to be n − 4. By definition of ALL, we know there exist c 0 Z 0 and d 0 Z 0 with n − 4 = 7 c 0 + 4 d 0 . Adding 4 to both sides, we get n = 7 c 0 + 4 d 0 + 4 = 7 c 0 + 4( d 0 + 1). Choose a = c 0 and b = d 0 + 1, which are both nonnegative integers by closure (since c 0 and d 0 are in Z 0 ). Then n = 7 a + 4 b holds, completing the proof of Case 5. [Now, at the end of the proof, we understand why the given cases were used. The point is that we can go from a solution for n − 4 to a solution for n by adding one more 4-cent stamp. But n − 4 falls outside the allowable range of m ’s when n is 18, 19, 20, or 21. So these values of n must be treated in separate base cases.]

情况 5. 假设 n ≥ 22。关键观察是,对于这样的 n,整数 m = n − 4 满足 18 ≤ m < n。因此,我们可以将归纳假设中的 m 取为 n − 4。根据 ALL 的定义,我们知道存在 c0∈Z≥0 和 d0∈Z≥0,使得 n − 4 = 7c 0 + 4d 0 。两边同时加 4,得到 n = 7c 0 + 4d 0 + 4 = 7c 0 + 4(d 0 + 1)。选择 a = c 0 和 b = d 0 + 1,根据闭包定理(因为 c 0 和 d 0 都在 Z≥0 中),它们都是非负整数。那么 n = 7a + 4b 成立,完成了情况 5 的证明。[现在,在证明的最后,我们理解了为什么使用给定的情况。关键在于,我们可以通过增加一张 4 分邮票,从 n − 4 的解过渡到 n 的解。但是,当 n 为 18、19、20 或 21 时,n − 4 超出了 m 的允许范围。因此,这些 n 值必须在单独的基本情况下处理。]

Proof of the Strong Induction Theorem (Optional)

强归纳定理的证明(可选)

We now prove the Strong Induction Theorem. Fix an integer b and an arbitrary open sentence P ( n ). Assume that we have proved the following statement:

现在我们来证明强归纳定理。固定一个整数 b 和一个任意的开语句 P(n)。假设我们已经证明了以下命题:

We must prove the statement n Z b , P ( n ) . The idea is to begin by proving a related statement by ordinary induction starting at b . Define a new open sentence Q ( n ), which is the statement “for all integers m in the range b m < n , P ( m ) is true.” We will prove n Z b , Q ( n ) by induction on n .

我们需要证明命题 ∀n∈Z≥b,P(n)。思路是先用归纳法证明一个相关的命题,从 b 开始。定义一个新的开语句 Q(n),即命题“对于所有在 b ≤ m < n 范围内的整数 m,P(m) 为真”。我们将用归纳法证明 ∀n∈Z≥b,Q(n)。

For the base case, we must prove the statement Q ( b ). This statement says “for all integers m in the range b m < b , P ( m ) is true.” There are no integers m satisfying b m < b , so this statement is vacuously true.

对于基本情况,我们必须证明命题 Q(b)。该命题表示“对于所有满足 b ≤ m < b 的整数 m,P(m) 为真”。不存在满足 b ≤ m < b 的整数 m,因此该命题显然为真。

For the induction step, fix an integer n b , assume Q ( n ) is true, and prove Q ( n + 1) is true. We have assumed the following induction hypothesis:

在归纳步骤中,固定一个整数 n ≥ b,假设 Q(n) 为真,并证明 Q(n + 1) 为真。我们假设了以下归纳前提:

We must prove:

我们必须证明:

Fix an integer m 0 , and assume b m 0 < n + 1. Since m 0 is an integer, we must have b m 0 < n or m 0 = n . In the case where b m 0 < n , the induction hypothesis and the Inference Rule for ALL tell us that P ( m 0 ) is true. In the case where m 0 = n , the induction hypothesis coincides with the hypothesis of the IF-statement that we assumed at the start of the proof. By the Inference Rule for IF, we see that P ( n ) is true. Because m 0 = n , P ( m 0 ) is true in this case as well.

固定一个整数 m 0 ,并假设 b ≤ m 0 < n + 1。由于 m 0 是一个整数,因此必定有 b ≤ m 0 < n 或 m 0 = n。当 b ≤ m 0 < n 时,归纳假设和所有情况的推理规则表明 P(m 0 ) 为真。当 m 0 = n 时,归纳假设与我们在证明开始时假设的 IF 语句的假设一致。根据 IF 语句的推理规则,我们可以看出 P(n) 为真。因为 m 0 = n,所以 P(m 0 ) 在这种情况下也成立。

We have completed the induction proof of the statement n Z b , Q ( n ) . We use this to prove n Z b , P ( n ) . Fix an arbitrary integer n b . Then n + 1 is also an integer in Z b , so Q ( n + 1) is true. So for all integers m in the range b m < n + 1, P ( m ) is true. Since n is an integer satisfying b n < n + 1, we can take m = n to conclude that P ( n ) is true, as needed.

我们已经完成了命题 ∀n∈Z≥b,Q(n) 的归纳证明。我们利用这个证明来证明 ∀n∈Z≥b,P(n)。固定任意整数 n ≥ b。那么 n + 1 也是 Z≥b 中的一个整数,所以 Q(n + 1) 为真。因此,对于所有满足 b ≤ m < n + 1 的整数 m,P(m) 为真。由于 n 是满足 b ≤ n < n + 1 的整数,我们可以取 m = n 来得出 P(n) 为真,这正是我们需要的结论。

Section Summary

章节概要

  1. Strong Induction. To prove n Z b , P ( n ) by strong induction:

    强归纳法。用强归纳法证明对于所有 n∈Z≥b,P(n):

    Step 1. Fix an arbitrary integer n b .

    步骤 1. 固定任意整数 n ≥ b。

    Step 2a. Assume that for all integers m in the range b m < n , P ( m ) is true.

    步骤 2a. 假设对于所有在 b ≤ m < n 范围内的整数 m,P(m) 为真。

    Step 2b. Aided by this assumption, prove P ( n ) is true.

    步骤 2b. 借助此假设,证明 P(n) 为真。

  2. Remarks on Induction Proofs. Ordinary induction can be used when the truth of P ( n + 1) can be deduced assuming only that the previous statement P ( n ) is true. Strong induction is needed when the truth of P ( n + 1) depends on several of the preceding statements P ( n ), P ( n − 1), …. Both types of induction proofs can be justified using the ordinary Induction Axiom. Although the strong induction proof template does not specifically mention a base case, we often need to treat small values of n as separate cases. The required cases in the proof are often based on corresponding cases in the recursive definitions of concepts appearing in the statement P ( n ).
    关于归纳证明的说明。当仅假设前一个命题 P(n) 为真即可推导出 P(n + 1) 的真值时,可以使用普通归纳法。当 P(n + 1) 的真值取决于多个前一个命题 P(n)、P(n − 1)、… 时,则需要使用强归纳法。这两种归纳证明都可以用普通归纳公理来证明。虽然强归纳证明模板没有明确提及基本情况,但我们通常需要将较小的 n 值视为单独的情况。证明中所需的情况通常基于命题 P(n) 中出现的概念的递归定义中的对应情况。

Exercises

练习

  1. Recursively define a 0 = −1, a 1 = 1, and a n = 8 a n −1 − 15 a n −2 for all n Z 2 . Prove: for all n Z 0 , a n = 2 · 5 n − 3 n +1 .
    递归定义 a 0 = −1,a 1 = 1,以及 a n = 8a n −1 − 15a n −2 ,对于所有 n∈Z≥2。证明:对于所有 n∈Z≥0,a n = 2 · 5 n − 3 n +1
  2. Define a sequence recursively by setting a 1 = 2, a 2 = 1, and a n = 2 a n −1 a n −2 for all integers n ≥ 3. (a) Compute a n for 1 ≤ a n ≤ 6, and then guess an explicit (non-recursive) formula for a n . (b) Use strong induction to prove your formula for a n holds for all integers n ≥ 1.
    递归地定义一个序列,令 a 1 = 2,a 2 = 1,且对于所有整数 n ≥ 3,a n = 2a n −1 − a n −2 。(a)计算 1 ≤ a n ≤ 6 的 a n ,然后猜测 a n 的一个显式(非递归)公式。(b)使用强归纳法证明你给出的 a n 公式对于所有整数 n ≥ 1 都成立。
  3. Define a 0 = 4, a 1 = 32, and a n = 12 a n −1 − 20 a n −2 for all integers n ≥ 2. Guess a non-recursive formula for a n (valid for all n ≥ 0), and prove your guess by strong induction.
    定义 a 0 = 4,a 1 = 32,以及 a n = 12a n −1 − 20a n −2 ,其中 n ≥ 2 为整数。猜测 a n 的一个非递归公式(对所有 n ≥ 0 都有效),并用强归纳法证明你的猜测。
  4. Use ordinary induction or strong induction to prove the following summation formulas for Fibonacci numbers (see Example 4.21), valid for all integers n ≥ 0. (a) k = 0 n F k = F n + 2 1 . (b) k = 0 n 1 F 2 k + 1 = F 2 n . (c) k = 0 n F 2 k = F 2 n + 1 1 .
    使用普通归纳法或强归纳法证明以下斐波那契数的求和公式(参见例 4.21),对于所有整数 n ≥ 0 都有效。 (a) ∑k=0nFk=Fn+2−1. (b) ∑k=0n−1F2k+1=F2n. (c) ∑k=0nF2k=F2n+1−1.
  5. Prove by strong induction: n Z 11 , a Z > 0 , b Z > 0 , n = 2 a + 5 b .
    用强归纳法证明:∀n∈Z≥11,∃a∈Z>0,∃b∈Z>0,n=2a+5b。
  6. Consider this statement: n Z n 0 , a Z 0 , b Z 0 , n = 12 a + 5 b . Find the smallest value of n 0 for which the statement is true, and then prove it.
    考虑以下命题:∀n∈Z≥n0,∃a∈Z≥0,∃b∈Z≥0,n=12a+5b。求使命题成立的最小n值,并证明该命题成立。
  7. Define a 0 = 1, a 1 = 3, a 2 = 33, and a n = 7 a n −1 − 8 a n −2 − 16 a n −3 for all n Z 3 . Prove: for all n Z 0 , a n = ( − 1) n + n 4 n .
    定义 a 0 = 1,a 1 = 3,a 2 = 33,且对于所有 n∈Z≥3,a n = 7a n −1 − 8a n −2 − 16a n −3 。证明:对于所有 n∈Z≥0,a n = ( − 1) n + n4 n
  8. Define a 0 = 1, a 1 = 2, a 2 = 5, and a n = 3 a n −1 − 3 a n −2 + a n −3 for all n Z 3 . Guess an explicit formula for a n and prove your formula holds for all n Z 0 .
    定义 a 0 = 1,a 1 = 2,a 2 = 5,且对于所有 n∈Z≥3,a n = 3a n −1 − 3a n −2 + a n −3 。猜测 a n 的一个显式公式,并证明该公式对于所有 n∈Z≥0 都成立。
  9. Define b 1 = 1, b 2 = 2, b 3 = 3, and b n = b n −1 + b n −2 + b n −3 for all n ≥ 4. Prove by strong induction: for all n Z 1 , b n < 2 n .
    定义 b 1 = 1,b 2 = 2,b 3 = 3,且对于所有 n ≥ 4,b n = b n −1 + b n −2 + b n −3 。用强归纳法证明:对于所有 n∈Z≥1,b n < 2 n
  10. Let F n be the Fibonacci sequence defined in Example 4.21. Find and prove a formula for F n 2 F n + 1 F n 1 , valid for all n Z 1 .
    设 F n 为例 4.21 中定义的斐波那契数列。求并证明 Fn2−Fn+1Fn−1 的公式,该公式对所有 n∈Z≥1 都成立。
  11. Prove the following exact formula for the Fibonacci numbers F n (see Example 4.21): for all n Z 0 , F n = [ ( 1 + 5 ) n ( 1 5 ) n ] / ( 2 n 5 ) .
    证明斐波那契数列 F n 的以下精确公式(参见例 4.21):对于所有 n∈Z≥0,Fn=[(1+5)n−(1−5)n]/(2n5)。
  12. In Example 4.26, prove a n = 2 n for all integers n ≥ 0 by ordinary induction (not strong induction).
    在例 4.26 中,用普通归纳法(而不是强归纳法)证明对于所有整数 n ≥ 0,a n = 2 n
  13. Define c 0 = 1 and c n + 1 = 3 k = 0 n c k for all integers n ≥ 0. Guess an explicit formula for c n and then prove it.
    定义 c 0 = 1 且 cn+1=3∏k=0nck 对于所有整数 n ≥ 0。猜测 c n 的一个显式公式,然后证明它。
  14. Define d 0 = 1 and d n + 1 = k = 0 n 2 k d k for all n Z 0 . Prove: for all integers n ≥ 2, d n = k = 1 n 1 ( 2 k + 1 ) .
    定义 d 0 = 1 且 dn+1=∑k=0n2kdk 对于所有 n∈Z≥0。证明:对于所有整数 n ≥ 2,dn=∏k=1n−1(2k+1)。
  15. Give just the structural outline of a proof by strong induction of this statement:
    n Z 1 , a Z 0 , b Z 0 , c Z 0 , d Z 0 , n = a 2 + b 2 + c 2 + d 2 .

    仅给出以下命题的强归纳证明的结构概要:∀n∈Z≥1,∃a∈Z≥0,∃b∈Z≥0,∃c∈Z≥0,∃d∈Z≥0,n=a2+b2+c2+d2。
  16. (a) Prove that for every integer n ≥ 0, there exists k Z such that n = 4 k or n = 4 k + 1 or n = 4 k + 2 or n = 4 k + 3. (b) Prove that the result in (a) also holds for all negative integers n .
    (a)证明对于每个整数 n ≥ 0,存在 k∈Z,使得 n = 4k 或 n = 4k + 1 或 n = 4k + 2 或 n = 4k + 3。(b)证明(a)中的结果对于所有负整数 n 也成立。
  17. Using the theorem that every integer is even or odd, prove by strong induction: for all n Z 1 , there exist k Z 1 and distinct exponents e 1 , e 2 , , e k Z 0 such that n = 2 e 1 + 2 e 2 + + 2 e k .
    利用每个整数都是偶数或奇数的定理,用强归纳法证明:对于所有 n∈Z≥1,存在 k∈Z≥1 和不同的指数 e1,e2,…,ek∈Z≥0,使得 n=2e1+2e2+⋯+2ek。
  18. (a) Prove: for all n Z 3 , F n + 1 < 2 F n , where F n is a Fibonacci number (Example 4.21). (b) Use strong induction to prove that every positive integer n can be expressed as a sum of one or more distinct Fibonacci numbers. Suggestion: Consider the largest Fibonacci number not exceeding n .
    (a) 证明:对于所有 n∈Z≥3,Fn+1<2Fn,其中 F n 是一个斐波那契数(例 4.21)。(b) 使用强归纳法证明每个正整数 n 都可以表示为一个或多个不同的斐波那契数之和。提示:考虑不大于 n 的最大斐波那契数。
  19. Consider the following template for proving n Z 0 , P ( n ) : Step 1. Prove P (0). Prove P (1). Step 2. Fix n Z 0 . Assume P ( n ) and P ( n + 1). Prove P ( n + 2). Prove the correctness of this template using ordinary induction (not strong induction).
    考虑以下证明 ∀n∈Z≥0,P(n) 的模板: 步骤 1. 证明 P(0)。证明 P(1)。 步骤 2. 固定 n∈Z≥0。假设 P(n) 和 P(n + 1)。证明 P(n + 2)。 使用普通归纳法(而非强归纳法)证明此模板的正确性。
  20. (a) Create a strong induction proof template to prove statements of the form n Z c , Q ( n ) . (b) Use the Strong Induction Theorem 4.22 to prove that your template in (a) works.
    (a) 创建一个强归纳证明模板,用于证明形如 ∀n∈Z≤c,Q(n) 的命题。(b) 使用强归纳定理 4.22 证明 (a) 中的模板有效。
  21. Let P ( x ) be the open sentence “0 ≤ x ≤ 1,” where x is a real variable. (a) Disprove: x R 0 , P ( x ) . (b) Prove: For all x R 0 , if (for all real numbers y in the range 0 ≤ y < x , P ( y ) is true) then P ( x ) is true. (c) Conclude that the Strong Induction Theorem 4.22 is not valid if we replace integers by real numbers.
    设 P(x) 为开语句“0 ≤ x ≤ 1”,其中 x 为实变量。(a) 反证:∀x∈R≥0,P(x) 成立。(b) 证明:对于所有 x∈R≥0,如果对于所有实数 y(0 ≤ y < x),P(y) 为真,则 P(x) 为真。(c) 由此得出结论:如果将整数替换为实数,强归纳定理 4.22 不成立。
  22. Suppose we know: “for all n Z , if (for all integers m < n , P ( m ) is true), then P ( n ) is true.” Must it follow that “ n Z , P ( n ) ” is true? Explain.
    假设我们知道:“对于所有 n∈Z,如果(对于所有小于 n 的整数 m,P(m) 为真),则 P(n) 为真。” 是否必然得出“∀n∈Z,P(n)”为真?请解释。
  23. Suppose we know that every nonempty subset of Z 1 has a least element (see Exercise 18 in § 4.1 ). Use this fact to prove the Strong Induction Theorem 4.22, taking b = 1. [ Suggestion: Let S be the set of positive integers n such that P ( n ) is false; prove S = .]
    假设我们已知 Z≥1 的每个非空子集都存在最小元素(参见 §4.1 中的练习 18)。利用这一事实证明强归纳定理 4.22,取 b = 1。[提示:设 S 为满足 P(n) 为假的正整数 n 的集合;证明 S=∅。]

4.4 Prime Numbers and Integer Division

4.4 质数与整数除法

Now that we have strong induction, we can investigate the divisibility properties of the positive integers more closely. We begin by defining prime numbers and showing that every integer n > 1 can be written as a product of primes. Then we study integer division, which is a process for dividing an integer a by a nonzero integer b to produce a unique quotient q and remainder r .

现在我们有了强归纳法,就可以更深入地研究正整数的整除性质了。首先,我们定义素数,并证明每个大于 1 的整数 n 都可以表示为素数的乘积。然后,我们研究整数除法,即用非零整数 b 除以整数 a,得到唯一的商 q 和余数 r 的过程。

Primes and Prime Factorizations

质数和质因数分解

A composite number is a positive integer that can be expressed as the product of two smaller positive integers, whereas a prime number cannot be written in this way. Formally, we have the following definition.

合数是可以表示为两个较小正整数乘积的正整数,而质数则不能这样表示。正式的定义如下。

#

4.28. Definition: Prime and Composite Integers.

4.28. 定义:质数和合数。

(a) For all n Z > 1 , n is composite iff a Z , b Z , ( 1 < a < n ) ( 1 < b < n ) n = a b .

(a)对于所有 n∈Z>1,nis 是复合的当且仅当 ∃a∈Z,∃b∈Z,(1<a<n)∧(1<b<n)∧n=ab。

(b) For all n Z > 1 , n is prime iff n is not composite .

(b)对于所有 n∈Z>1,nis 为素数当且仅当 nis 不是合数。

Note that we give these definitions only for integers n > 1. By convention, 1 is not prime and 1 is not composite . Similarly, 0 is neither prime nor composite. It can be proved from the above definition that an integer p > 1 is prime iff the only positive divisors of p are 1 and p . The first few primes are

注意,我们仅对大于 1 的整数 n 给出这些定义。按照惯例,1 不是素数,1 也不是合数。类似地,0 既不是素数也不是合数。根据上述定义可以证明,大于 1 的整数 p 是素数当且仅当 p 的正因子只有 1 和 p。前几个素数是

2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , .

Our next theorem states that every integer in Z > 1 can be factored into primes.

我们的下一个定理指出,Z>1 中的每个整数都可以分解成素数。

4.29. Theorem on Existence of Prime Factorizations. Every integer n > 1 can be written as a product of one or more primes. More precisely, for all n Z > 1 , there exist k Z 1 and primes p 1 , p 2 , …, p k with n = p 1 p 2 · · · p k . (The primes appearing here need not be distinct.)

4.29. 素因子分解存在性定理。每个大于 1 的整数 n 都可以表示为一个或多个素数的乘积。更准确地说,对于所有 n∈Z>1,存在 k∈Z≥1 和素数 p 1 , p 2 , …, p k ,使得 n = p 1 p 2 · · · p k 。(这里出现的素数不必互不相同。)

Proof. We use strong induction on n . Fix an arbitrary integer n Z > 1 . Assume: for all integers m in the range 1 < m < n , m can be written as a product of one or more primes. Prove n can be written as a product of one or more primes. We know n is prime or n is not prime, so use cases.

证明。我们对 n 使用强归纳法。固定任意整数 n∈Z>1。假设:对于所有 1 < m < n 范围内的整数 m,m 都可以表示为一个或多个素数的乘积。证明 n 也可以表示为一个或多个素数的乘积。我们知道 n 要么是素数,要么不是素数,因此使用不同的情况。

Case 1. Assume n is prime. Then n is a product of one prime, namely itself. (In other words, we can choose k = 1 and p 1 = n above, to obtain n = p 1 .)

情况 1. 假设 n 是素数。那么 n 就是一个素数的乘积,即它本身。(换句话说,我们可以选择 k = 1 和 p 1 = n,从而得到 n = p 1 。)

Case 2. Assume n is not prime. Then n is composite, so there exist a , b Z with 1 < a < n and 1 < b < n and n = ab . Since a and b are both larger than 1 and less than n , we can apply the induction hypothesis to both a and b to see that a is a product of one or more primes, and b is a product of one or more primes. Say a = p 1 p 2 · · · p k and b = q 1 q 2 · · · q m where every p i and q j is prime. Then n = ab = p 1 p 2 · · · p k q 1 q 2 · · · q m , which expresses n as a product of k + m primes.

情况 2. 假设 n 不是素数。那么 n 是合数,所以存在 a, b∈Z,满足 1 < a < n 且 1 < b < n,并且 n = ab。由于 a 和 b 都大于 1 且小于 n,我们可以对 a 和 b 应用归纳假设,得出 a 是一个或多个素数的乘积,b 也是一个或多个素数的乘积。例如,a = p 1 p 2 · · · p k ,b = q 1 q 2 · · · q m ,其中每个 p i 和 q j 都是素数。那么 n = ab = p 1 p 2 · · · p k q 1 q 2 · · · q m ,这表示 n 是 k + m 个素数的乘积。

Later, we strengthen this theorem by proving that the prime factorization of n is unique in a certain respect. More specifically, if we have written n = p 1 p 2 · · · p k and n = q 1 q 2 · · · q m where all p i and q j are prime, then we must have k = m and q 1 , …, q k is a reordering of p 1 , …, p k . For now, we use the existence of prime factorizations to prove a famous theorem of Euclid about prime numbers.

稍后,我们将通过证明 n 的素因数分解在某种意义上是唯一的来加强这个定理。更具体地说,如果我们记 n = p 1 p 2 · · · p k 和 n = q 1 q 2 · · · q m ,其中所有 p i 和 q j 都是素数,那么我们必然有 k = m,并且 q 1 , …, q k 是 p 1 , …, p k 的重排。现在,我们利用素因数分解的存在性来证明欧几里得关于素数的一个著名定理。

4.30. Theorem on Infinitude of Primes There are infinitely many prime integers. More precisely, for any k Z 1 and any finite list of primes p 1 , p 2 , …, p k , there exists a prime q different from every p i .

4.30. 素数无穷定理 素数有无穷多个。更准确地说,对于任意 k∈Z≥1 和任意有限素数列表 p 1 , p 2 , …, p k ,存在一个素数 q 与每个 p i 都不同。

Proof. Fix k Z 1 and primes p 1 , p 2 , …, p k . Consider the integer n = 1 + r = 1 k p r . We know n is prime or n is not prime.

证明。固定 k∈Z≥1 和素数 p 1 , p 2 , …, p k 。考虑整数 n=1+∏r=1kpr。我们知道 n 是素数或 n 不是素数。

Case 1. Assume n is prime. Since n is larger than every p i , n is a prime number distinct from all p i .

情况 1. 假设 n 是素数。由于 n 大于所有 p i ,因此 n 是一个素数,且与所有 p i 都不同。

Case 2. Assume n is not prime. Then we can write n as a product of primes, say n = q 1 q 2 · · · q k with all q j prime. We prove q 1 is a prime different from every p i , using proof by contradiction. Assume, to get a contradiction, that there exists i with q 1 = p i . Then p i divides n since n = q 1 q 2 · · · q k = p i ( q 2 · · · q k ). Also p i divides r = 1 k p r . It follows that p i divides 1 = n r = 1 k p r . This is impossible, since p i > 1. This contradiction completes the proof of Case 2.

情况 2. 假设 n 不是素数。那么我们可以将 n 表示为素数的乘积,例如 n = q 1 q 2 · · · q k ,其中所有 q j 都是素数。我们用反证法证明 q 1 是一个素数,且与每个 p i 都不同。为了得到矛盾,假设存在 i 使得 q 1 = p i 。那么,p i 整除 n,因为 n = q 1 q 2 · · · q k = p i (q 2 · · · q k )。此外,p i 整除 ∏r=1kpr。由此可知,p i 整除 1=n−∏r=1kpr。这是不可能的,因为 p i > 1。这个矛盾完成了情形 2 的证明。

Integer Division

整数除法

In grade school, we learn to divide one integer by another, producing a quotient and a remainder . For example, if we divide 100 by 7, the quotient is 14 and the remainder is 2. One way to present this information is by writing 100/7 = 14 + (2/7). When studying integers, it is convenient to avoid fractions by multiplying this equation by the denominator 7. We obtain the integer equation 100 = 7 · 14 + 2.

在小学,我们学习用一个整数除以另一个整数,得到商和余数。例如,100 除以 7,商是 14,余数是 2。一种表示方法是写成 100/7 = 14 + (2/7)。学习整数时,为了避免分数,我们可以将这个等式乘以分母 7。这样就得到了整数等式 100 = 7 × 14 + 2。

In general, we want to be able to divide any nonnegative integer a by any positive integer b to produce a quotient q and a remainder r . In other words, given a ≥ 0 and b > 0, we seek integers q ≥ 0 and r ≥ 0 satisfying a / b = q + ( r / b ), or equivalently, a = bq + r . The remainder r should be small in the sense that r / b should be less than 1, which holds iff 0 ≤ r < b . Our next result shows that this division process is always possible. To motivate the proof, consider the following example.

一般来说,我们希望能够用任意非负整数 a 除以任意正整数 b,得到商 q 和余数 r。换句话说,给定 a ≥ 0 和 b > 0,我们寻找满足 a/b = q + (r/b) 的整数 q ≥ 0 和 r ≥ 0,或者等价地,a = bq + r。余数 r 应该很小,即 r/b 应该小于 1,这当且仅当 0 ≤ r < b。我们的下一个结果表明,这种除法过程总是可能的。为了更好地理解证明,请考虑以下示例。

4.31. Example. Let a = 25 and b = 7; find the quotient and remainder for this a and b . Solution. The answer can be found by the long division algorithm learned in school. But here is a more basic solution relying on the idea that we can perform a division by repeated subtraction. Start with 25 and subtract 7 repeatedly. First, 25 − 7 = 18. Second, 18 − 7 = 11. Third, 11 − 7 = 4. The current value 4 is less than 7, so we have found the remainder r = 4. We subtracted 7 three times to reach the remainder, so the quotient is q = 3.

4.31. 例题。设 a = 25,b = 7;求 a 和 b 的商和余数。解:答案可以用学校里学过的长除法算法求得。但这里提供一个更简单的解法,利用重复减法的思想。从 25 开始,重复减去 7。首先,25 − 7 = 18。其次,18 − 7 = 11。再次,11 − 7 = 4。当前值 4 小于 7,所以我们求得余数 r = 4。我们减去 7 三次才得到余数,所以商为 q = 3。

The idea of the induction proof below is to recast this iterative computation in a recursive way. Specifically, in a single subtraction, we pass from the initial input 25 to 25 − 7 = 18. We could recursively calculate the quotient and remainder when 18 is divided by 7 (namely, q 1 = 2 and r 1 = 4), and then obtain the quotient and remainder for 25 by adding 1 to the quotient (giving q = 3) and keeping the old remainder (giving r = 4). The initial condition for the recursive calculation occurs when the initial number a is already less than b . Then we can take q = 0 and r = a . For example, 4 divided by 7 has quotient 0 and remainder 4.

以下归纳证明的思路是将这种迭代计算转化为递归计算。具体来说,在一次减法运算中,我们从初始输入 25 过渡到 25 − 7 = 18。我们可以递归地计算 18 除以 7 的商和余数(即 q = 2 和 r = 4),然后通过将商加 1(得到 q = 3)并保留原余数(得到 r = 4)来得到 25 的商和余数。递归计算的初始条件是初始数 a 小于 b。此时我们可以取 q = 0 和 r = a。例如,4 除以 7 的商为 0,余数为 4。

Let us see how the intuition in the preceding example translates into a formal proof.

让我们看看前面例子中的直觉是如何转化为正式证明的。

4.32. Integer Division Theorem (Preliminary Version) For all integers a ≥ 0 and b > 0, there exist integers q , r with a = bq + r and 0 ≤ r < b .

4.32. 整数除法定理(初步版本)对于所有整数 a ≥ 0 和 b > 0,存在整数 q、r,使得 a = bq + r 且 0 ≤ r < b。

Proof. To better understand how the induction proof handles quantifiers, we write the statement to be proved in formal symbols as follows:

证明。为了更好地理解归纳证明如何处理量词,我们将待证明的语句用形式符号写成如下形式:

b b Z Z > 0 , a 一个 Z Z 0 , q q Z Z , r r Z Z , a 一个 = b b q q + r r ( 0 r r < b b ) .

Fix b Z > 0 . We prove the inner statement by strong induction on a . Fix an arbitrary integer a Z 0 . Assume: for all integers c in the range 0 ≤ c < a ,

固定 b∈Z>0。我们用强归纳法证明内层语句。固定任意整数 a∈Z≥0。假设:对于所有整数 c,其值在 0 ≤ c < a 范围内,

q q 1 Z Z , r r 1 Z Z , c c = b b q q 1 + r r 1 and 0 r r 1 < b b .

We must prove:

我们必须证明:

q q Z Z , r r Z Z , a 一个 = b b q q + r r and 0 r r < b b .

We know a < b or a b , so use cases.

我们知道 a < b 或 a ≥ b,因此有用例。

Case 1. Assume a < b . Choose q = 0 and r = a . We know q , r Z 0 , a = bq + r (since a = b · 0 + a ) and 0 ≤ a < b , so our choice proves the existence statement.

情况 1. 假设 a < b。选择 q = 0 和 r = a。我们知道 q,r∈Z≥0,a = bq + r(因为 a = b · 0 + a)且 0 ≤ a < b,因此我们的选择证明了存在性陈述。

Case 2. Assume a b . In the induction hypothesis, let us take c = a b . This is a legitimate value of c , since c = a b ≥ 0 and c = a b < a (because b > 0). Now, the induction hypothesis gives us integers q 1 , r 1 ≥ 0 with c = bq 1 + r 1 and 0 ≤ r 1 < b . Adding b to both sides, a = c + b = bq 1 + r 1 + b = b ( q 1 + 1) + r 1 . So, choosing q = q 1 + 1 and r = r 1 (which are in Z 0 ), we have a = bq + r and 0 ≤ r < b .

情况 2. 假设 a ≥ b。在归纳假设中,我们取 c = a − b。这是一个合理的 c 值,因为 c = a − b ≥ 0 且 c = a − b < a(因为 b > 0)。现在,归纳假设给出整数 q 1 , r 1 ≥ 0,其中 c = bq 1 + r 1 且 0 ≤ r 1 < b。两边同时加上 b,得到 a = c + b = bq 1 + r 1 + b = b(q 1 + 1) + r 1 。因此,选择 q = q 1 + 1 和 r = r 1 (它们在 Z≥0 中),我们有 a = bq + r 和 0 ≤ r < b。

Integer Division for Negative Integers (Optional)

负整数除法(可选)

We now improve the Integer Division Theorem in two ways. First, we extend the result to the case where a and b might be negative integers (although b can never be zero). Second, we prove that the quotient and remainder are uniquely determined by a and b . The next several lemmas deal with these issues. We begin by allowing the dividend a to be negative; in this case, the quotient q might be negative.

我们现在从两个方面改进整数除法定理。首先,我们将结果推广到 a 和 b 可以为负整数的情况(尽管 b 永远不能为零)。其次,我们证明商和余数由 a 和 b 唯一确定。接下来的几个引理将讨论这些问题。我们首先允许被除数 a 为负数;在这种情况下,商 q 也可能为负数。

4.33. Lemma: Extension of Integer Division to Negative a.

4.33. 引理:整数除法推广到负数 a。

b b Z Z > 0 , a 一个 Z Z , q q Z Z , r r Z Z , a 一个 = b b q q + r r and 0 r r < b b .

Proof. Fix b Z > 0 . Fix a Z . We know a ≥ 0 or a < 0, so use cases.

证明。固定 b∈Z>0。固定 a∈Z。我们知道 a ≥ 0 或 a < 0,因此有用例。

Case 1. Assume a ≥ 0. Then the needed result follows from the previous theorem.

情况 1. 假设 a ≥ 0。则所需的结果可由前面的定理得出。

Case 2. Assume a < 0. Then − a > 0, so the previous theorem says there exist q 2 , r 2 Z with − a = bq 2 + r 2 and 0 ≤ r 2 < b . Now, a = − bq 2 r 2 . [We are tempted to choose q = − q 2 and r = − r 2 , but this produces a negative remainder most of the time. This suggests the following subcases.] We know r 2 = 0 or r 2 > 0, so use cases.

情况 2. 假设 a < 0。则 −a > 0,因此根据之前的定理,存在 q2,r2∈Z,使得 −a = bq 2 + r 2 且 0 ≤ r 2 < b。现在,a = −bq 2 − r 2 。[我们倾向于选择 q = −q 2 和 r = −r 2 ,但这在大多数情况下会产生负余数。这提示了以下子情况。] 我们知道 r 2 = 0 或 r 2 > 0,因此存在以下用例。

Case 2a. Assume r 2 = 0. Then a = b ( − q 2 ) + 0, so we can choose q = − q 2 and r = 0.

情况 2a. 假设 r 2 = 0。则 a = b( − q 2 ) + 0,因此我们可以选择 q = −q 2 和 r = 0。

Case 2b. Assume r 2 > 0. Then − b < − r 2 < 0, so (adding b ) 0 < b r 2 < b . So we can manipulate a = − bq 2 r 2 as follows: we know

情况 2b。假设 r 2 > 0。则 −b < −r 2 < 0,因此(加上 b)0 < b − r 2 < b。所以我们可以对 a = −bq 2 − r 2 进行如下运算:我们知道

a 一个 = b b q q 2 r r 2 = b b q q 2 b b + b b r r 2 = b b ( q q 2 1 ) + ( b b r r 2 ) .

Now choose q = q 2 1 Z and r = b r 2 Z ; observe that 0 ≤ r < b for this choice of r .

现在选择 q=−q2−1∈Z 和 r=b−r2∈Z;观察可知,对于此 r 的选择,0 ≤ r < b。

4.34. Example. To illustrate the above proof, first suppose a = −21 and b = 7. Then − a = 21 = 7 · 3 + 0 (so q 2 = 3, r 2 = 0), and we see that a = −21 = 7 · ( − 3) + 0 (so q = −3, r = 0). On the other hand, suppose a = −26 and b = 7. Then − a = 26 = 7 · 3 + 5 (so q 2 = 3, r 2 = 5). Negating gives a = −26 = 7 · ( − 3) − 5, which has a forbidden negative remainder. We fix this by writing a = −26 = 7 · ( − 3) − 7 + 7 − 5 = 7 · ( − 4) + 2, so q = −4 = − q 2 − 1 and r = 2 = b r 2 , in accordance with the proof above.

4.34. 示例。为了说明上述证明,首先假设 a = −21 且 b = 7。那么 −a = 21 = 7 · 3 + 0(因此 q 2 = 3,r 2 = 0),由此可知 a = −21 = 7 · ( − 3) + 0(因此 q = −3,r = 0)。另一方面,假设 a = −26 且 b = 7。那么 −a = 26 = 7 · 3 + 5(因此 q 2 = 3,r 2 = 5)。取反后得到 a = −26 = 7 · ( − 3) − 5,余数为负数,这是不允许的。我们通过令 a = −26 = 7 · ( − 3) − 7 + 7 − 5 = 7 · ( − 4) + 2 来解决这个问题,因此 q = −4 = −q 2 − 1 和 r = 2 = b − r 2 ,根据上面的证明。

For the next extension, we allow the divisor b to be negative. We still want nonnegative remainders, so we now require r to satisfy 0 ≤ r < | b |.

接下来,我们允许除数 b 为负数。我们仍然希望余数为非负数,因此现在要求 r 满足 0 ≤ r < |b|。

4.35. Lemma: Extension of Integer Division to Negative b

4.35. 引理:整数除法推广到负数 b

b b Z Z 0 , a 一个 Z Z , q q Z Z , r r Z Z , a 一个 = b b q q + r r and 0 r r < | b b | .

Proof. Fix b Z 0 . We know b > 0 or b < 0, so use cases.

证明。固定 b∈Z≠0。我们知道 b > 0 或 b < 0,因此满足用例。

Case 1. Assume b > 0. In this case, we already proved the needed result in the previous lemma.

情况 1. 假设 b > 0。在这种情况下,我们已经在前一个引理中证明了所需的结果。

Case 2. Assume b < 0. Then b = −| b | where | b | Z > 0 . Applying the previous results to divide a by | b |, we see there exist integers q 3 , r 3 Z 0 with

情况 2. 假设 b < 0。则 b = −|b|,其中 |b|∈Z>0。应用前面的结果将 a 除以 |b|,我们发现存在整数 q3,r3∈Z≥0,使得

a 一个 = | b b | q q 3 + r r 3 and 0 r r 3 < | b b | .

Now, | b | = − b , so a = ( − b ) q 3 + r 3 = b ( − q 3 ) + r 3 . Choose q = − q 3 and r 3 = r to prove the existence assertion in the lemma.

现在,|b| = −b,所以 a = ( − b)q 3 + r 3 = b( − q 3 ) + r 3 。选择 q = −q 3 和 r 3 = r 来证明引理中的存在性断言。

For example, dividing a = −26 by b = −7 gives quotient q = 4 and remainder r = 2.

例如,a = −26 除以 b = −7,得到商 q = 4,余数 r = 2。

Uniqueness of the Quotient and Remainder

商和余数的唯一性

The final, crucial improvement to our results so far is to prove that the quotient and remainder are uniquely determined by a and b . Here is the full-fledged statement of the Integer Division Theorem. The proof assumes some basic facts about absolute value are known; these facts are discussed in more detail in § 8.4 .

对我们目前结果的最终也是至关重要的改进是证明商和余数由 a 和 b 唯一确定。以下是整数除法定理的完整表述。证明的前提是已知一些关于绝对值的基本性质;这些性质将在 §8.4 中详细讨论。

4.36. Integer Division Theorem (Final Version). For all integers a and all nonzero integers b , there exist unique integers q , r with a = bq + r and 0 ≤ r < | b |.

4.36. 整数除法定理(最终版)。对于所有整数 a 和所有非零整数 b,存在唯一的整数 q、r,使得 a = bq + r 且 0 ≤ r < |b|。

Proof. First we restate the theorem in formal symbols to better understand the uniqueness assertion. We must prove:

证明。首先,我们用形式符号重述该定理,以便更好地理解唯一性断言。我们必须证明:

b b Z Z 0 , a 一个 Z Z , ! ( q q , r r ) Z Z × × Z Z , a 一个 = b b q q + r r ( 0 r r < | b b | ) .

Fix arbitrary b Z 0 and a Z . We have already proved the existence of ( q , r ) above. To prove uniqueness, we use the uniqueness proof template (recall this arises from the rule for eliminating the uniqueness symbol). Fix arbitrary ( q 1 , r 1 ) Z × Z and arbitrary ( q 2 , r 2 ) Z × Z . Assume a = bq 1 + r 1 and 0 ≤ r 1 < | b | and a = bq 2 + r 2 and 0 ≤ r 2 < | b |. We must prove ( q 1 , r 1 ) = ( q 2 , r 2 ), which means q 1 = q 2 and r 1 = r 2 (by the Ordered Pair Axiom). We first prove q 1 = q 2 by contradiction. Assume q 1 q 2 , and derive a contradiction. Combining the two equations for a assumed earlier, we get bq 1 + r 1 = bq 2 + r 2 . Some algebra transforms this to b ( q 1 q 2 ) = r 2 r 1 . Since q 1 and q 2 are unequal integers, | q 1 q 2 | ≥ 1, and hence | r 2 r 1 | = | b || q 1 q 2 | ≥ | b | · 1 = | b |. So | r 2 r 1 | ≥ | b |. On the other hand, the assumed inequalities for r 1 and r 2 combine to show that r 1 r 2 < | b | and r 2 r 1 < | b |. Since | r 2 r 1 | equals r 2 r 1 or r 1 r 2 , we therefore have | r 2 r 1 | < | b |. We have reached the contradiction “| r 2 r 1 | ≥ | b | and | r 2 r 1 | < | b |.” Therefore, q 1 = q 2 . Now, to prove r 1 = r 2 , rewrite the assumed equations for a to get r 1 = a bq 1 = a bq 2 = r 2 .

固定任意 b∈Z≠0 和 a∈Z。我们已经在上面证明了 (q, r) 的存在性。为了证明唯一性,我们使用唯一性证明模板(回想一下,这源于消除唯一性符号的规则)。固定任意 (q1,r1)∈Z×Z 和任意 (q2,r2)∈Z×Z。假设 a = bq 1 + r 1 且 0 ≤ r 1 < |b|,并且 a = bq 2 + r 2 且 0 ≤ r 2 < |b|。我们必须证明 (q 1 , r 1 ) = (q 2 , r 2 ),这意味着 q 1 = q 2 且 r 1 = r 2 (根据有序对公理)。我们首先用反证法证明 q 1 = q 2 。假设 q 1 ≠ q 2 ,并由此得出矛盾。将之前假设的两个方程合并,得到 bq 1 + r 1 = bq 2 + r 2 。经过一些代数运算,可将其转化为 b(q 1 − q 2 ) = r 2 − r 1 。由于 q 1 和 q 2 是不相等的整数,因此 |q 1 − q 2 | ≥ 1,进而 |r 2 − r 1 | = |b||q 1 − q 2 | ≥ |b| · 1 = |b|。因此,|r 2 − r 1 | ≥ |b|。另一方面,假设 r 1 和 r 2 的不等式结合起来表明 r 1 − r 2 < |b| 且 r 2 − r 1 < |b|。由于 |r 2 − r 1 |等于 r 2 − r 1 或 r 1 − r 2 ,因此我们有 |r 2 − r 1 | < |b|。我们得到了矛盾“|r 2 − r 1 | ≥ |b| 且 |r 2 − r 1 | < |b|”。因此,q 1 = q 2 。现在,为了证明 r 1 = r 2 ,重写假设的 a 的方程,得到 r 1 = a − bq 1 = a − bq 2 = r 2

4.37. Remark. Consider the special case of the Division Theorem where b = 2. On one hand, the existence part of the theorem says that any a Z can be written in the form a = 2 q + r where 0 ≤ r < 2. In other words, for every a Z , either there exists q Z with a = 2 q + 0, or there exists q Z with a = 2 q + 1. This says that every integer a is even or odd. On the other hand, one consequence of the uniqueness of ( q , r ) for a is that the remainders 0 and 1 cannot both work for the same fixed integer a . So, no integer a is both even and odd. This gives another proof of the theorem that every integer is either even or odd, but not both , which we used frequently in earlier sections.

4.37. 备注。考虑除法定理中 b = 2 的特殊情况。一方面,定理的存在性部分表明,任意整数 a∈Z 都可以写成 a = 2q + r 的形式,其中 0 ≤ r < 2。换句话说,对于每个整数 a∈Z,要么存在 q∈Z 使得 a = 2q + 0,要么存在 q∈Z 使得 a = 2q + 1。这表明每个整数 a 要么是偶数,要么是奇数。另一方面,(q, r) 对于 a 的唯一性的一个推论是,余数 0 和 1 不能同时适用于同一个固定的整数 a。因此,不存在既是偶数又是奇数的整数 a。这给出了每个整数要么是偶数要么是奇数,但不能既是偶数又是奇数的定理的另一个证明,我们在前面的章节中经常用到这个证明。

We can obtain similar results for other specific choices of b . For example, letting b = 3, we see that every integer a can be written in exactly one of the forms a = 3 q , a = 3 q + 1, or a = 3 q + 2 for some integer q . Similarly, taking b = 10, we see that every integer a has the form a = 10 q + r for unique q , r Z with r ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. This observation is the first step in rigorously proving the existence and uniqueness of the decimal expansions (or more generally, the base- b expansions) of integers. More details on this process appear in some of the exercises below.

对于其他特定的 b 值,我们可以得到类似的结果。例如,令 b = 3,我们发现每个整数 a 都可以写成 a = 3q、a = 3q + 1 或 a = 3q + 2 的形式,其中 q 为某个整数。类似地,令 b = 10,我们发现每个整数 a 都可以写成 a = 10q + r 的形式,其中 q, r ∈ Z,r ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}。这一观察是严格证明整数的十进制展开式(或更一般地,以 b 为底的展开式)的存在性和唯一性的第一步。关于此过程的更多细节将在下面的练习中给出。

Section Summary

章节概要

  1. Prime and Composite Numbers. An integer n > 1 is composite iff n = ab for some integers a , b with 1 < a < n and 1 < b < n . An integer p > 1 is prime iff p is not composite iff the only positive divisors of p are 1 and p .
    质数和合数。一个大于 1 的整数 n 是合数,当且仅当存在整数 a 和 b,使得 n = ab,其中 1 < a < n 且 1 < b < n。一个大于 1 的整数 p 是质数,当且仅当 p 不是合数,当且仅当 p 的正约数只有 1 和 p 两个。
  2. Facts about Primes. Every integer n > 1 can be written as a product of one or more primes. There are infinitely many primes.
    关于素数的事实。每个大于1的整数n都可以表示为一个或多个素数的乘积。素数有无穷多个。
  3. Integer Division Theorem. For all integers a and b with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < | b |. We call q the quotient and r the remainder when a is divided by b . Taking b = 2, we see that every integer is either even or odd, but not both.
    整数除法定理。对于任意整数 a 和 b(b ≠ 0),存在唯一的整数 q 和 r,使得 a = bq + r 且 0 ≤ r < |b|。我们称 q 为 a 除以 b 的商,r 为 a 除以 b 的余数。取 b = 2,我们可以看出,每个整数要么是偶数,要么是奇数,但不会同时是偶数和奇数。

Exercises

练习

  1. (a) List all primes less than 100. (b) Write each integer as a product of primes: 91, 96, 111, 3003, and 8000000.
    (a)列出所有小于 100 的质数。(b)将每个整数写成质数的乘积:91、96、111、3003 和 80000。
  2. For each a and b below, find the quotient and remainder when a is divided by b . (a) a = 58, b = 11 (b) a = 58, b = −11 (c) a = −58, b = 11 (d) a = −58, b = −11 (e) a = 0, b = 11 (f) a = 6, b = 11 (g) a = −6, b = 11 (h) a = 6, b = −11.
    对于以下每个 a 和 b,求 a 除以 b 的商和余数。 (a) a = 58, b = 11 (b) a = 58, b = −11 (c) a = −58, b = 11 (d) a = −58, b = −11 (e) a = 0, b = 11 (f) a = 6, b = 11 (g) a = −6, b = 11 (h) a = 6, b = −11.
  3. For each a and b below, find the quotient and remainder when a is divided by b . (a) a = 300, b = 27 (b) a = 9103, b = 11 (c) a = 4574, b = 19 (d) a = 100000, b = 123.
    对于以下每个 a 和 b,求 a 除以 b 的商和余数。 (a) a = 300, b = 27 (b) a = 9103, b = 11 (c) a = 4574, b = 19 (d) a = 10000, b = 123.
  4. Use Definition 4.28 to prove that an integer p > 1 is prime iff the only positive divisors of p are 1 and p .
    利用定义 4.28 证明,一个大于 1 的整数 p 是素数当且仅当 p 的唯一正因子是 1 和 p。
  5. Prove: n Z > 0 , m Z > 0 , if m is composite, then nm is composite.
    证明:对于 ∀n∈Z>0,∀m∈Z>0,如果 m 是合数,则 nm 也是合数。
  6. True or false? Prove your answers. (a) The product of any two positive integers is always composite. (b) The sum of two prime integers is never prime. (c) The sum of any two composite integers is composite. (d) The product of any two composite integers is composite. (e) Every composite integer has an even number of positive divisors.
    判断正误,并证明你的答案。 (a) 任意两个正整数的乘积总是合数。 (b) 两个质数的和永远不是质数。 (c) 任意两个合数的和是合数。 (d) 任意两个合数的乘积是合数。 (e) 每个合数都有偶数个正约数。
  7. Suppose k Z > 0 and p 1 , …, p k are integers. Give a careful proof that for all i between 1 and k , p i divides r = 1 k p r . (This fact was used in the proof of Theorem 4.30.) Suggestion: Prove p i divides r = 1 s p r for all s in the range i s k by induction.
    假设 k∈Z>0 且 p 1 , …, p k 为整数。请给出严谨的证明,对于所有介于 1 和 k 之间的 i,p i 整除 ∏r=1kpr。(此事实已用于定理 4.30 的证明。)提示:用数学归纳法证明对于所有 i ≤ s ≤ k 的 s,p i 整除 ∏r=1spr。
  8. (a) Prove that every prime integer other than 2 is odd. (b) Prove that every positive integer is either a power of 2 or is divisible by an odd prime.
    (a)证明除2以外的所有质数都是奇数。(b)证明每个正整数要么是2的幂,要么能被某个奇质数整除。
  9. Use the uniqueness assertion in the Integer Division Theorem to prove carefully that 5 does not divide 22, without assuming in advance that 22/5 is not an integer.
    利用整数除法定理中的唯一性断言,仔细证明 5 不能整除 22,不要预先假设 22/5 不是整数。
  10. Use the technique in the previous exercise to give a fully rigorous proof that 7 is prime. (Use proof by exhaustion to show that for all b ∈ {2, 3, 4, 5, 6}, b does not divide 7.)
    利用上一题中的方法,给出7是素数的完全严格的证明。(使用穷举法证明对于所有b∈{2, 3, 4, 5, 6},b都不整除7。)
  11. The following statements resemble the Integer Division Theorem, but they are all false. Negate and disprove each statement. (a) a Z , b Z , ! ( q , r ) Z × Z , ( a = b q + r 0 r < | b | ) . (b) a Z , b Z 0 , ! ( q , r ) Z × Z , ( a = b q + r 0 r < b ) . (c) a Z , b Z 0 , ! ( q , r ) Z × Z , ( a = b q + r 0 r | b | ) . (d) a Z , b Z 0 , ! ( q , r ) Z 0 × Z 0 , ( a = b q + r 0 r < | b | ) .
    以下陈述与整数除法定理相似,但它们都是错误的。请否定并反驳每个陈述。 (a) ∀a∈Z,∀b∈Z,∃!(q,r)∈Z×Z,(a=bq+r∧0≤r<|b|). (b) ∀a∈Z,∀b∈Z≠0,∃!(q,r)∈Z×Z,(a=bq+r∧0≤r<b). (c) ∀a∈Z,∀b∈Z≠0,∃!(q,r)∈Z×Z,(a=bq+r∧0≤r≤|b|). (d) ∀a∈Z,∀b∈Z≠0,∃!(q,r)∈Z≥0×Z≥0,(a=bq+r∧0≤r<|b|).
  12. What are the possible remainders when a prime integer p > 10 is divided by 10? Prove your answer.
    当一个大于 10 的素数 p 除以 10 时,可能有哪些余数?请证明你的答案。
  13. What are the possible remainders when a square integer n 2 is divided by 8? Prove your answer.
    当一个平方整数 n 2 除以 8 时,可能的余数有哪些?证明你的答案。
  14. (a) Prove or disprove: for all open sentences P ( q , r ),
    [ ! ( q , r ) Z × Z , P ( q , r ) ] [ ! q Z , ! r Z , P ( q , r ) ] .
    (b) Prove or disprove: for all open sentences P ( q , r ),
    [ ! q Z , ! r Z , P ( q , r ) ] [ ! ( q , r ) Z × Z , P ( q , r ) ] .

    (a) 证明或反驳:对于所有开语句 P(q, r), [∃!(q,r)∈Z×Z,P(q,r)]⇒[∃!q∈Z,∃!r∈Z,P(q,r)]. (b) 证明或反驳:对于所有开语句 P(q, r), [∃!q∈Z,∃!r∈Z,P(q,r)]⇒[∃!(q,r)∈Z×Z,P(q,r)].
  15. (a) For a , b Z > 0 , how are the quotient and remainder when a is divided by b related to the quotient and remainder when a + 1 is divided by b ? (b) Prove the preliminary version of the Integer Division Theorem by fixing b > 0 and using ordinary induction (not strong induction) on a .
    (a) 对于 a,b∈Z>0,a 除以 b 的商和余数与 a + 1 除以 b 的商和余数有何关系? (b) 通过固定 b > 0 并对 a 使用普通归纳法(而非强归纳法)来证明整数除法定理的初步版本。
  16. Prove this variation of the Integer Division Theorem: for all a , b Z with b ≠ 0, there exist unique integers q , r with a = bq + r and −| b |/2 < r ≤ | b |/2.
    证明整数除法定理的这个变体:对于所有 a,b∈Z 且 b ≠ 0,存在唯一的整数 q, r,使得 a = bq + r 且 −|b|/2 < r ≤ |b|/2。
  17. Fix r 0 Z . State and prove a version of the Integer Division Theorem where the remainder r is required to satisfy r 0 r < r 0 + | b |.
    固定 r0∈Z。陈述并证明整数除法定理的一个版本,其中余数 r 需要满足 r 0 ≤ r < r 0 + |b|。
  18. For a Z , let a mod 4 { 0 , 1 , 2 , 3 } be the remainder when a is divided by 4. (a) Prove by induction: for all n , a 1 , a 2 , , a n Z > 0 , if a mod 4 = 1 for all i , then ( i = 1 n a i ) mod 4 = 1 . (b) Prove there exist infinitely many primes p such that p mod 4 = 3 . [Adapt the proof of Theorem 4.30.]
    对于 a∈Z,令 amod4∈{0,1,2,3} 表示 a 除以 4 的余数。(a) 用数学归纳法证明:对于所有 n,a1,a2,…,an∈Z>0,如果对于所有 i,amod4=1,则 (∏i=1nai)mod4=1。(b) 证明存在无穷多个素数 p 使得 pmod4=3。[改编定理 4.30 的证明。]
  19. Decimal Expansions of Integers. (a) Prove by induction that every positive integer n can be written in the form n = k = 0 m d k 10 k where m ≥ 0, each d k ∈ {0, 1, 2, …, 9}, and d m ≠ 0. (b) Prove that m and d 0 , d 1 , …, d m in part (a) are uniquely determined by n .
    整数的十进制展开式。(a) 用数学归纳法证明每个正整数 n 都可以写成 n=∑k=0mdk10k 的形式,其中 m ≥ 0,每个 d k ∈ {0, 1, 2, …, 9},且 d m ≠ 0。(b) 证明 (a) 部分中的 m 和 d 0 , d 1 , …, d m 由 n 唯一确定。
  20. State and prove a generalization of the previous exercise in which 10 is replaced by an arbitrary base b Z > 1 .
    陈述并证明前一个练习的推广,其中 10 被任意基数 b∈Z>1 所取代。
  21. (a) Prove: for all x , c , d Z > 0 , x c d 1 = ( x c 1 ) k = 0 d 1 x c k . (b) Suppose a ≥ 0, b > 0, and dividing a by b yields quotient q and remainder r . For fixed x Z > 1 , find (with proof) the quotient and remainder when x a − 1 is divided by x b − 1.
    (a) 证明:对于所有 x,c,d∈Z>0,xcd−1=(xc−1)∑k=0d−1xck。(b) 假设 a ≥ 0,b > 0,且 a 除以 b 的商为 q,余数为 r。对于固定的 x∈Z>1,求(并证明)x a − 1 除以 x b − 1 的商和余数。

4.5 Greatest Common Divisors

4.5 最大公约数

In this section, we study the greatest common divisor (gcd) of two integers a and b , denoted gcd( a , b ). We present Euclid’s algorithm for computing gcd( a , b ) based on repeated division. One consequence of this algorithm is the fact that gcd( a , b ) can always be written in the form ax + by for some integers x and y . This fact has important applications in modern cryptography.

本节研究两个整数 a 和 b 的最大公约数(gcd),记为 gcd(a, b)。我们介绍基于重复除法的欧几里得算法来计算 gcd(a, b)。该算法的一个推论是,gcd(a, b) 总可以写成 ax + by 的形式,其中 x 和 y 为整数。这一结论在现代密码学中有着重要的应用。

Greatest Common Divisors

最大公约数

Before discussing greatest common divisors, we recall the definition of divisibility.

在讨论最大公约数之前,我们先回顾一下整除性的定义。

4.38. Definition: Divisors. For all integers a and c , c is a divisor of a iff u Z , a = c u . When this condition holds, we also say that c divides a and a is a multiple of c . The notation c | a means that c divides a .

4.38. 定义:除数。对于任意整数 a 和 c,c 是 a 的除数当且仅当存在 u∈Z,a=cu。当此条件成立时,我们也称 c 整除 a 且 a 是 c 的倍数。记号 c|a 表示 c 整除 a。

Note that c | a stands for the statement c divides a ,” whereas c / a denotes the rational number obtained when c is divided by a . Take care not to confuse these two symbols!

请注意,c|a 表示“c 能整除 a”,而 c/a 表示 c 除以 a 所得的有理数。切勿混淆这两个符号!

4.39. Definition: Greatest Common Divisors. Given integers a , b that are not both zero, the greatest common divisor gcd( a , b ) is the largest integer d such that d divides a and d divides b . We also define gcd(0, 0) = 0.

4.39. 定义:最大公约数。给定两个不同时为零的整数 a 和 b,最大公约数 gcd(a, b) 是最大的整数 d,使得 d 能整除 a 且 d 能整除 b。我们还定义 gcd(0, 0) = 0。

As a technical aside, it can be shown that gcd( a , b ) exists and is unique, using the fact that all divisors d of a nonzero integer a satisfy d ≤ | a |; see Theorem 8.58(b).

顺便提一下,可以证明 gcd(a, b) 存在且唯一,利用非零整数 a 的所有因子 d 满足 d ≤ |a| 的事实;参见定理 8.58(b)。

4.40. Example. Let a = 88 and b = 60. The positive divisors of a are 1, 2, 4, 8, 11, 22, 44, 88. The positive divisors of b are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Comparing these lists, we see that the positive integers d dividing both a and b are 1, 2, and 4. So gcd(88, 60) = 4. Another way to compute greatest common divisors is to compare prime factorizations. Here, 88 = 2 3 · 11 and 60 = 2 2 · 3 · 5. Both 88 and 60 are divisible by two copies of the prime 2 but have no other common prime factors, so we conclude that gcd(88, 60) = 2 2 = 4. (See Theorem 4.60 for a justification of this technique.)

4.40. 例。设 a = 88,b = 60。a 的正约数有 1, 2, 4, 8, 11, 22, 44, 88。b 的正约数有 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60。比较这些列表,我们发现能同时整除 a 和 b 的正整数 d 是 1、2 和 4。因此,gcd(88, 60) = 4。计算最大公约数的另一种方法是比较质因数分解。这里,88 = 2 × 11,60 = 2 × 3 × 5。88 和 60 都能被两个质数 2 整除,但没有其他公共质因数,因此我们得出结论:gcd(88, 60) = 2 × 4 = 4。(参见定理 4.60 以了解此方法的证明。)

It turns out that the two methods of finding greatest common divisors just discussed — listing all the divisors, or comparing prime factorizations — are horribly inefficient for large integers a and b . A much faster algorithm was discovered thousands of years ago by Euclid, the author of the famous geometry textbook Elements (which is also a text on number theory). The basis for this algorithm is the following result relating integer division to greatest common divisors.

事实证明,刚才讨论的两种求最大公约数的方法——列出所有因子或比较质因数分解——对于较大的整数 a 和 b 来说效率极低。欧几里得早在几千年前就发现了一种速度快得多的算法,他是著名几何教科书《几何原本》(同时也是一本数论教材)的作者。该算法的基础是以下将整数除法与最大公约数联系起来的结果。

4.41. Theorem on Greatest Common Divisors (a) For all a Z , gcd( a , 0) = | a |.

4.41. 最大公约数定理 (a) 对于所有 a∈Z,gcd(a, 0) = |a|。

(b) For all a , b , q , r Z , if b ≠ 0 and a = bq + r , then gcd( a , b ) = gcd( b , r ).

(b)对于所有 a、b、q、r∈Z,如果 b ≠ 0 且 a = bq + r,则 gcd(a, b) = gcd(b, r)。

Proof. (a) Fix a Z . We know a = 0 or a ≠ 0, so consider cases.

证明。(a) 固定 a∈Z。我们知道 a = 0 或 a ≠ 0,所以考虑以下情况。

Case 1. Assume a = 0. Here, gcd( a , 0) = 0 = | a | by definition.

情况 1. 假设 a = 0。根据定义,gcd(a, 0) = 0 = |a|。

Case 2. Assume a ≠ 0. We know d |0 for all integers d , since 0 = d · 0. Therefore, the common divisors of a and 0 are exactly the same integers as the divisors of a itself. We know the largest divisor of a is | a |, so gcd( a , 0) = | a | in this case.

情况 2. 假设 a ≠ 0。我们知道对于所有整数 d,d|0,因为 0 = d · 0。因此,a 和 0 的公约数与 a 自身的公约数完全相同。我们知道 a 的最大公约数是 |a|,所以在这种情况下,gcd(a, 0) = |a|。

(b) Fix a , b , q , r Z . Assume b ≠ 0 and a = bq + r . Prove gcd( a , b ) = gcd( b , r ). Our strategy is to prove that the set of common divisors of a and b coincides with the set of common divisors of b and r ; since it then follows that the largest element of the first set (namely gcd( a , b )) must equal the largest element of the second set (namely gcd( b , r )).

(b) 固定 a, b, q, r∈Z。假设 b ≠ 0 且 a = bq + r。证明 gcd(a, b) = gcd(b, r)。我们的策略是证明 a 和 b 的公约数集合与 b 和 r 的公约数集合重合;因为由此可知,第一个集合(即 gcd(a, b))中的最大元素必定等于第二个集合(即 gcd(b, r))中的最大元素。

To show these two sets are equal, fix an integer d ; we prove that d | a and d | b iff d | b and d | r . For one direction, assume d | a and d | b , so a = ds and b = dt for some s , t Z . We must prove d | b and d | r . We already assumed d | b . Also, r = a bq = ds dtq = d ( s tq ) where s tq is an integer, so d | r . For the other direction, assume d | b and d | r , so b = dt and r = du for some t , u Z . We must prove d | a and d | b . We already assumed d | b . Also, a = bq + r = dtq + du = d ( tq + u ) where tq + u is an integer, so d | a .

为了证明这两个集合相等,固定一个整数 d;我们证明 d|a 且 d|b 当且仅当 d|b 且 d|r。对于一个方向,假设 d|a 且 d|b,则对于某些 s,t∈Z,有 a = ds 且 b = dt。我们需要证明 d|b 且 d|r。我们已经假设了 d|b。此外,r = a − bq = ds − dtq = d(s − tq),其中 s − tq 为整数,因此 d|r。对于另一个方向,假设 d|b 且 d|r,则对于某些 t,u∈Z,有 b = dt 且 r = du。我们需要证明 d|a 且 d|b。我们已经假设了 d|b。此外,a = bq + r = dtq + du = d(tq + u),其中 tq + u 为整数,因此 d|a。

Euclid’s GCD Algorithm

欧几里得的最大公约数算法

The previous theorem justifies the following recursive algorithm for computing gcd( a , b ), called Euclid’s Algorithm .

前面的定理证明了以下用于计算 gcd(a, b) 的递归算法的合理性,该算法称为欧几里得算法。

4.42. Euclid’s Recursive GCD Algorithm.

4.42. 欧几里得递归最大公约数算法。

Input: a , b Z 0 .

输入:a,b∈Z≥0。

Output: gcd( a , b ).

输出:gcd(a, b)。

Procedure: (a) If b = 0, return a .

过程:(a)如果 b = 0,则返回 a。

(b) If b > 0, use integer division to write a = bq + r for some q , r Z with 0 ≤ r < b ; return gcd( b , r ).

(b)如果 b > 0,则使用整数除法写出 a = bq + r,其中 q,r∈Z,且 0 ≤ r < b;返回 gcd(b, r)。

It can be proved, using strong induction on b , that this algorithm terminates with the correct answer after finitely many steps. The key observation is that the recursive call in step (b) replaces the second input b by the smaller nonnegative integer r .

利用强归纳法对 b 进行证明,可以得出该算法在有限步后终止并得到正确答案。关键在于,步骤 (b) 中的递归调用将第二个输入 b 替换为较小的非负整数 r。

4.43. Example. Let us recompute gcd(88, 60) using Euclid’s algorithm. Initially a = 88 and b = 60, so we are in case (b) of the algorithm. We divide a by b , getting:

4.43. 示例。让我们使用欧几里得算法重新计算 gcd(88, 60)。初始时 a = 88 且 b = 60,因此我们处于算法的 (b) 情况。我们将 a 除以 b,得到:

88 = 60 1 + 28 ; gcd 最大公约数 ( 88 , 60 ) = gcd 最大公约数 ( 60 , 28 ) .

In the first recursive call, the inputs are 60 and 28. Dividing again gives

在第一次递归调用中,输入为 60 和 28。再次除法得到

60 = 28 2 + 4 ; gcd 最大公约数 ( 60 , 28 ) = gcd 最大公约数 ( 28 , 4 ) .

The new inputs are 28 and 4; dividing gives

新的输入值是 28 和 4;相除得到

28 = 4 7 + 0 ; gcd 最大公约数 ( 28 , 4 ) = gcd 最大公约数 ( 4 , 0 ) .

By case (a) in the algorithm, gcd(4, 0) = 4. Working our way back up through the recursive calls, we see gcd(88, 60) = 4.

根据算法中的 (a) 情况,gcd(4, 0) = 4。通过递归调用向上回溯,我们看到 gcd(88, 60) = 4。

Although we described the gcd algorithm recursively, the previous example shows how to unravel this recursion into an iterative procedure consisting of a series of integer divisions. Specifically, assuming we start with positive inputs a and b , the algorithm performs the following divisions:

虽然我们用递归的方式描述了最大公约数算法,但前面的例子展示了如何将这种递归分解为一系列整数除法的迭代过程。具体来说,假设我们从正数 a 和 b 开始,该算法执行以下除法:

a 一个 = b b q q 1 + r r 1 , 0 < r r 1 < b b ; b b = r r 1 q q 2 + r r 2 , 0 < r r 2 < r r 1 ; r r 1 = r r 2 q q 3 + r r 3 , 0 < r r 3 < r r 2 ; r r 2 = r r 3 q q 4 + r r 4 , 0 < r r 4 < r r 3 ; r r k k 2 = r r k k 1 q q k k + r r k k , 0 < r r k k < r r k k 1 ; r r k k 1 = r r k k q q k k + 1 + 0 , (remainder (余 r r k k + 1 is zero). 为零)。

(4.1)

When we finally reach a remainder r k +1 = 0, we return gcd( a , b ) = r k , the last nonzero remainder. This must eventually occur, since b > r 1 > r 2 > r 3 > · · · and all remainders are nonnegative integers.

当最终余数 r k +1 = 0 时,我们返回 gcd(a, b) = r k ,即最后一个非零余数。这必然会发生,因为 b > r 1 > r 2 > r 3 > · · ·,并且所有余数都是非负整数。

4.44. Example. Let us use the algorithm to compute gcd(171, 51):

4.44. 示例。让我们使用该算法计算 gcd(171, 51):

171 = 51 3 + 18 ,

(4.2)

51 = 18 2 + 15 ,

(4.3)

18 = 15 1 + 3 ,

(4.4)

15 = 3 5 + 0.

(4.5)

So gcd(171, 51) = 3.

所以 gcd(171, 51) = 3。

Linear Combination Property of GCDs

最大公约数的线性组合性质

One of the most important properties of greatest common divisors is that for all a , b Z , gcd( a , b ) can be written in the form ax + by for some integers x and y . In fact, we can enhance Euclid’s algorithm with some extra computations that find x and y explicitly. The idea is to work backwards through the chain of divisions to express gcd( a , b ) (which is the last nonzero remainder) in terms of quantities appearing earlier. We illustrate this idea by revisiting the two previous examples.

最大公约数最重要的性质之一是,对于任意整数 a, b∈Z,gcd(a, b) 可以写成 ax + by 的形式,其中 x 和 y 为整数。事实上,我们可以通过一些额外的计算来增强欧几里得算法,从而显式地找到 x 和 y。其思路是逆向推导除法链,用前面出现的量来表示 gcd(a, b)(即最后一个非零余数)。我们通过回顾前面的两个例子来说明这个思路。

4.45. Example. We computed gcd(88, 60) = 4; let us now find x , y Z with 4 = 88 x + 60 y . Recall the divisions executed by the recursive algorithm to find gcd(88, 60):

4.45. 示例。我们计算出 gcd(88, 60) = 4;现在让我们找到满足 4 = 88x + 60y 的 x, y∈Z。回顾一下递归算法在求 gcd(88, 60) 时执行的除法运算:

88 = 60 1 + 28 ; 60 = 28 2 + 4 ; 28 = 4 7 + 0.

The gcd 4 appears as the remainder in the second equation. Isolating the 4, we see that 4 = 60 + 28 · ( − 2). Now, from the first equation above, 28 = 88 − 60. Substituting this expression for 28 into our expression for 4, we find that

最大公约数 4 出现在第二个方程的余数中。分离出 4,我们看到 4 = 60 + 28 · ( − 2)。现在,由上面的第一个方程可知,28 = 88 − 60。将 28 的表达式代入 4 的表达式,我们发现

4 = 60 + 28 ( 2 ) = 60 + ( 88 60 ) ( 2 ) = 88 ( 2 ) + 60 3.

The last step used the distributive law to collect terms; note that (88 − 60) · ( − 2) contributes ( − 1)( − 2) = 2 copies of 60, and there is one more copy of 60 added in front, for a total of 3. We now see that 4 = 88 x + 60 y holds if we choose x = −2 and y = 3.

最后一步利用分配律合并项;注意 (88 − 60) · ( − 2) 贡献了 ( − 1)( − 2) = 2 个 60,前面又加了一个 60,总共 3 个。现在我们看到,如果选择 x = −2 和 y = 3,则 4 = 88x + 60y 成立。

Before looking at the next example, let us introduce some terminology.

在看下一个例子之前,让我们先介绍一些术语。

4.46. Definition: Linear Combinations. For all d , s , t Z , d is a linear combination of s and t iff x , y Z , d = s x + t y .

4.46. 定义:线性组合。对于所有 d,s,t∈Z,d 是 s 和 t 的线性组合当且仅当存在 x,y∈Z,d=sx+ty。

4.47 Example. Let us find x , y Z with 3 = gcd(171, 51) = 171 x + 51 y . Our strategy is to use the divisions ( 4.4 ), ( 4.3 ), ( 4.2 ) to express 3 as a linear combination of 18 and 15, then as a linear combination of 51 and 18, and finally as a linear combination of 171 and 51. First, from ( 4.4 ),

4.47 例。求 x,y∈Z,使得 3 = gcd(171, 51) = 171x + 51y。我们的策略是利用除法式 (4.4)、(4.3) 和 (4.2) 将 3 表示为 18 和 15 的线性组合,然后表示为 51 和 18 的线性组合,最后表示为 171 和 51 的线性组合。首先,由 (4.4) 可知,

3 = 18 1 + 15 ( 1 ) .

Now from ( 4.3 ), 15 = 51 + 18 · ( − 2). Substituting into the previous equation and collecting terms,

现在由(4.3)式可知,15 = 51 + 18 · ( − 2)。代入前式并合并项,

3 = 18 1 + [ 51 + 18 ( 2 ) ] ( 1 ) = 18 3 + 51 ( 1 ) .

Note that the new coefficient of 18 was computed from the distributive law as 1 + ( − 2)( − 1) = 3. Now from ( 4.2 ), 18 = 171 + 51 · ( − 3). Substituting into the previous equation and collecting terms,

注意,新的系数 18 是根据分配律计算得出的,即 1 + ( − 2)( − 1) = 3。现在,根据 (4.2) 式,18 = 171 + 51 · ( − 3)。将其代入前面的方程并合并各项,

3 = [ 171 + 51 ( 3 ) ] 3 + 51 ( 1 ) = 171 3 + 51 ( 10 ) .

Note that the new coefficient of 51 was computed from the distributive law as ( − 3) · 3 + ( − 1) = −10. So 3 = 171 x + 51 y holds if we choose x = 3 and y = −10.

注意,新的系数 51 是根据分配律计算出来的,即 ( − 3) · 3 + ( − 1) = −10。因此,如果我们选择 x = 3 和 y = −10,则 3 = 171x + 51y 成立。

In summary, to find x and y with gcd( a , b ) = ax + by : first find the gcd using repeated divisions as in ( 4.1 ); then work backwards to express the gcd r k as a linear combination of r k −2 and r k −1 , then as a linear combination of r k −3 and r k −2 , and so on, until finally r k has been expressed as a linear combination of the original inputs a and b . We do one more complete example to illustrate the full process.

总之,要找到满足 gcd(a, b) = ax + by 的 x 和 y,首先使用重复除法(如式 (4.1) 所示)找到最大公约数;然后反向推导,将最大公约数 r k 表示为 r k −2 和 r k −1 的线性组合,再表示为 r k −3 和 r k −2 的线性组合,依此类推,直到最终将 r k 表示为原始输入 a 和 b 的线性组合。我们再举一个完整的例子来说明整个过程。

4.48. Example. Find gcd(999, 101), and find x , y Z with gcd(999, 101) = 999 x + 101 y . Solution. First find the gcd by repeated divisions:

4.48. 例题。求 gcd(999, 101),并找出满足 gcd(999, 101) = 999x + 101y 的 x, y∈Z。解:首先通过重复除法求出 gcd:

999 = 101 9 + 90 , 101 = 90 1 + 11 , 90 = 11 8 + 2 , 11 = 2 5 + 1 , 2 = 2 1 + 0.

So gcd(999, 101) = 1. Next, work backwards through the equations, taking care to collect common terms at each stage , as shown in the far right column below:

所以 gcd(999, 101) = 1。接下来,反向推导方程,注意在每个阶段收集公因数,如下图最右侧一栏所示:

1 = 11 + 2 ( 5 ) = 11 + [ 90 + 11 ( 8 ) ] ( 5 ) = 11 41 + 90 ( 5 ) = [ 101 + 90 ( 1 ) ] 41 + 90 ( 5 ) = 101 41 + 90 ( 46 ) = 101 41 + ( 999 + 101 ( 9 ) ) ( 46 ) = 101 455 + 999 ( 46 ) .

So 1 = 999 x + 101 y holds with x = −46 and y = 455.

所以当 x = −46 且 y = 455 时,1 = 999x + 101y 成立。

We prove the linear combination property of gcds in § 4.6 .

我们在第 4.6 节中证明了最大公约数的线性组合性质。

Matrix Reduction Algorithm for Computing GCDs (Optional)

In this optional section, we present another algorithm for computing d = gcd( a , b ) and finding integers x , y such that d = ax + by . This algorithm proceeds by performing certain elementary row operations on matrices with integer entries. There are three allowable row operations: (i) switch two rows in a matrix; (ii) multiply one row in a matrix by −1; (iii) add any integer multiple of one row of a matrix to a different row.

You may have studied similar elementary row operations for real-valued matrices in linear algebra. In the real case, we can add any real multiple of one row to another row in operation (iii), and we can multiply a given row by any nonzero real number in operation (ii). These row operations are used in the Gaussian elimination algorithm to solve a system of linear equations by row-reducing the augmented matrix for the system.

#

Our new algorithm for computing gcd( a , b ) is analogous to Gaussian elimination, but we are only allowed to use the integer versions of the elementary row operations. Given nonzero integers a and b as input, we begin with the matrix

[ 1 0 a 0 1 b ] .

Next we perform a sequence of row operations whose goal is to produce a zero in the rightmost column. For example, we could start by dividing a by b , obtaining a quotient q 1 and a remainder r 1 such that a = bq 1 + r 1 and 0 ≤ r 1 < | b |. Adding − q 1 times row 2 to row 1 produces the new matrix

[ 1 q 1 r 1 0 1 b ] .
If r 1 is nonzero, we could continue by dividing b by r 1 , obtaining q 2 and r 2 with b = r 1 q 2 + r 2 and 0 ≤ r 2 < | r 1 |. Adding − q 2 times row 1 to row 2 leads to the matrix
[ 1 q 1 r 1 q 2 1 + q 1 q 2 r 2 ] .

We can continue in this way, following the same sequence of division steps used in ( 4.1 ) to row-reduce the matrix. We ultimately arrive at a matrix of the form

[ x y r k u v 0 ] ,

where a zero has appeared in the rightmost column. Remarkably, the entries x , y , and r k in the other row now satisfy r k = gcd( a , b ) = ax + by . A proof of this assertion is outlined in the exercises.

4.49. Example. We illustrate the matrix reduction algorithm and its connection to the previous algorithm by solving Example 4.48 once again. The row-reduction steps are shown here:

[ 1 0 999 0 1 101 ] R 1 9 R 2 [ 1 9 90 0 1 101 ] R 1 R 2 [ 1 9 90 1 10 11 ] R 1 8 R 2 [ 9 89 2 1 10 11 ] R 1 5 R 2 [ 9 89 2 46 455 1 ] R 1 2 R 2 [ 101 999 0 46 455 1 ]

Looking at row 2, we read off 1 = gcd(999, 101) = 999 · (−46) + 101 · 455.

The matrix reduction algorithm is really the same as Euclid’s algorithm based on repeated division, but the matrix formulation has several advantages. First, the algorithm both computes the gcd and writes it as a linear combination of the original inputs in a single computation; there is no need to backtrack through a chain of divisions to find x and y . Second, when executing the original algorithm by hand, it is easy to make arithmetic errors when substituting expressions into other expressions and using the distributive law. The matrix algorithm executes the same arithmetic via row operations, where mistakes are less likely. Third, we are free to use any row operations we like, not just those based on the divisions in ( 4.1 ). By choosing row operations wisely, we can sometimes reach the goal in fewer steps, as seen in the next example.

4.50. Example. We compute gcd(119, 21) as follows:

[ 1 0 119 0 1 21 ] R 1 6 R 2 [ 1 6 7 0 1 21 ] R 2 3 R 1 [ 1 6 7 3 17 0 ] R 1 × 1 [ 1 6 7 3 17 0 ]

We see that gcd(119, 21) = 7 = 119 · (−1) + 21 · 6. By subtracting six copies of R 2 at stage 1 instead of five copies, we made the gcd appear more quickly (up to a sign, which is easily removed). As additional savings, the row operation adding three copies of row 1 to row 2 is not really needed, once we notice that −7 divides 21 in the last column.

计算最大公约数的矩阵简化算法(可选) 在本可选章节中,我们将介绍另一种计算 d = gcd(a, b) 并找到整数 x, y 使得 d = ax + by 的算法。该算法通过对元素为整数的矩阵执行一些初等行变换来实现。允许的行变换有三种:(i) 交换矩阵中的两行;(ii) 将矩阵中的一行乘以 -1;(iii) 将矩阵中某一行的任意整数倍加到另一行。您可能在线性代数中学习过实值矩阵的类似初等行变换。在实数情况下,我们可以在变换 (iii) 中将某一行的任意实数倍加到另一行,并且在变换 (ii) 中可以将给定的行乘以任意非零实数。这些行变换在 Gaussian 消元算法中用于通过对增广矩阵进行行简化来求解线性方程组。我们计算 gcd(a, b) 的新算法类似于高斯消元法,但我们只能使用初等行变换的整数版本。给定非零整数 a 和 b 作为输入,我们从矩阵 [10a01b] 开始。接下来,我们执行一系列行变换,目标是在最右列生成零。例如,我们可以先将 a 除以 b,得到商 q 1 和余数 r 1 ,使得 a = bq 1 + r 1 且 0 ≤ r 1 < |b|。将第 2 行乘以 −q 1 的结果加到第 1 行,得到新矩阵 [1−q1r101b]。如果 r 1 非零,我们可以继续用 b 除以 r 1 ,得到 q 2 和 r 2 ,其中 b = r 1 q 2 + r 2 ,且 0 ≤ r 2 < |r 1 |。将第 1 行乘以 −q 2 得到矩阵 [1−q1r1−q21+q1q2r2]。我们可以继续这样做,遵循与 (4.1) 中相同的除法步骤顺序来简化矩阵。最终,我们得到形如 [xyrkuv0] 的矩阵,其中最右列出现了一个零。值得注意的是,另一行中的元素 x、y 和 r k 现在满足 r k = gcd(a, b) = ax + by。此断言的证明在练习中概述。4.49. 示例。我们再次求解示例 4.48,以说明矩阵简化算法及其与先前算法的联系。行简化步骤如下所示:[1099901101]→R1−9R2[1−99001101]→R1−R2[1−990−11011]→R1−8R2[9−892−11011]→R1−5R2[9−892−464551]→R1−2R2[101−9990−464551] 观察第2行,我们读出 1 = gcd(999, 101) = 999 · (−46) + 101 · 455。矩阵简化算法实际上与基于重复除法的欧几里得算法相同,但矩阵形式有几个优点。首先,该算法在一次计算中既计算了最大公约数,又将其表示为原始输入的线性组合;无需回溯一系列除法运算来找到 x 和 y。其次,手动执行原始算法时,在代入表达式和使用分配律时很容易出现算术错误。矩阵算法通过行运算执行相同的算术运算,出错的可能性更小。第三,我们可以自由使用任何行运算,而不仅仅局限于基于 (4.1) 中除法的运算。通过明智地选择行运算,我们有时可以用更少的步骤达到目标,如下例所示。4.50. 示例。我们按如下方式计算 gcd(119, 21):[101190121]→R1−6R2[1−6−70121]→R2−3R1[1−6−73−170]→R1×−1[−1673170]。我们看到 gcd(119, 21) = 7 = 119 · (−1) + 21 · 6。通过在第一阶段减去六个 R 2 而不是五个,我们更快地得到了 gcd(仅差一个符号,很容易去除)。此外,由于我们注意到最后一列中的 −7 可以整除 21,因此将第一行的三个副本添加到第二行的行操作实际上并不需要。

Section Summary

章节概要

  1. Facts about Divisors. For integers a and c , the notation c | a means c divides a (i.e., u Z , a = c u ). For all a , b , c , x , y Z , c divides a iff | c | divides | a |; if c divides a and a > 0, then c a ; if c | a and c | b , then c |( ax + by ).
    关于除数的事实。对于整数 a 和 c,记号 c|a 表示 c 整除 a(即,∃u∈Z,a=cu)。对于所有 a,b,c,x,y∈Z,c 整除 a 当且仅当 |c| 整除 |a|;如果 c 整除 a 且 a > 0,则 c ≤ a;如果 c|a 且 c|b,则 c|(ax + by)。
  2. Greatest Common Divisors. For integers a and b (not both zero), gcd( a , b ) is the largest integer d such that d | a and d | b . We have gcd( a , b ) = gcd(| a |, | b |); gcd( a , 0) = | a |; and if a = bq + r for some integers q , r , then gcd( a , b ) = gcd( b , r ). We can always write gcd( a , b ) = ax + by for some integers x , y ; we say that gcd( a , b ) is a linear combination of a and b .
    最大公约数。对于整数 a 和 b(不能同时为零),gcd(a, b) 是满足 d|a 且 d|b 的最大整数 d。我们有 gcd(a, b) = gcd(|a|, |b|);gcd(a, 0) = |a|;并且如果对于某些整数 q 和 r,a = bq + r,则 gcd(a, b) = gcd(b, r)。我们总可以将 gcd(a, b) 写成 ax + by,其中 x 和 y 为某些整数;我们称 gcd(a, b) 是 a 和 b 的线性组合。
  3. Euclid’s GCD Algorithm. To find the gcd of positive integers a and b , divide a by b to get a remainder r 1 ; divide b by r 1 to get a remainder r 2 ; divide r 1 by r 2 to get a remainder r 3 ; and so on. The last nonzero remainder r k is gcd( a , b ). Working backwards through the divisions, we can express r k as a linear combination of r k −1 and r k −2 , then as a linear combination of r k −2 and r k −3 , until eventually r k has been written as a linear combination of the original inputs a and b .
    欧几里得最大公约数算法。要求正整数 a 和 b 的最大公约数,首先将 a 除以 b 得到余数 r 1 ;将 b 除以 r 1 得到余数 r 2 ;将 r 1 除以 r 2 得到余数 r 3 ;依此类推。最后一个非零余数 r k 即为 gcd(a, b)。通过反向除法,我们可以将 r k 表示为 r k −1 和 r k −2 的线性组合,然后表示为 r k −2 和 r k −3 的线性组合,直到最终将 r k 表示为原始输入 a 和 b 的线性组合。
  4. Matrix Reduction Algorithm for GCDs. To find the gcd of positive integers a and b , start with the matrix [ 1 0 a 0 1 b ] . Repeatedly add integer multiples of one row to the other row (or multiply one row by −1) until a zero appears in the last column. If the other row has entries x , y , d with d > 0, then d = gcd( a , b ) = ax + by .
    求最大公约数的矩阵简化算法。要找到正整数 a 和 b 的最大公约数,首先得到矩阵 [10a01b]。重复地将一行的整数倍加到另一行(或将一行乘以 -1),直到最后一列出现零。如果另一行的元素为 x、y、d,且 d > 0,则 d = gcd(a, b) = ax + by。

Exercises

练习

  1. For each pair a , b below, compute gcd( a , b ) using repeated division.

    对于下面的每一对 a、b,使用重复除法计算 gcd(a, b)。

    (a) a = 98, b = 21 (b) a = 228, b = 168 (c) a = 100, b = 39 (d) a = 513, b = 252.

    (a)a = 98,b = 21 (b)a = 228,b = 168 (c)a = 100,b = 39 (d)a = 513,b = 252。

  2. For each pair a , b in the previous problem, find integers x , y with gcd( a , b ) = ax + by by working backwards through the chain of divisions.
    对于上一题中的每一对 a、b,通过逆向除法链找到整数 x、y,使得 gcd(a, b) = ax + by。
  3. For each pair a , b below, compute d = gcd( a , b ) and integers x , y with d = ax + by using the matrix reduction algorithm.

    对于下面的每一对 a、b,使用矩阵简化算法计算 d = gcd(a, b) 和整数 x、y,使得 d = ax + 。

    (a) a = 144, b = 89 (b) a = 516, b = 215 (c) a = 111111, b = 117845.

    (a)a = 144,b = 89 (b)a = 516,b = 215 (c)a = 1111,b = 117845。

  4. Prove: for all a , b Z , gcd( a , b ) = gcd(| a |, | b |).
    证明:对于所有 a,b∈Z,gcd(a, b) = gcd(|a|, |b|)。
  5. Explain how Euclid’s algorithm computes gcd( a , b ) in each of these special cases: (a) a = b = 0; (b) a = 0 ≠ b ; (c) b divides a ; (d) a divides b .
    解释欧几里得算法在下列每种特殊情况下如何计算 gcd(a, b):(a)a = b = 0;(b)a = 0 ≠ b;(c)b 整除 a;(d)a 整除 b。
  6. Prove by strong induction on b that the recursive Algorithm 4.42 always terminates in finitely many steps with the correct answer.
    用强归纳法证明,递归算法 4.42 总是在有限步内终止并得到正确答案。
  7. Use induction on n to prove: for all integers n > 0, d , x 1 , …, x n , a 1 , …, a n , if d | x i for all i ∈ {1, 2, …, n }, then d |( a 1 x 1 + · · · + a n x n ).
    使用归纳法证明:对于所有整数 n > 0,d,x 1 ,…,x n ,a 1 ,…,a n ,如果对于所有 i ∈ {1, 2, …, n},d|x i ,则 d|(a 1 x 1 + · · · + a n x n )。
  8. (a) Prove: for all positive integers a , b , gcd( b , a ) = gcd( a , b ) = gcd( a b , b ).

    (a)证明:对于所有正整数 a、b,gcd(b, a) = gcd(a, b) = gcd(a − b, b)。

    (b) Use part (a) to formulate a new recursive algorithm for computing gcds. Compare this algorithm to Euclid’s gcd algorithm.

    (b) 利用 (a) 部分的结果,设计一个新的递归算法来计算最大公约数。将该算法与欧几里得的最大公约数算法进行比较。

  9. Prove or disprove: for all positive integers a , b , gcd( a , b ) = gcd( a + b , a b ).
    证明或反证:对于所有正整数 a、b,gcd(a, b) = gcd(a + b, a − b)。
  10. Suppose p is prime and a Z . What are the possible values of gcd( a , p )? Under what conditions on a does each value occur?
    假设 p 是素数,a∈Z。gcd(a, p) 的可能取值有哪些?a 取每个值时需要满足什么条件?
  11. Prove: for all p , a , b Z , if p is prime and p divides ab , then p divides a or p divides b . [Use the previous exercise and the fact that gcd( a , p ) is a linear combination of a and p .]
    证明:对于任意 p, a, b∈Z,如果 p 是素数且 p 整除 ab,则 p 整除 a 或 p 整除 b。[利用前面的练习以及 gcd(a, p) 是 a 和 p 的线性组合这一事实。]
  12. For fixed n Z > 0 , find all possible values of gcd(8 n + 1, 5 n − 3) and determine when each value occurs. Also write the gcd as a linear combination of 8 n + 1 and 5 n − 3.
    对于固定的 n∈Z>0,求 gcd(8n + 1, 5n − 3) 的所有可能值,并确定每个值出现的时机。此外,将 gcd 表示为 8n + 1 和 5n − 3 的线性组合。
  13. (a) Prove: for all n Z > 0 , gcd( n 3 , n 2 + n + 1) = 1. (b) Write 1 as an explicit linear combination of n 3 and n 2 + n + 1.
    (a)证明:对于所有 n∈Z>0,gcd(n 3 , n 2 + n + 1) = 1。(b)将 1 写成 n 3 和 n 2 + n + 1 的显式线性组合。
  14. Prove: for all positive integers a , b , c , gcd( ca , cb ) = c · gcd( a , b ). [ Hint: Examine how Euclid’s Algorithm acts on inputs ca , cb compared to inputs a , b by multiplying each line of ( 4.1 ) by c .]
    证明:对于任意正整数 a、b、c,gcd(ca, cb) = c · gcd(a, b)。[提示:通过将式 (4.1) 的每一行乘以 c,来考察欧几里得算法在输入 ca、cb 和输入 a、b 上的表现。]
  15. Define the Fibonacci numbers by F 0 = 0, F 1 = 1, and F n = F n −1 + F n −2 for all n ≥ 2. (a) Prove by induction: for all n ≥ 1, gcd( F n , F n −1 ) = 1. (b) For 1 ≤ n ≤ 6, express 1 as a linear combination of F n and F n −1 . (c) Based on your answers to (b), conjecture and then prove a formula (valid for all n ≥ 1) expressing 1 as a linear combination of F n and F n −1 .
    定义斐波那契数列为 F 0 = 0,F 1 = 1,且对于所有 n ≥ 2,F n = F n −1 + F n −2 。(a)用数学归纳法证明:对于所有 n ≥ 1,gcd(F n , F n −1 ) = 1。(b)对于 1 ≤ n ≤ 6,将 1 表示为 F n 和 F n −1 的线性组合。(c)基于你对 (b) 的回答,猜想并证明一个公式(对所有 n ≥ 1 都成立),该公式将 1 表示为 F n 和 F n −1 的线性组合。
  16. Suppose a and b are fixed integers, A = [ x y z u v w ] is a matrix with integer entries such that ax + by = z and au + bv = w , and B = [ x y z u v w ] is obtained from A by performing one of the elementary row operations (i), (ii), or (iii) described on page 188. Prove that the entries of B are all integers, ax ′ + by ′ = z ′, au ′ + bv ′ = w ′, and gcd( z , w ) = gcd( z ′, w ′).
    假设 a 和 b 是固定的整数,A=[xyzuvw] 是一个元素均为整数的矩阵,满足 ax + by = z 且 au + bv = w,B=[x′y′z′u′v′w′] 是通过对 A 执行第 188 页所述的初等行变换 (i)、(ii) 或 (iii) 之一得到的。证明 B 的所有元素均为整数,ax′ + by′ = z′,au′ + bv′ = w′,且 gcd(z, w) = gcd(z′, w′)。
  17. Suppose a and b are fixed integers, A 0 = [ 1 0 a 0 1 b ] , and A 0 , A 1 , A 2 , …, A k is any finite sequence of matrices such that for 0 ≤ i < k , A i +1 is obtained from A i by an elementary row operation for integer matrices. Write A i = [ x i y i z i u i v i w i ] . Use induction on k and the previous exercise to prove that for all i between 0 and k , ax i + by i = z i , au i + bv i = w i and gcd( z i , w i ) = gcd( a , b ).
    假设 a 和 b 是固定的整数,A0=[10a01b],A 0 , A 1 , A 2 , …, A k 是任意有限矩阵序列,使得对于 0 ≤ i < k,A i +1 可以通过对整数矩阵进行初等行变换从 A i 得到。记 Ai=[xiyiziuiviwi]。利用对 k 的归纳法和前面的练习证明,对于 0 到 k 之间的所有 i,ax i + by i = z i ,au i + bv i = w i 并且 gcd(z i , w i ) = gcd(a, b)。
  18. Prove that the matrix reduction algorithm for finding gcd( a , b ) always terminates in finitely many steps with the correct answer, assuming that at each step we perform any elementary row operation that reduces the maximum absolute value n of the two entries in the third column. (Use strong induction on n .) Also explain why, at each step before the end, at least one such row operation must be possible.
    证明求 gcd(a, b) 的矩阵约简算法总能在有限步内终止并得到正确答案,假设每一步都执行一个初等行变换,使得第三列中两个元素的绝对值最大为 n。(对 n 使用强归纳法。)并解释为什么在终止前的每一步中,至少存在一个这样的行变换。
  19. Given a , b Z > 0 , write a = bq + r where q , r Z and 0 ≤ r < b . Prove: for all x Z > 1 , gcd( x a − 1, x b − 1) = gcd( x b − 1, x r − 1). ( Hint: See Exercise 21 in § 4.4 .)
    给定 a, b∈Z>0,令 a = bq + r,其中 q, r∈Z 且 0 ≤ r < b。证明:对于所有 x∈Z>1,gcd(x a − 1, x b − 1) = gcd(x b − 1, x r − 1)。(提示:参见 §4.4 中的练习 21。)
  20. Use the previous exercise to prove: for all positive integers a , b , x with x > 1, gcd( x a − 1, x b − 1) = x gcd( a , b ) − 1.
    利用前面的练习证明:对于所有正整数 a、b、x,其中 x > 1,gcd(x a − 1, x b − 1) = x gcd( a , b ) − 1。

4.6 GCDs and Uniqueness of Prime Factorizations

4.6 最大公约数与质因数分解的唯一性

In the last section, we looked at specific examples where gcd( a , b ) could be expressed in the form ax + by for certain integers x and y . Here we prove that such expressions always exist. We then use this result to obtain further facts about prime integers. In particular, we prove that the factorization of a positive integer into a product of primes is unique except for rearranging the order of the factors.

上一节中,我们考察了一些具体例子,其中 gcd(a, b) 可以表示为 ax + by 的形式,其中 x 和 y 为某些整数。本节证明这种表达式总是存在的。然后,我们利用这一结果推导出关于素数的更多结论。特别地,我们证明了正整数分解为素数乘积的唯一性(除了改变因子的顺序之外)。

Proof of the Linear Combination Property of GCDs

最大公约数线性组合性质的证明

We now prove the linear combination property of gcds for all integers a , b ≥ 0; an exercise asks you to generalize to the case where a or b could be negative.

现在我们证明对于所有整数 a, b ≥ 0,最大公约数的线性组合性质;一个练习要求你将其推广到 a 或 b 为负数的情况。

4.51. Theorem on Writing GCDs as Linear Combinations. For all integers a , b ≥ 0, there are integers x and y such that gcd( a , b ) = ax + by . Proof. To clarify the structure of the induction proof, we write the theorem statement in formal symbols as follows:

4.51. 将最大公约数表示为线性组合的定理。对于任意整数 a, b ≥ 0,存在整数 x 和 y,使得 gcd(a, b) = ax + by。证明。为了清晰地展示归纳证明的结构,我们将定理用形式符号写成如下形式:

b b Z Z 0 , a 一个 Z Z 0 , x x Z Z , y Z Z , gcd 最大公约数 ( a 一个 , b b ) = a 一个 x x + b b y .

We prove this statement by strong induction on b . Fix an arbitrary integer b Z 0 . [Next we must state the strong induction hypothesis. We change to new letters here to avoid a notation conflict between the assumed statement and the statement to be proved.] Assume: for all integers m in the range 0 ≤ m < b ,

我们用强归纳法证明这个命题。固定任意整数 b∈Z≥0。[接下来我们必须陈述强归纳假设。这里我们改用新的字母,以避免假设命题和待证明命题之间出现符号冲突。] 假设:对于所有在 0 ≤ m < b 范围内的整数 m,

c c Z Z 0 , s s Z Z , t t Z Z , gcd 最大公约数 ( c c , m ) = c c s s + m t t .

(4.6)

We must prove

我们必须证明

a 一个 Z Z 0 , x x Z Z , y Z Z , gcd 最大公约数 ( a 一个 , b b ) = a 一个 x x + b b y .

(4.7)

We know b = 0 or b > 0, so consider two cases.

我们知道 b = 0 或 b > 0,所以考虑两种情况。

Case 1. Assume b = 0; prove ( 4.7 ). Fix an arbitrary integer a Z 0 ; prove x Z , y Z , gcd ( a , 0 ) = a x + 0 y . We know gcd( a , 0) = a , so choose x = 1 and y = 0. Note 1 and 0 are in Z , and gcd( a , 0) = a = a · 1 + 0 · 0.

情况 1. 假设 b = 0;证明 (4.7)。固定任意整数 a∈Z≥0;证明 ∃x∈Z,∃y∈Z,gcd(a,0)=ax+0y。我们知道 gcd(a, 0) = a,所以选择 x = 1 和 y = 0。注意 1 和 0 都在 Z 中,且 gcd(a, 0) = a = a · 1 + 0 · 0。

Case 2. Assume b > 0; prove ( 4.7 ). Fix an arbitrary integer a Z 0 ; prove x Z , y Z , gcd ( a , b ) = a x + b y . Because b > 0, we can divide a by b , obtaining integers q , r with a = bq + r and 0 ≤ r < b . By the Theorem on Greatest Common Divisors, we know gcd( a , b ) = gcd( b , r ). The key observation is that 0 ≤ r < b , so that we are allowed to choose m = r in the induction hypothesis. By the Inference Rule for ALL, we can take c = b in ( 4.6 ), so that the induction hypothesis tells us there are integers s and t with gcd( b , r ) = bs + rt . We also know r = a bq , and hence

情况 2. 假设 b > 0;证明 (4.7)。固定任意整数 a∈Z≥0;证明存在 x∈Z, y∈Z, gcd(a,b)=ax+by。因为 b > 0,我们可以将 a 除以 b,得到整数 q 和 r,使得 a = bq + r 且 0 ≤ r < b。根据最大公约数定理,我们知道 gcd(a, b) = gcd(b, r)。关键在于 0 ≤ r < b,因此我们可以在归纳假设中选择 m = r。根据全集推理规则,我们可以在 (4.6) 中取 c = b,因此归纳假设告诉我们存在整数 s 和 t,使得 gcd(b, r) = bs + rt。我们还知道 r = a − bq,因此……

gcd 最大公约数 ( a 一个 , b b ) = gcd 最大公约数 ( b b , r r ) = b b s s + r r t t = b b s s + ( a 一个 b b q q ) t t = a 一个 t t + b b ( s s q q t t ) .

So, choosing x = t and y = s qt (which are integers since s , q , t Z ), we do have gcd( a , b ) = ax + by , as needed. This completes the induction proof.

因此,令 x = t 和 y = s − qt(由于 s, q, t∈Z,它们都是整数),我们确实有 gcd(a, b) = ax + by,符合要求。这就完成了归纳证明。

We can translate this proof into a recursive algorithm that takes inputs a , b Z 0 and returns d , x , y Z such that d = gcd( a , b ) = ax + by . The base case of the algorithm occurs when b = 0; then we return d = a , x = 1, and y = 0. Otherwise, when b > 0, divide a by b to get q , r Z with a = bq + r and 0 ≤ r < b . Recursively compute d , s , t with d = gcd( b , r ) = bs + rt ; and return d as the gcd, x = t , and y = s qt . Tracing through all the recursive calls, you can check that the net effect of this recursive algorithm agrees with the iterative computation presented in the last section, where we worked backwards through the chain of divisions to find x and y .

我们可以将此证明转化为一个递归算法,该算法接受输入 a, b∈Z≥0,并返回 d, x, y∈Z,其中 d = gcd(a, b) = ax + by。算法的基本情况是 b = 0;此时我们返回 d = a,x = 1,y = 0。否则,当 b > 0 时,将 a 除以 b 得到 q, r∈Z,其中 a = bq + r,0 ≤ r < b。递归地计算 d, s, t,其中 d = gcd(b, r) = bs + rt;并返回 d 作为最大公约数,x = t,y = s − qt。通过追踪所有递归调用,您可以验证此递归算法的最终结果与上一节中介绍的迭代计算一致,在上一节中,我们通过反向执行除法链来找到 x 和 y。

Properties of Prime Numbers

质数的性质

In § 4.4 , we proved that for all n Z > 1 , there exist factorizations of n into products of one or more primes. Our next major goal is to prove a uniqueness theorem about prime factorizations. The uniqueness of the factorizations is harder to prove than their existence, so we must first develop some facts about prime numbers. Recall that p Z > 1 is prime iff the only positive divisors of p are 1 and p .

在第 4.4 节中,我们证明了对于所有 n∈Z>1,n 都存在分解,可以分解成一个或多个素数的乘积。我们的下一个主要目标是证明关于素数分解的唯一性定理。分解的唯一性比分解的存在性更难证明,因此我们必须首先推导一些关于素数的性质。回想一下,p∈Z>1 是素数当且仅当 p 的正因子只有 1 和 p 两个。

4.52 Euclid’s Lemma on Divisibility by Primes For all p , a , b Z , if p is prime and p divides ab , then p divides a or p divides b .

4.52 欧几里得关于素数整除性的引理 对于所有 p,a,b∈Z,如果 p 是素数并且 p 整除 ab,那么 p 整除 a 或 p 整除 b。

Proof. Fix p , a , b Z . Assume p is prime and p divides ab . Prove p divides a or p divides b . [Use the OR-template to continue.] Assume p does not divide a ; prove p divides b . We have assumed p divides ab , which means ab = pc for some integer c . We must prove d Z , b = p d . [How do we proceed? We need to use our other assumptions: p is prime and p does not divide a . A key idea is that these assumptions allow us to compute gcd( p , a ).] Since p is prime, the only positive divisors of p are 1 and p . Thus the greatest common divisor of p and a could only be 1 or p . Since we assumed p is not a divisor of a , we must have gcd( p , a ) = 1. We now invoke Theorem 4.51 to conclude there exist integers x and y with 1 = ax + py . [We need one more trick that does not come from the proof templates: multiply both sides by b , to make ab appear on the right side.] Multiplying by b , using algebra, and recalling that ab = pc , we get

证明。固定 p, a, b∈Z。假设 p 是素数且 p 整除 ab。证明 p 整除 a 或 p 整除 b。[使用 OR 模板继续。] 假设 p 不整除 a;证明 p 整除 b。我们假设 p 整除 ab,这意味着对于某个整数 c,ab = pc。我们必须证明 ∃d∈Z, b=pd。[我们该如何进行?我们需要利用其他假设:p 是素数且 p 不整除 a。一个关键思想是,这些假设允许我们计算 gcd(p, a)。] 由于 p 是素数,p 的唯一正约数是 1 和 p。因此,p 和 a 的最大公约数只能是 1 或 p。由于我们假设 p 不整除 a,因此 gcd(p, a) 必须等于 1。现在我们利用定理 4.51 得出结论:存在整数 x 和 y,使得 1 = ax + py。我们还需要一个证明模板之外的技巧:两边同时乘以 b,使 ab 出现在右边。两边同时乘以 b,利用代数运算,并记住 ab = pc,我们得到

b b = ( a 一个 x x + p p y ) b b = a 一个 x x b b + p p y b b = ( a 一个 b b ) x x + p p y b b = ( p p c c ) x x + p p y b b = p p ( c c x x + y b b ) .

Choosing d to be the integer cx + yb , we have b = pd , as needed.

选择 d 为整数 cx + yb,则有 b = pd,符合要求。

4.53. Theorem on Divisibility by Primes For all k Z 1 and all p , a 1 , a 2 , , a k Z , if p is prime and p divides a 1 a 2 · · · a k , then for some i ∈ {1, 2, …, k }, p divides a i .

4.53. 素数整除定理 对于所有 k∈Z≥1 和所有 p,a1,a2,…,ak∈Z,如果 p 是素数并且 p 整除 a 1 a 2 · · · a k ,则对于某个 i ∈ {1, 2, …, k},p 整除 a i

Proof. We use ordinary induction on k .

证明。我们对 k 使用普通归纳法。

Base Case. Prove that for all p , a 1 Z , if p is prime and p divides a 1 , then for some i ∈ {1}, p divides a i . This statement is readily proved (choose i = 1).

基本情况。证明对于所有 p,a1∈Z,如果 p 是素数且 p 整除 a 1 ,则对于某个 i ∈ {1},p 整除 a i 。此命题很容易证明(选择 i = 1)。

Induction Step. Fix an arbitrary k Z 1 ; assume the theorem statement holds for this k ; prove the theorem statement holds for k + 1. Fix p , a 1 , , a k , a k + 1 Z ; assume p is prime and p divides a 1 · · · a k a k +1 ; prove there exists i ∈ {1, 2, …, k + 1} such that p divides a i . The key realization is that the number a 1 · · · a k a k +1 can be regarded as the product of the two integers a = a 1 · · · a k and b = a k +1 . Since p divides ab by assumption, the previous lemma applies to show that p divides a or p divides b . Now use cases.

归纳步骤。固定任意 k∈Z≥1;假设定理对该 k 成立;证明定理对 k+1 也成立。固定 p,a1,…,ak,ak+1∈Z;假设 p 为素数且 p 整除 a 1 · · · a k a k +1 ;证明存在 i ∈ {1, 2, …, k + 1} 使得 p 整除 a i 。关键在于,数 a 1 · · · a k a k +1 可以看作是两个整数 a = a 1 · · · a k 和 b = a k +1 的乘积。根据假设,p 整除 ab,因此前面的引理可以用来证明 p 整除 a 或 p 整除 b。现在用实例来说明。

Case 1. Assume p divides a . This means p divides a 1 a 2 · · · a k . By induction hypothesis, there exists i ∈ {1, 2, …, k } such that p divides a i , as needed.

情况 1. 假设 p 整除 a。这意味着 p 整除 a 1 a 2 · · · a k 。根据归纳假设,存在 i ∈ {1, 2, …, k} 使得 p 整除 a i ,满足要求。

Case 2. Assume p divides b , so p divides a k +1 . Choose i = k + 1; then p divides a i , as needed.

情况 2. 假设 p 整除 b,则 p 整除 a k +1 。选择 i = k + 1;则 p 整除 a i ,根据需要。

Uniqueness of Prime Factorizations

质因数分解的唯一性

Now we are ready to prove the uniqueness of prime factorizations. The next result is sometimes called the Fundamental Theorem of Arithmetic.

现在我们可以证明质因数分解的唯一性了。下一个结果有时被称为算术基本定理。

4.54. Theorem on Existence and Uniqueness of Prime Factorizations.

4.54. 素因子分解的存在性和唯一性定理。

(a) Every integer n > 1 can be written as the product of one or more primes.

(a)每个大于 1 的整数 n 都可以写成一个或多个素数的乘积。

(b) For all n Z > 1 , all s , t Z 1 , and all primes p 1 , …, p s , q 1 , …, q t , if n = p 1 p 2 · · · p s = q 1 q 2 · · · q t , then s = t and we can reorder the q j so that for all i ∈ {1, 2, …, s }, p i = q i . Proof. Part (a) was proved in § 4.4 . We prove part (b) by strong induction on n . Fix an integer n Z > 1 , and assume part (b) holds for all n ′ in the range 1 < n ′ < n . Prove the statement in part (b) holds for n . Fix s , t Z 1 and primes p 1 , …, p s , q 1 , …, q t , and assume

(b) 对于所有 n∈Z>1,所有 s,t∈Z≥1,以及所有素数 p 1 , …, p s , q 1 , …, q t ,如果 n = p 1 p 2 · · · p s = q 1 q 2 · · · q t ,则 s = t,我们可以重新排列 q j ,使得对于所有 i ∈ {1, 2, …, s},p i = q i 。证明。(a) 部分已在 §4.4 中证明。我们对 n 进行强归纳法证明 (b) 部分。固定整数 n∈Z>1,并假设 (b) 部分对所有 n′ 在 1 < n′ < n 的范围内成立。证明 (b) 部分的结论对 n 成立。固定 s,t∈Z≥1 和素数 p 1 , …, p s , q 1 , …, q t ,并假设

n n = p p 1 p p 2 p p s s = q q 1 q q 2 q q t t .

We know s = 1 or t = 1 or ( s > 1 and t > 1), so use cases.

我们知道 s = 1 或 t = 1 或 (s > 1 且 t > 1),因此有用例。

Case 1. Assume s = 1. Then n = p 1 = q 1 · · · q t . In this case, t must be 1, since otherwise p 1 = q 1 ( q 2 · · · q t ) expresses p 1 as a product of the integers q 1 and q 2 · · · q t , which are both strictly between 1 and p 1 . This is impossible, since p 1 is prime. Knowing that t = 1, we now have s = t = 1 and p 1 = n = q 1 .

情况 1. 假设 s = 1。则 n = p 1 = q 1 · · · q t 。在这种情况下,t 必须为 1,否则 p 1 = q 1 (q 2 · · · q t ) 会将 p 1 表示为整数 q 1 和 q 2 · · · q t 的乘积,而这两个整数都严格介于 1 和 p 1 之间。这是不可能的,因为 p 1 是素数。已知 t = 1,则 s = t = 1,且 p 1 = n = q 1

Case 2 . Assume t = 1. As in case 1, we see that s = 1 (because q 1 is prime), so s = t and p 1 = n = q 1 .

情况 2. 假设 t = 1。与情况 1 一样,我们看到 s = 1(因为 q 1 是素数),所以 s = t 且 p 1 = n = q 1

Case 3. Assume s > 1 and t > 1. On one hand, since n = p 1 ( p 2 · · · p s ), we see that p 1 divides n . On the other hand, n = q 1 · · · q t , so the prime p 1 divides the product q 1 · · · q t . By the Theorem on Divisibility by Primes, p 1 divides some q j where 1 ≤ j t . By reordering the q i if needed, we can arrange that p 1 divides q 1 . But q 1 is prime, so its only positive divisors are 1 and q 1 . As p 1 > 1, we must have p 1 = q 1 . Now we know

情况 3. 假设 s > 1 且 t > 1。一方面,由于 n = p 1 (p 2 · · · p s ),我们可以看出 p 1 整除 n。另一方面,n = q 1 · · · q t ,因此素数 p 1 整除乘积 q 1 · · · q t 。根据素数整除定理,p 1 整除某个 q j ,其中 1 ≤ j ≤ t。如果需要,我们可以重新排列 q i ,使 p 1 整除 q 1 。但 q 1 是质数,所以它的正约数只有 1 和 q 1 。由于 p 1 > 1,我们必须有 p 1 = q 1 。现在我们知道

p p 1 ( p p 2 p p s s ) = n n = q q 1 q q 2 q q t t = p p 1 ( q q 2 q q t t ) .

Dividing both sides by the nonzero integer p 1 , we obtain a new integer

两边同时除以非零整数 p 1 ,我们得到一个新的整数

n n = p p 2 p p s s = q q 2 q q t t .

Note n ′ < n since p 1 > 1, whereas n ′ > 1 since s > 1 and p 2 > 1. Thus, n ′ is an integer in the range 1 < n ′ < n , so we can apply the induction hypothesis to n ′. We have written n ′ as the product of s − 1 primes p 2 · · · p s and also as the product of t − 1 primes q 2 · · · q t . By the induction hypothesis, we conclude that s − 1 = t − 1 and q 2 , …, q t is a rearrangement of p 2 , …, p s . Since also p 1 = q 1 , we see that s = t and q 1 , …, q t is a rearrangement of p 1 , …, p s . This completes the induction proof.

注意,由于 p 1 > 1,所以 n′ < n;而由于 s > 1 且 p 2 > 1,所以 n′ > 1。因此,n′ 是 1 < n′ < n 范围内的整数,所以我们可以将归纳假设应用于 n′。我们将 n′ 写成 s − 1 个素数 p 2 · · · p s 的乘积,以及 t − 1 个素数 q 2 · · · q t 的乘积。根据归纳假设,我们得出 s − 1 = t − 1,且 q 2 , …, q t 是 p 2 , …, p s 的重排。由于 p 1 = q 1 ,我们得出 s = t,且 q 1 , …, q t 是 p 1 , …, p s 的重排。归纳证明完毕。

The Fundamental Theorem of Arithmetic can be rephrased as follows.

算术基本定理可以改写如下。

4.55. Theorem on Prime Factorization of Integers. Let p 1 < p 2 < p 3 < · · · be a list of all the distinct primes. For every nonzero integer n , there exist unique integers s ∈ { + 1, − 1} and e 1 , e 2 , , e j , Z 0 such that all but finitely many e j are zero, and

4.55. 整数素因数分解定理。设 p 1 < p 2 < p 3 < · · · 为所有不同素数的列表。对于每个非零整数 n,存在唯一的整数 s ∈ { + 1, − 1} 和 e1,e2,…,ej,…∈Z≥0,使得除有限个 e j 外,其余所有 e 均为零,并且

n n = s s p p 1 e e 1 p p 2 e e 2 p p j j e e j j = s s j j 1 p p j j e e j j .

Proof. Let n be a fixed nonzero integer. We know n > 1 or n = 1 or n < 0, so use cases.

证明。设 n 为一个固定的非零整数。我们知道 n > 1 或 n = 1 或 n < 0,因此存在这些用例。

Case 1. Assume n > 1. To prove existence of the factorization, use the earlier theorem to write n as a product of one or more primes, say n = q 1 q 2 · · · q k . Now choose s = +1, and for all j ≥ 1, let e j be the number of times the prime p j appears in the list q 1 , q 2 , …, q k . Here, all but finitely many e j are zero, and n = s j 1 p j e j since multiplication of integers is commutative. To prove uniqueness, suppose we also had n = s j 1 p j e j where s ′ and e j ′ satisfy the conditions in the theorem statement. We must have s ′ = +1 = s since n > 0 and every p j > 0. If some e j ′ were unequal to e j , then we would have two prime factorizations n = j 1 p j e j = j 1 p j e j in which the two lists of primes were not rearrangements of each other, because p j occurs a different number of times in the two lists. This violates the uniqueness property in the earlier theorem. So e j ′ = e j for all j , completing the proof of Case 1.

情况 1. 假设 n > 1。为了证明因式分解的存在性,利用之前的定理将 n 表示为一个或多个素数的乘积,例如 n = q 1 q 2 · · · q k 。现在选择 s = +1,对于所有 j ≥ 1,令 e j 表示素数 p j 在列表 q 1 , q 2 , …, q k 中出现的次数。这里,除了有限个 e j 之外,其余的都为零,并且由于整数乘法满足交换律,所以 n=s∏j≥1pjej。为了证明唯一性,假设我们还有 n=s′∏j≥1pjej′,其中 s′ 和 e j ′ 满足定理陈述中的条件。由于 n > 0 且每个 p j > 0,因此 s′ = +1 = s。如果某个 e j ′ 不等于 e j ,那么我们将有两个质因数分解 n=∏j≥1pjej=∏j≥1pjej′,其中两个质数列表并非彼此的重排,因为 p j 在两个列表中出现的次数不同。这违反了先前定理中的唯一性性质。因此,对于所有 j,e j ′ = e j ,从而完成了情况 1 的证明。

Case 2. Assume n = 1. For existence, choose s = 1 and every e j = 0. For uniqueness, note that this is the only choice of s and e j that produces a positive product less than 2.

情况 2. 假设 n = 1。为了存在性,选择 s = 1 和每个 e j = 0。为了唯一性,注意这是 s 和 e j 的唯一选择,使得乘积小于 2。

Case 3. Assume n < 0. This case follows by applying the existence and uniqueness results already proved to − n > 0; the only required modification is to take s = −1 instead of s = +1.

#

情况 3. 假设 n < 0。这种情况可以通过将已证明的存在性和唯一性结果应用于 −n > 0 得到;唯一需要修改的是将 s = −1 代替 s = +1。

Section Summary

章节概要

  1. Linear Combination Property of GCDs. For all a , b Z , there are integers x , y Z with d = gcd( a , b ) = ax + by . To find d , x , y when a > 0 and b = 0, let d = a , x = 1, y = 0. When a , b > 0, write a = bq + r for q , r Z with 0 ≤ r < b ; recursively find d , s , t with d = gcd( b , r ) = bs + rt ; then gcd( a , b ) = d , x = t , and y = s qt .
    最大公约数的线性组合性质。对于任意整数 a, b∈Z,存在整数 x, y∈Z,使得 d = gcd(a, b) = ax + by。当 a > 0 且 b = 0 时,为了找到 d、x、y,令 d = a,x = 1,y = 0。当 a, b > 0 时,令 a = bq + r,其中 q, r∈Z,且 0 ≤ r < b;递归地找到 d、s、t,使得 d = gcd(b, r) = bs + rt;然后 gcd(a, b) = d,x = t,y = s − qt。
  2. Properties of Primes. For prime p and a Z , gcd( a , p ) is p if p divides a and is 1 if p does not divide a . Whenever a prime p divides a product a 1 a 2 · · · a k of integers, p must divide some a i . Every integer n > 1 can be written as a product of prime integers. The prime factorization is unique in the following sense: if n = p 1 · · · p s = q 1 · · · q t with all p i and q j prime, then s = t and q 1 , …, q t is a rearrangement of p 1 , …, p s .
    素数的性质。对于素数 p 和 a∈Z,如果 p 整除 a,则 gcd(a, p) 为 p;如果 p 不整除 a,则 gcd(a, p) 为 1。当素数 p 整除整数乘积 a 1 a 2 · · · a k 时,p 必定整除某个整数 a i 。每个大于 1 的整数 n 都可以表示为若干个素数的乘积。质因数分解的唯一性体现在以下方面:如果 n = p 1 · · · p s = q 1 · · · q t ,其中所有 p i 和 q j 均为质数,则 s = t,并且 q 1 , …, q t 是 p 1 , …, p s 的重新排列。
  3. Fundamental Theorem of Arithmetic. Existence and uniqueness of prime factorizations can be reformulated as follows. Any nonzero integer n can be expressed uniquely in the form s j 1 p j e j , where s is +1 or −1, the e j are nonnegative integers, all but finitely many e j are zero, and p 1 < p 2 < · · · < p j < · · · is a complete list of prime integers.
    算术基本定理。质因数分解的存在性和唯一性可以重新表述如下。任何非零整数 n 都可以唯一地表示为 s∏j≥1pjej 的形式,其中 s 为 +1 或 −1,e j 为非负整数,除有限个 e j 外,其余均为零,且 p 1 < p 2 < · · · < p j < · · · 是质数的完备列表。

Exercises

练习

  1. For each a , b , find d = gcd( a , b ) and integers x , y with d = ax + by using the recursive algorithm from this section. (a) a = 693, b = 525 (b) a = 999, b = 629 (c) a = 34, b = 21.
    对于每个 a、b,求 d = gcd(a, b) 以及满足 d = ax + 的整数 x、y,使用本节中的递归算法。(a) a = 693,b = 525 (b) a = 999,b = 629 (c) a = 34,b = 21。
  2. Extend Theorem 4.51 to the case where a or b could be negative.
    将定理 4.51 推广到 a 或 b 为负数的情况。
  3. (a) Use the known identity gcd( a , b ) = gcd( a b , b ) and strong induction on a + b to give another proof of Theorem 4.51. (b) Translate this proof into a recursive algorithm for finding d , x , y with d = gcd( a , b ) = ax + by , and illustrate this algorithm on inputs a = 30, b = 8.
    (a)利用已知的恒等式 gcd(a, b) = gcd(a − b, b) 和对 a + b 的强归纳法,给出定理 4.51 的另一个证明。(b)将此证明转化为一个递归算法,用于找到 d、x、y,其中 d = gcd(a, b) = ax + by,并用输入 a = 30、b = 8 来说明该算法。
  4. Prove: For all a , b Z , gcd( a , b ) = 1 if and only if there exist x , y Z with ax + by = 1.
    证明:对于所有 a,b∈Z,gcd(a, b) = 1 当且仅当存在 x,y∈Z 使得 ax + by = 1。
  5. For all a , b , d Z , we proved: if gcd( a , b ) = d , then x , y Z , ax + by = d . Is the converse of this statement always true? Prove your answer, and compare to the previous exercise.
    对于任意 a, b, d∈Z,我们证明了:如果 gcd(a, b) = d,则存在 x, y∈Z,ax + by = d。这个命题的逆命题是否总是成立?证明你的答案,并与前面的练习进行比较。
  6. Prove: for all a , b , c Z , c is a common divisor of a and b iff c divides gcd( a , b ). [This conclusion strengthens the original definition of gcd( a , b ), which initially tells us only that c is less than or equal to gcd( a , b ).]
    证明:对于任意 a, b, c∈Z,c 是 a 和 b 的公约数当且仅当 c 整除 gcd(a, b)。[此结论强化了 gcd(a, b) 的原始定义,该定义最初仅表明 c 小于或等于 gcd(a, b)。]
  7. Suppose p Z > 1 is a fixed integer with the following property: for all a , b Z , if p divides ab , then p divides a or p divides b . Must p be prime? Prove your answer.
    假设 p∈Z>1 是一个固定的整数,且满足以下性质:对于任意 a,b∈Z,如果 p 整除 ab,则 p 也整除 a 或 p 也整除 b。p 一定是素数吗?请证明你的答案。
  8. Prove: for all a , b , c Z , if gcd( b , c ) = 1 and c | ab , then c | a .
    证明:对于所有 a、b、c∈Z,如果 gcd(b, c) = 1 且 c|ab,则 c|a。
  9. Prove: for all q Q , there exists unique ( m , n ) Z × Z with n > 0 and q = m / n and gcd( m , n ) = 1. [This proves that every rational number has a unique representation as a fraction in lowest terms .]
    证明:对于所有 q∈Q,存在唯一的 (m,n)∈Z×Z,其中 n > 0,q = m/n,且 gcd(m, n) = 1。[这证明了每个有理数都可以唯一地表示为最简分数。]
  10. (a) Prove: for all r , s , t Z , if gcd( r , t ) = 1 = gcd( s , t ), then gcd( rs , t ) = 1. (b) Prove by induction on n > 0: for all integers r 1 , r 2 , …, r n , t , if gcd( r i , t ) = 1 for all i ∈ {1, 2, …, n }, then gcd ( i = 1 n r i , t ) = 1 .
    (a) 证明:对于所有 r,s,t∈Z,如果 gcd(r, t) = 1 = gcd(s, t),则 gcd(rs, t) = 1。(b) 对 n > 0 进行归纳证明:对于所有整数 r 1 , r 2 , …, r n , t,如果对于所有 i ∈ {1, 2, …, n},gcd(r i , t) = 1,则 gcd(∏i=1nri,t)=1。
  11. Give a new proof of Theorem 4.51 based on Exercise 18 of § 4.1 , as follows. Fix a , b Z > 0 , and define S = { a x + b y : x , y Z } Z > 0 . Show that S has a least element d , and d = gcd( a , b ).
    根据 §4.1 的练习 18,给出定理 4.51 的新证明如下。固定 a,b∈Z>0,并定义 S={ax+by:x,y∈Z}∩Z>0。证明 S 有最小元素 d,且 d = gcd(a, b)。
  12. (a) Given a , b Z , show that the integers x , y Z such that gcd( a , b ) = ax + by are not uniquely determined by the pair ( a , b ). In fact, show that for each pair ( a , b ) Z × Z , there are infinitely many pairs ( x , y ) Z × Z with d = gcd( a , b ) = ax + by . (b) Given fixed integers a , b , x 0 , y 0 with 1 = gcd( a , b ) = ax 0 + by 0 , explicitly describe (with proof) the set of all pairs of integers ( x , y ) solving 1 = ax + by .
    (a) 给定 a, b∈Z,证明满足 gcd(a, b) = ax + by 的整数对 x, y∈Z 并非由 (a, b) 唯一确定。事实上,证明对于每一对 (a, b)∈Z×Z,都存在无穷多对 (x, y)∈Z×Z 满足 gcd(a, b) = ax + by。(b) 给定固定的整数 a, b, x 0 , y 0 ,且满足 gcd(a, b) = ax 0 + by 0 ,显式地描述(并证明)所有满足 gcd(a, b) = ax + by 的整数对 (x, y) 的集合。
  13. Prove that the iterative method for finding x , y with gcd( a , b ) = ax + by (based on working backwards through the divisions in ( 4.1 )) always gives the same x , y as the recursive algorithm described after the proof of Theorem 4.51.
    证明:对于满足 gcd(a, b) = ax + by 的 x, y,通过 (4.1) 中的除法逆向推导,迭代法总是能得到与定理 4.51 证明之后描述的递归算法相同的 x, y。
  14. The gcd of a finite list of positive integers a 1 , …, a n is the largest integer d such that d | a i for all i between 1 and n . (It can be shown that such a d exists.) Use induction to prove: for all n Z 2 and all a 1 , , a n Z > 0 , gcd( a 1 , …, a n ) = gcd(gcd( a 1 , …, a n −1 ), a n ).
    有限个正整数列表 a 1 , …, a n 的最大公约数是最大的整数 d,使得对于所有介于 1 和 n 之间的 i,d|a i 。(可以证明这样的 d 存在。)使用归纳法证明:对于所有 n∈Z≥2 和所有 a1,…,an∈Z>0,gcd(a 1 , …, a n ) = gcd(gcd(a 1 , …, a n −1 ), a n )。
  15. Continuing the previous exercise, prove that for all positive integers n , a 1 , …, a n , there exist integers x 1 , …, x n with gcd( a 1 , …, a n ) = a 1 x 1 + · · · + a n x n .
    继续前面的练习,证明对于所有正整数 n,a 1 , …, a n ,存在整数 x 1 , …, x n ,使得 gcd(a 1 , …, a n ) = a 1 x 1 + · · · + a n x n
  16. Describe an algorithm that takes a 1 , , a n Z > 0 as input and computes d , x 1 , , x n Z with d = gcd( a 1 , …, a n ) = a 1 x 1 + · · · + a n x n . Prove that your algorithm terminates with the correct answer.
    描述一个算法,该算法以 a1,…,an∈Z>0 作为输入,并计算 d,x1,…,xn∈Z,其中 d = gcd(a 1 , …, a n ) = a 1 x 1 + · · · + a n x n 。证明你的算法终止并得到正确答案。

4.7 Consequences of Prime Factorization (Optional)

4.7 质因数分解的后果(可选)

This optional section explores some consequences of the Fundamental Theorem of Arithmetic. Recall that this theorem expresses each nonzero integer n in the form n = s j 1 p j e j , where p 1 < p 2 < · · · < p j < · · · is the list of all prime numbers, and the integers s ∈ { + 1, − 1} and e 1 , e 2 , , e j , Z 0 are uniquely determined by n . Furthermore, only finitely many e j are nonzero. To emphasize that s and all e j depend on n and are uniquely determined by n , we write s = s ( n ) and e j = e j ( n ). So for all nonzero integers n ,

本可选章节探讨算术基本定理的一些推论。回顾一下,该定理将每个非零整数 n 表示为 n=s∏j≥1pjej 的形式,其中 p 1 < p 2 < · · · < p j < · · · 是所有素数的列表,整数 s ∈ { + 1, − 1} 和 e1,e2,…,ej,…∈Z≥0 由 n 唯一确定。此外,只有有限个 e j 非零。为了强调 s 和所有 e j 都依赖于 n 并且由 n 唯一确定,我们记 s = s(n) 和 e j = e j (n)。因此,对于所有非零整数 n,

n n = s s ( n n ) j j 1 p p j j e e j j ( n n ) .

Multiplication Rule for Prime Factorizations

质因数分解的乘法法则

The following result shows how the prime factorization of the product of two nonzero integers is related to the factorizations of each separate integer. Intuitively, the result reduces multiplication of nonzero integers to the addition of the vectors of exponents in the prime factorizations.

以下结果展示了两个非零整数乘积的质因数分解与每个整数的质因数分解之间的关系。直观地说,该结果将非零整数的乘法简化为质因数分解中指数向量的加法。

4.56. Theorem on Products of Integers. For all nonzero n , m Z , s ( nm ) = s ( n ) s ( m ) and e j ( nm ) = e j ( n ) + e j ( m ) for all j Z 1 . Proof. Fix nonzero n , m Z . We have the factorizations

4.56. 整数乘积定理。对于所有非零的 n,m∈Z,有 s(nm) = s(n)s(m) 且对于所有 j∈Z≥1,有 e j (nm) = e j (n) + e j (m)。证明。固定非零的 n,m∈Z。我们有分解式。

n n = s s ( n n ) j j 1 p p j j e e j j ( n n ) , m = s s ( m ) j j 1 p p j j e e j j ( m ) .

Multiplying these together, we obtain

将它们相乘,我们得到

n n m = s s ( n n ) s s ( m ) j j 1 p p j j e e j j ( n n ) + e e j j ( m ) .

(4.8)

On the other hand, we know the nonzero integer nm has a unique expression

另一方面,我们知道非零整数 nm 具有唯一表达式。

n n m = s s ( n n m ) j j 1 p p j j e e j j ( n n m )

(4.9)

where s ( nm ) ∈ { + 1, − 1}, each e j ( n m ) Z 0 , and only finitely many values e j ( nm ) are nonzero. Since s ( n ) s ( m ) ∈ { + 1, − 1}, every e j ( n ) + e j ( m ) is a nonnegative integer, and at most finitely many numbers e j ( n ) + e j ( m ) are nonzero, ( 4.8 ) satisfies the conditions required in ( 4.9 ). By uniqueness, we must therefore have s ( nm ) = s ( n ) s ( m ) and e j ( nm ) = e j ( n ) + e j ( m ) for all integers j ≥ 1.

其中 s(nm) ∈ { + 1, − 1},每个 ej(nm)∈Z≥0,且只有有限个值 e j (nm) 非零。由于 s(n)s(m) ∈ { + 1, − 1},每个 e j (n) + e j (m) 都是非负整数,且至多有有限个数 e j (n) + e j (m) 非零,因此 (4.8) 满足 (4.9) 中要求的条件。根据唯一性,我们必须有 s(nm) = s(n)s(m) 且 e j (nm) = e j (n) + e j (m) 对于所有整数 j ≥ 1。

Characterization of Divisors

除数的特征

The next theorem shows how to use prime factorizations to detect when one integer is a divisor (or multiple) of another integer.

下一个定理展示了如何使用质因数分解来检测一个整数何时是另一个整数的约数(或倍数)。

4.57. Theorem on Factorization of Divisors. For all nonzero integers a and b , a is a divisor of b iff for all j Z 1 , e j ( a ) ≤ e j ( b ).

4.57. 除数分解定理。对于所有非零整数 a 和 b,a 是 b 的除数当且仅当对于所有 j∈Z≥1,e j (a) ≤ e j (b)。

Proof. Fix a , b Z 0 . First assume a is a divisor of b . Then there exists an integer c with b = ac ; we have c ≠ 0 since b ≠ 0. Fix j Z 1 . By the multiplication rule for prime factorizations, e j ( b ) = e j ( a ) + e j ( c ) ≥ e j ( a ) + 0, so e j ( a ) ≤ e j ( b ).

证明。固定 a, b∈Z≠0。首先假设 a 是 b 的一个因子。那么存在一个整数 c 使得 b = ac;由于 b ≠ 0,所以 c ≠ 0。固定 j∈Z≥1。根据质因数分解的乘法规则,e j (b) = e j (a) + e j (c) ≥ e j (a) + 0,所以 e j (a) ≤ e j (b)。

Conversely, now assume that for all j Z 1 , e j ( a ) ≤ e j ( b ). Prove a | b , which means d Z , b = a d . Choose

反之,现在假设对于所有 j∈Z≥1,e j (a) ≤ e j (b)。证明 a|b,这意味着存在 d∈Z,b=ad。选择

d d = s s j j 1 p p j j e e j j ( b b ) e e j j ( a 一个 ) ,

where s = 1 if a and b have the same sign, and s = −1 otherwise. Note that d is an integer, since the exponent of each p j is the nonnegative integer e j ( b ) − e j ( a ), and at most finitely many of these exponents can be nonzero. On one hand, the choice of s shows that b and ad have the same sign. On the other hand, for all j Z 1 , e j ( ad ) = e j ( a ) + e j ( d ) = e j ( a ) + ( e j ( b ) − e j ( a )) = e j ( b ). So b = ad , as needed.

其中,当 a 和 b 同号时,s = 1;否则,s = −1。注意,d 为整数,因为每个 p j 的指数是非负整数 e j (b) − e j (a),且这些指数至多有有限个非零。一方面,s 的选择表明 b 和 ad 同号。另一方面,对于所有 j∈Z≥1,e j (ad) = e j (a) + e j (d) = e j (a) + (e j (b) − e j (a)) = e j (b)。所以,b = ad,根据需要。

Prime Factorization of GCDs and LCMs

最大公约数和最小公倍数的质因数分解

Our next result shows how to compute the greatest common divisor of two nonzero integers using their prime factorizations. First, we need one new definition.

我们的下一个结果展示了如何利用两个非零整数的质因数分解来计算它们的最大公约数。首先,我们需要一个新的定义。

4.58. Definition: Minimum. For all x , y R , define min ( x , y ) = x if x y , and min ( x , y ) = y if y < x .

4.58. 定义:最小值。对于所有 x,y∈R,定义 min(x, y) = x 当 x ≤ y 时,min(x, y) = y 当 y < x 时。

This definition says that min ( x , y ) is the smaller of the two numbers x and y . You are asked to prove the next lemma in the exercises.

这个定义表明 min(x, y) 是 x 和 y 中较小的那个。练习中要求你证明下一个引理。

4.59. Lemma on the Minimum For all x , y , z R , z ≤ min ( x , y ) iff z x and z y .

4.59. 最小引理 对于所有 x,y,z∈R,z ≤ min (x, y) 当且仅当 z ≤ x 且 z ≤ y。

The next theorem shows that we can calculate greatest common divisors by comparing prime factorizations and taking the minimum exponent of each prime.

下一个定理表明,我们可以通过比较质因数分解并取每个质数的最小指数来计算最大公约数。

4.60. Theorem on Prime Factorization of GCDs For all nonzero integers a and b , gcd ( a , b ) = j 1 p j min ( e j ( a ) , e j ( b ) ) .

4.60. 关于最大公约数素因数分解的定理 对于所有非零整数 a 和 b,gcd(a,b)=∏j≥1pjmin(ej(a),ej(b))。

Proof. Fix nonzero integers a and b , and let d = j 1 p j min ( e j ( a ) , e j ( b ) ) . First we show d is a common divisor of a and b . For every j Z 1 , e j ( d ) = min ( e j ( a ), e j ( b )) ≤ e j ( a ), so d | a by the Theorem on Factorization of Divisors. Similarly, since e j ( d ) ≤ e j ( b ) for all j , d | b follows. To finish the proof, we show that d is the largest common divisor of a and b . Fix any integer f such that f | a and f | b . We prove that f d by showing the stronger statement f | d . Because f | a and f | b , e j ( f ) ≤ e j ( a ) and e j ( f ) ≤ e j ( b ) for all j ≥ 1. By the Lemma on the Minimum, e j ( f ) ≤ min ( e j ( a ), e j ( b )) = e j ( d ) for all j . This implies f | d by the Theorem on Factorization of Divisors. We have now proved that d = gcd( a , b ).

证明。固定非零整数 a 和 b,令 d=∏j≥1pjmin(ej(a),ej(b))。首先,我们证明 d 是 a 和 b 的公约数。对于任意 j∈Z≥1,e j (d) = min (e j (a), e j (b)) ≤ e j (a),因此根据除数分解定理,d|a。类似地,由于对于所有 j,e j (d) ≤ e j (b),因此 d|b。为了完成证明,我们证明 d 是 a 和 b 的最大公约数。固定任意整数 f,使得 f|a 且 f|b。我们通过证明更强的结论 f|d 来证明 f ≤ d。因为 f|a 且 f|b,所以对于所有 j ≥ 1,e j (f) ≤ e j (a) 且 e j (f) ≤ e j (b)。根据最小值引理,对于所有 j,e j (f) ≤ min (e j (a), e j (b)) = e j (d)。根据除数分解定理,这蕴含 f|d。至此,我们已证明 d = gcd(a, b)。

We can extend the definitions of gcd and min to apply to k inputs instead of two inputs (see Exercise 14 of § 4.6 ). The proof above extends to show

我们可以将最大公约数和最小值的定义扩展到 k 个输入而不是两个输入(参见 §4.6 的练习 14)。上述证明可以推广到以下情况:

gcd 最大公约数 ( a 一个 1 , , a 一个 k k ) = j j 1 p p j j min 最小 ( e e j j ( a 一个 1 ) , , e e j j ( a 一个 k k ) ) .

A similar analysis is possible for the least common multiple (lcm) of two or more integers. By definition, 1cm( a 1 , … , a k ) is the least positive integer d such that every a i divides d (we assume every a i is nonzero here). The proof for gcds can be adapted to show

对于两个或多个整数的最小公倍数(lcm),也可以进行类似的分析。根据定义,1cm(a 1 , … ,a k ) 是最小的正整数 d,使得每个 a i 都能整除 d(这里我们假设每个 a i 都非零)。最大公约数的证明方法可以稍作修改来证明

1cm 1厘米 ( a 一个 1 , , a 一个 k k ) = j j 1 p p j j max 最大限度 ( e e j j ( a 一个 1 ) , , e e j j ( a 一个 k k ) ) ,

where max ( z 1 , …, z k ) is the largest of the real numbers z 1 , …, z k .

其中 max (z 1 , …, z k ) 是实数 z 1 , …, z k 中的最大值。

Prime Factorization of Rational Numbers

有理数的质因数分解

The prime factorization theorem for nonzero integers can be extended to a prime factorization theorem for nonzero rational numbers by allowing negative exponents. We continue to assume that p 1 < p 2 < · · · < p j < · · · is the list of all prime integers.

通过允许负指数,可以将非零整数的质因数分解定理推广到非零有理数的质因数分解定理。我们继续假设 p 1 < p 2 < · · · < p j < · · · 是所有质数的列表。

4.61. Theorem on Unique Prime Factorization of Rational Numbers. For every nonzero q Q , there exist unique integers s ( q ) ∈ { + 1, − 1} and e j ( q ) for j Z 1 , such that all but finitely many e j ( q ) are zero, and

4.61. 有理数素因数分解唯一性定理。对于每个非零 q∈Q,存在唯一的整数 s(q) ∈ { + 1, − 1} 和 e j (q),其中 j∈Z≥1,使得除有限个 e j (q) 外,其余所有 e j (q) 均为零,并且

q q = s s ( q q ) j j 1 p p j j e e j j ( q q ) .

Proof. Fix nonzero q Q . First we prove existence of the claimed factorization. Since q is rational, we know there are integers n and m such that m ≠ 0 and q = n / m . Note n ≠ 0, since q ≠ 0. By the Theorem on Prime Factorization of Integers, we know n = s ( n ) j 1 p j e j ( n ) and m = s ( m ) j 1 p j e j ( m ) . Dividing n by m ,

证明。固定非零整数 q∈Q。首先,我们证明所声称的因式分解的存在性。由于 q 是有理数,我们知道存在整数 n 和 m,使得 m ≠ 0 且 q = n/m。注意 n ≠ 0,因为 q ≠ 0。根据整数素因数分解定理,我们知道 n=s(n)∏j≥1pjej(n) 且 m=s(m)∏j≥1pjej(m)。将 n 除以 m,

q q = n n / m = [ s s ( n n ) / s s ( m ) ] j j 1 p p j j e e j j ( n n ) e e j j ( m ) .

Choosing s ( q ) = s ( n )/ s ( m ), which is in { + 1, − 1}, and (for each j ) choosing e j ( q ) = e j ( n ) − e j ( m ), which is in Z since e j ( n ) , e j ( m ) Z 0 , we have found a factorization of q of the required type.

选择 s(q) = s(n)/s(m),它在 { + 1, − 1} 中,并且(对于每个 j)选择 e j (q) = e j (n) − e j (m),它在 Z 中,因为 ej(n),ej(m)∈Z≥0,我们找到了 q 的所需类型的分解。

Proving uniqueness of the factorization is a little tricky, since n and m are not uniquely determined by q . Assume

证明因式分解的唯一性有点棘手,因为 n 和 m 并非由 q 唯一确定。假设

q q = s s j j 1 p p j j e e j j = s s j j 1 p p j j e e j j

(4.10)

where the integers s and e j as well as the integers s ′ and e j ′ satisfy all the conditions in the theorem statement. We must prove s = s ′ and e j = e j ′ for all j Z 1 . If q is positive, we must have s = +1 = s ′; while if q is negative, we must have s = −1 = s ′. To deal with the exponents, we divide the index set Z 1 into three subsets. Let A = { j Z 1 : e j = e j } , let B = { j Z 1 : e j < e j } , and let C = { j Z 1 : e j > e j } . Our goal is to prove B = C = . To begin, we can divide both expressions for q in ( 4.10 ) by the common factor s j A p j e j , which leaves

其中整数 s 和 e j 以及整数 s′ 和 e j ′ 满足定理陈述中的所有条件。我们必须证明对于所有 j∈Z≥1,s = s′ 且 e j = e j ′。如果 q 为正,则必须有 s = +1 = s′;如果 q 为负,则必须有 s = −1 = s′。为了处理指数,我们将索引集 Z≥1 分为三个子集。令 A={j∈Z≥1:ej=ej′},令 B={j∈Z≥1:ej<ej′},令 C={j∈Z≥1:ej>ej′}。我们的目标是证明 B=C=∅。首先,我们可以将 (4.10) 中 q 的两个表达式除以公因子 s∏j∈Apjej,从而得到

j j B B p p j j e e j j j j C C p p j j e e j j = j j B B p p j j e e j j j j C C p p j j e e j j .

(Here and below, we interpret a product over an empty set of indices as 1.) Next, divide both sides by j B p j e j j C p j e j , to get

(此处及下文中,我们将空索引集上的乘积解释为 1。)接下来,两边同时除以 ∏j∈Bpjej∏j∈Cpjej′,得到

j j C C p p j j e e j j e e j j = j j B B p p j j e e j j e e j j .

(4.11)

By the definitions of B and C , the left side is either 1 or a product of strictly positive powers of primes p j with j C ; and the right side is either 1 or a product of strictly positive powers of primes p j with j B . Consider various cases that might occur.

根据 B 和 C 的定义,左侧要么是 1,要么是素数 p j 的严格正幂的乘积,其中 j ∈ C;右侧要么是 1,要么是素数 p j 的严格正幂的乘积,其中 j ∈ B。考虑可能出现的各种情况。

Case 1. Assume B and C . Note that the primes p j for j B are all different from the primes p j for j C , since B and C cannot overlap. Now ( 4.11 ) displays two different prime factorizations of a positive integer, violating uniqueness of prime factorization of integers.

情况 1. 假设 B≠∅ 且 C≠∅。注意,对于 j ∈ B,所有素数 p j 都与 j ∈ C,所有素数 p j 不同,因为 B 和 C 不能重叠。现在,(4.11) 式展示了一个正整数的两种不同的素因数分解,这违反了整数素因数分解的唯一性。

Case 2. Assume B = and C . Then the right side of ( 4.11 ) is 1, whereas the left side is larger than 1, which is impossible.

情况 2. 假设 B=∅ 且 C≠∅。则 (4.11) 式右侧为 1,而左侧大于 1,这是不可能的。

Case 3. Assume B and C = . Then the left side of ( 4.11 ) is 1, whereas the right side is larger than 1, which is impossible. $Case 4.$ Assume B = and C = . This is the only possible case, and in this case we must have e j = e j ′ for all j Z 1 , as needed. This completes the uniqueness proof.

情况 3. 假设 B≠∅ 且 C=∅。则 (4.11) 式左侧为 1,而右侧大于 1,这是不可能的。情况 4. 假设 B=∅ 且 C=∅。这是唯一可能的情况,在这种情况下,对于所有 j∈Z≥1,我们必须有 e j = e j ′,这符合要求。至此,唯一性证明完成。

Proving Irrationality of n th Roots

证明n次方根的无理性

In § 4.4 , we proved that 2 is irrational using proof by contradiction. Now that we have the unique factorization theorem for rational numbers, we can quickly prove a much more general result that tells us exactly which numbers q n (where q Q > 0 and n Z > 1 ) are rational. In using the notation q n , we are tacitly assuming the following fact about real numbers: for every positive real number x and every positive integer n , there exists a unique positive real number w such that w n = x ;by definition, x n is the unique such w solving w n = x . This can be proved using the Intermediate Value Theorem from calculus; the case of square roots ( n = 2) is proved in Theorem 8.61. Informally, our main result says that the only positive rational numbers q with rational n th roots are those for which the exponents of all primes in the factorization of q are multiples of n .

在第 4.4 节中,我们用反证法证明了 2 是无理数。现在我们有了有理数的唯一因式分解定理,就可以快速证明一个更一般的结果,该结果精确地告诉我们哪些数 qn(其中 q∈Q>0 且 n∈Z>1)是有理数。在使用符号 qn 时,我们隐含地假设了关于实数的以下事实:对于每个正实数 x 和每个正整数 n,存在唯一的正实数 w 使得 w n = x;根据定义,xn 是满足 w n = x 的唯一 w。这可以用微积分中的介值定理来证明;平方根的情况(n = 2)在定理 8.61 中已证明。通俗地说,我们的主要结果表明,唯一具有有理数 n 次根的正有理数 q 是那些因式分解中所有素数的指数都是 n 的倍数的数。

4.62 Theorem on Rationality of Roots. For all q Q > 0 and all n Z > 1 , q n is rational iff for all j Z 1 , n divides e j ( q ); and in this case,

4.62 根的有理性定理。对于所有 q∈Q>0 和所有 n∈Z>1,qn 是有理数当且仅当对于所有 j∈Z≥1,n 整除 e j (q);并且在这种情况下,

q q n n = j j 1 p p j j e e j j ( q q ) / n n .

Proof. Fix q Q > 0 and n Z > 1 . By the fact quoted above, let w R > 0 be the unique real number such that w n = q . $Part 1.$ Assume w Q , and prove n divides every e j ( q ). Since w is positive and rational, it has a prime factorization w = j 1 p j e j ( w ) . Raising this expression to the power n ,

证明。固定 q∈Q>0 和 n∈Z>1。根据上述事实,设 w∈R>0 为唯一实数,使得 w n = q。$第一部分。$ 假设 w∈Q,并证明 n 整除所有 e j (q)。由于 w 是正有理数,因此它有质因数分解 w=∏j≥1pjej(w)。将此表达式取 n 次方,

j j 1 p p j j n n e e j j ( w 西 ) = w 西 n n = q q = j j 1 p p j j e e j j ( q q ) .

Since the prime factorization of q is unique, we conclude that e j ( q ) = ne j ( w ) for every j ≥ 1. Since e j ( w ) Z , we see that n divides every e j ( q ). Furthermore, e j ( w ) = e j ( q )/ n , so the last formula in the theorem statement holds.

由于 q 的质因数分解是唯一的,我们得出结论:对于每个 j ≥ 1,都有 e j (q) = ne j (w)。由于 ej(w)∈Z,我们看到 n 整除所有 e j (q)。此外,e j (w) = e j (q)/n,因此定理陈述中的最后一个公式成立。

Part 2. Assume n divides every e j ( q ); prove w Q . Let w 1 = j 1 p j e j ( q ) / n , which is a well-defined positive rational number since every e j ( q )/ n is an integer. By direct computation, w 1 n = j 1 p j e j ( q ) = q = w n . By the fact quoted above, w is the unique positive real number whose n th power is q , so w = w 1 Q , as needed.

第二部分。假设 n 整除所有 e j (q);证明 w∈Q。令 w1=∏j≥1pjej(q)/n,这是一个定义明确的正有理数,因为所有 e j (q)/n 都是整数。直接计算可知,w1n=∏j≥1pjej(q)=q=wn。根据上述结论,w 是唯一一个 n 次方为 q 的正实数,因此 w=w1∈Q,满足要求。

4.63. Example. The rational number q = 0.729 = 2 −3 3 6 5 −3 has rational cube root q 3 = 0.9 = 2 1 3 2 5 1 . But q is not rational since e 1 ( q ) = −3 is not divisible by 2. More generally, for all n Z > 1 { 3 } , q n is not rational.

4.63. 例。有理数 q = 0.729 = 23522 有理数立方根 q3 = 0.9 = 2−1325−1。但 q 不是有理数,因为 e(q) = −3 不能被 2 整除。更一般地,对于所有 n∈Z>1−3,qn 不是有理数。

Similarly, it follows from the theorem that the only positive integers k such that k is rational are integers of the form k = j 2 for some positive integer j ; the only k Z > 0 such that k 3 Q are those of the form k = j 3 for some j Z > 0 ; and so on. In particular, for every prime integer p , the numbers p , p 3 , etc., are all irrational.

类似地,由定理可知,满足 k = j 2 形式的正整数 k 为有理数;满足 k3∈Q 形式的 k∈Z>0 为满足 k = j 3 形式的 k 为有理数;以此类推。特别地,对于每个素数 p,p、p3 等均为无理数。

Section Summary

章节概要

Let p 1 < p 2 < · · · < p j < · · · be the list of all prime integers.

令 p 1 < p 2 < · · · < p j < · · · 为所有素数的列表。

  1. Computing with Prime Factorizations. Given positive integers n and m with prime factorizations n = j 1 p j e j ( n ) and m = j 1 p j e j ( m ) , we have:
    n m = j 1 p j e j ( n ) + e j ( m ) ; n divides m iff e j ( n ) e j ( m ) for all j 1 ;
    gcd ( n , m ) = j 1 p j min ( e j ( n ) , e j ( m ) ) ; and ( n , m ) = j 1 p j max ( e j ( n ) , e j ( m ) ) .
    Similar formulas hold for the product, gcd, and lcm of a finite list of nonzero integers.
    利用质因数分解进行计算。给定正整数 n 和 m,它们的质因数分解分别为 n = ∏j≥1pjej(n) 和 m = ∏j≥1pjej(m),则有:nm = ∏j≥1pjej(n) + ej(m);n 整除 m 当且仅当对于所有 j≥1,ej(n) ≤ ej(m);gcd(n,m) = ∏j≥1pjmin(ej(n), ej(m));且 gcd(n,m) = ∏j≥1pjmax(ej(n), ej(m))。对于有限个非零整数,类似的公式也适用于它们的乘积、最大公约数和最小公倍数。
  2. Prime Factorization of Rational Numbers. For every nonzero rational number q , there exist unique s ∈ { + 1, − 1} and unique integers e j ( q ) for j ≥ 1, such that all but finitely many e j ( q ) are zero, and q = s ( q ) j 1 p j e j ( q ) .
    有理数的质因数分解。对于每个非零有理数 q,存在唯一的 s ∈ { + 1, − 1} 和唯一的整数 e j (q),其中 j ≥ 1,使得除有限多个 e j (q) 外,其余均为零,且 q=s(q)∏j≥1pjej(q)。
  3. n th Roots of Rational Numbers. For any positive rational q = j 1 p j e j ( q ) and n Z > 1 , q n is rational iff n divides e j ( q ) for all j ≥ 1.
    有理数的 n 次方根。对于任意正有理数 q=∏j≥1pjej(q) 和 n∈Z>1,qn 是有理数当且仅当对于所有 j ≥ 1,n 整除 e j (q)。

Exercises

练习

  1. Let n = 2 5 3 3 7 4 , m = 2 3 3 5 5 2 7 1 , and s = 3 4 5 4 7 2 . (a) Find gcd( n , m ), gcd( m , s ), gcd( n , s ), and gcd( n , m , s ). (b) Find ( n , m ) , ( m , s ) , ( n , s ) , and ( n , m , s ) . (c) Are any of n , m , or s rational?
    设 n = 2 3 7 2,m = 2 3 5 7 6,s = 3 5 7 9。 (a) 求 gcd(n, m)、gcd(m, s)、gcd(n, s) 和 gcd(n, m, s)。 (b) 求 (n, m)、(m, s)、(n, s) 和 (n, m, s)。 (c) n、m 或 s 中是否有任意的? 合理的?
  2. Find the prime factorizations of each rational number. (a) 0.625 (b) 212/121 (c) 2.64 (d) 3300/77077.
    求下列各有理数的质因数分解。 (a) 0.625 (b) 212/121 (c) 2.64 (d) 3300/77077.
  3. (a) Prove the Lemma on Minimums. (b) State and prove an analogous result for max ( x , y ).
    (a)证明最小值引理。(b)陈述并证明 max(x, y) 的类似结果。
  4. Suppose x 1 , …, x k are fixed real numbers. (a) Give a recursive definition of min ( x 1 , …, x k ) and max ( x 1 , …, x k ). (b) Generalize the results of the previous exercise to apply to this situation.
    假设 x 1 , …, x k 是固定的实数。(a) 给出 min(x 1 , …, x k ) 和 max(x 1 , …, x k ) 的递归定义。(b) 将上一题的结果推广到这种情况。
  5. Given integers a 1 , …, a k > 0, prove gcd ( a 1 , , a k ) = j 1 p j min ( e j ( a 1 ) , , e j ( a k ) ) .
    给定整数 a 1 , …, a k > 0,证明 gcd(a1,…,ak)=∏j≥1pjmin(ej(a1),…,ej(ak))。
  6. Given integers a 1 , …, a k > 0, prove 1cm ( a 1 , , a k ) = j 1 p j max ( e j ( a 1 ) , , e j ( a k ) ) .
    给定整数 a 1 , …, a k > 0,证明 1cm(a1,…,ak)=∏j≥1pjmax(ej(a1),…,ej(ak))。
  7. Extend Theorem 4.56 to the case where n and m are nonzero rational numbers.
    将定理 4.56 推广到 n 和 m 为非零有理数的情况。
  8. Given nonzero rational numbers q and r , find and prove a formula for the prime factorization of q / r .
    给定非零有理数 q 和 r,求出并证明 q/r 的质因数分解公式。
  9. Use prime factorizations to prove: for all r , s , t Z > 0 , if gcd( r , t ) = 1 = gcd( s , t ), then gcd( rs , t ) = 1.
    利用质因数分解证明:对于所有 r,s,t∈Z>0,如果 gcd(r, t) = 1 = gcd(s, t),则 gcd(rs, t) = 1。
  10. Use prime factorizations to prove: for all a , b , c Z > 0 , if gcd( a , b ) = 1 and a divides bc , then a divides c .
    利用质因数分解证明:对于所有 a,b,c∈Z>0,如果 gcd(a, b) = 1 且 a 整除 bc,则 a 整除 c。
  11. (a) Prove or disprove: for all m , n Z > 0 , if m n , then e j ( m ) ≥ e j ( n ) for all j ≥ 1. (b) Prove or disprove: for all m , n Z > 0 , if e j ( m ) ≥ e j ( n ) for all j ≥ 1 then m n .
    (a)证明或反证:对于所有 m,n∈Z>0,如果 m ≥ n,则对于所有 j ≥ 1,e j (m) ≥ e j (n)。(b)证明或反证:对于所有 m,n∈Z>0,如果对于所有 j ≥ 1,e j (m) ≥ e j (n),则 m ≥ n。
  12. Prove: for all nonzero a , b Q , there exists n Z with b = na iff for all j Z 1 , e j ( a ) ≤ e j ( b ).
    证明:对于所有非零 a,b∈Q,存在 n∈Z 使得 b = na 当且仅当对于所有 j∈Z≥1,e j (a) ≤ e j (b)。
  13. Prove: for all positive integers a , b , c , gcd( ca , cb ) = c · gcd( a , b ) using prime factorizations. (See Exercise 14 in § 4.5 for a different proof.)
    证明:对于任意正整数 a、b、c,gcd(ca, cb) = c · gcd(a, b),使用质因数分解法。(参见 §4.5 中的练习 14,了解另一种证明方法。)
  14. Prove: for all positive integers a , b , d , gcd( a , b ) = d iff d | a and d | b and gcd( a / d , b / d ) = 1.
    证明:对于所有正整数 a、b、d,gcd(a, b) = d 当且仅当 d|a 且 d|b 且 gcd(a/d, b/d) = 1。
  15. Suppose we define divisibility in Q as follows: for all q , r Q , q divides r iff s Q , r = s q . Characterize all pairs ( q , r ) such that q divides r under this definition.
    假设我们在有理数域 Q 中定义整除性如下:对于所有 q,r∈Q,q 整除 r 当且仅当 ∃s∈Q,r=sq。根据此定义,刻画所有满足 q 整除 r 的数对 (q, r)。
  16. (a) Prove: for all x , y R , x + y = min ( x , y ) + max ( x , y ). (b) Deduce that for all a , b Z 0 , a b = gcd ( a , b ) ( a , b ) .
    (a) 证明:对于所有 x,y∈R,x + y = min (x, y) + max (x, y)。 (b) 推导:对于所有 a,b∈Z≠0,ab=gcd(a,b)⋅(a,b)。
  17. Prove: for all m , n Z > 0 , if there exists q Q with q n = m , then there exists k Z with k n = m . [Thus if an integer has a rational n th root, then the root must actually be an integer.]
    证明:对于所有 m,n∈Z>0,如果存在 q∈Q 使得 q n = m,则存在 k∈Z 使得 k n = m。[因此,如果一个整数有 n 次有理根,则该根必然是一个整数。]
  18. Given a nonzero q Q , let P ( q ) be the set of primes p j such that e j ( q ) ≠ 0. (a) Prove: for all nonzero q Q , P ( − q ) = P ( q ). (b) Prove: for all nonzero q , r Q , P ( q r ) P ( q ) P ( r ) . Show that equality holds if P ( q ) P ( r ) = .
    给定一个非零素数 q∈Q,令 P(q) 为满足 e j (q) ≠ 0 的素数 p j 的集合。(a) 证明:对于所有非零素数 q∈Q,P( − q) = P(q)。(b) 证明:对于所有非零素数 q,r∈Q,P(qr)⊆P(q)∪P(r)。证明当 P(q)∩P(r)=∅ 时等号成立。
  19. (a) Given nonzero rational numbers q 1 , …, q k , define rational versions of gcd( q 1 , …, q k ) and 1cm( q 1 , … , q k ). (b) For nonzero q , r Q , can we always write gcd( q , r ) = qx + ry for some integers x and y ?
    (a) 给定非零有理数 q 1 , …, q k ,定义 gcd(q 1 , …, q k ) 和 1cm(q 1 , … ,q k ) 的有理数形式。(b) 对于非零有理数 q,r∈Q,我们能否总能写出 gcd(q, r) = qx + ry,其中 x 和 y 为整数?
#

Review of Set Theory and Integers

集合论与整数回顾

Tables 4.1–4.6 on the following pages review the main definitions, proof templates, and theorems covered in the preceding chapters on set theory and integers. All unquantified variables in the tables represent arbitrary objects, sets, real numbers, or integers (as required by context).

以下各页的表 4.1 至 4.6 回顾了前面章节中关于集合论和整数的主要定义、证明模板和定理。表中所有未量化的变量均代表任意对象、集合、实数或整数(视上下文而定)。

TABLE 4.1.
表 4.1. Set Definitions.
集合定义。

Concept

概念

Defined Term

定义术语

Definition Text

定义文本

Set Membership

成员资格

x A [undefined term]

x ∈ A [未定义项]

(informal) x is a member of A .

(非正式)x 是 A 的成员。

x A

x∉A

¬( x A ).

¬(x ∈ A)。

Subset

子集

A B

A⊆ B

x , x A x B .

∀x,x∈A⇒x∈B。

A B

A⊈B

x , x A x B .

∃x,x∈A∧x∉B。

Proper Subset

真子集

A B

A⊊B

A B and A B .

A⊆B 且 A≠B。

Set Equality

集合相等

A = B

A = B

x , x A x B

∀x,x∈A⇔x∈B

(equivalently: A B and B A ).

(等价地:A⊆B 且 B⊆A)。

Union

联盟

x A B

x∈A∪B

x A or x B .

x ∈ A 或 x ∈ B。

Intersection

路口

x A B

x∈A∩B

x A and x B .

x ∈ A 且 x ∈ B。

Set Difference

集合差异

x A B

x ∈ A − B

x A and x B .

x ∈ A 且 x∉B。

Empty Set

空集

x

x∈∅

x is false for all x .

对于所有 x,x∈∅ 为假。

B =

B=∅

x , x B .

∀x,x∉B。

B

B≠∅

x , x B .

∃x,x∈B。

Singleton Set

单件套装

z ∈ { a }

z ∈ {a}

z = a .

来自 = a。

Unordered Pair

无序对

z ∈ { a , b }

z ∈ {a, b}

z = a or z = b .

z = a 或 z = b。

Unordered Triple

无序三元组

z ∈ { a , b , c }

z ∈ {a, b, c}

z = a or z = b or z = c .

z = a 或 z = b 或 z = c。

Ordered Pair

有序对

( a , b ) [undefined term]

(a,b)[未定义术语]

Axiom: a , b , c , d ,

公理:∀a,∀b,∀c,∀d,

( a , b ) = ( c , d ) ( a = c b = d ) .

(a,b)=(c,d)⇔(a=c∧b=d)。

[or, define ( a , b ) = {{ a }, { a , b }}]

[或者,定义 (a, b) = {{a}, {a, b}}]

Open Interval

开放区间

x ∈ ( a , b )

x ∈ (a, b)

x R and a < x < b .

x∈R 且 a < x < b。

Closed Interval

闭区间

x ∈ [ a , b ]

x ∈ [a, b]

x R and a x b .

x∈R 且 a ≤ x ≤ b。

Half-Open Interval

半开区间

x ∈ [ a , b )

x ∈ [a, b)

x R and a x < b .

x∈R 且 a ≤ x < b。

Half-Open Interval

半开区间

x ∈ ( a , b ]

x ∈ (a, b]

x R and a < x b .

x∈R 且 a < x ≤ b。

Power Set

力量集

S P X

S∈PX

S X .

S⊆ X。

Product Set

产品套装

z X × Y

z ∈ X × Y

x , y , ( x X y Y ) z = ( x , y ).

∃x,∃y,(x∈X∧y∈Y)∧z=(x,y)。

( a , b ) ∈ X × Y

(a, b) ∈ X × Y

a X and b Y .

a ∈ X 且 b ∈ Y。

Indexed Union

索引联盟

z i I A i

z∈⋃i∈IAi

i I , z A i .

∃i∈I,z∈Ai。

Indexed Intersection

索引交集

z i I A i

z∈⋂i∈IAi

i I , z A i (need I ).

∀i∈I,z∈Ai(需要I≠∅)。

TABLE 4.2.
表 4.2. Proof Templates Involving Sets and Induction.
涉及集合和归纳法的证明模板。

Statement

陈述

Proof Template to Prove Statement

用于证明陈述的证明模板

A B

A⊆ B

Fix an arbitrary object x 0 .
固定任意对象 x 0 . Assume x 0 A .
假设 x 0 ∈ A. Prove x 0 B .
证明 x 0 ∈ B. [To continue, expand definitions of “ x 0 A ” and “ x 0 B .”]
[继续,扩展“x 0 ∈ A”和“x 0 ∈ B”的定义。]

S × T C

S × T⊆ C

Fix an arbitrary ordered pair ( a , b ). Assume ( a , b ) ∈ S × T , which means a S and b T . Prove ( a , b ) ∈ C [by expanding the definitions of membership in S , T , and C ].

固定任意有序对 (a, b)。假设 (a, b) ∈ S × T,这意味着 a ∈ S 且 b ∈ T。证明 (a, b) ∈ C [通过扩展 S、T 和 C 的成员定义]。

A = B (set equality)

A = B(设定相等)

Part 1. Prove A B . Part 2. Prove B A .

第一部分:证明 A⊆ B。第二部分:证明 B⊆ A。

A = B (set equality)

A = B(设定相等)

Fix an arbitrary object x 0 .
固定一个任意对象 x 0 Prove x 0 A x 0 B by a chain proof.
用链式证明证明 x 0 ∈ A ⇔ x 0 ∈ B。 [If A and B contain only ordered pairs, replace x 0 by ( x , y ).]
[如果 A 和 B 只包含有序对,则将 x 0 替换为 (x, y)。]

A =

A=∅

Assume, to get a contradiction, that A .
为了得出矛盾,假设 A≠∅。 We have assumed z , z A .
我们假设存在 z,z∈A。 Expand the definition of “ z A ,”
扩展“z ∈ A”的定义, and try to find a contradiction.
并尝试找出矛盾。

n Z b , P ( n )

∀n∈Z≥b,P(n)

We use (ordinary) induction on n .
我们对 n 使用(普通)归纳法。 1. Base Case. Prove P ( b ).
1. 基本情况。证明 P(b)。 2a. Induction Step. Fix an arbitrary integer n Z b .
2a. 归纳步骤。固定任意整数 n∈Z≥b。 2b. Induction Hypothesis. Assume P ( n ).
2b. 归纳假设。假设 P(n) 成立。 2c. Prove P ( n + 1).
2c. 证明 P(n + 1)。 [Use P ( n ) to help prove P ( n + 1).]
[利用 P(n) 帮助证明 P(n + 1)]。

n Z b , P ( n )

∀n∈Z≥b,P(n)

We use strong induction on n . 1. Fix an arbitrary integer n Z b . 2a. Strong Induction Hypothesis. Assume that for all integers m in the range b m < n , P ( m ) is true. 2b. Prove P ( n ). [Small values of n may require separate cases. Use P ( m ), for well-chosen values of m < n , to help prove P ( n ).]

我们对 n 使用强归纳法。 1. 固定任意整数 n∈Z≥b。 2a. 强归纳假设。假设对于所有介于 b ≤ m < n 之间的整数 m,P(m) 为真。 2b. 证明 P(n)。 [较小的 n 值可能需要特殊处理。对于 m < n 的合适值,可以使用 P(m) 来帮助证明 P(n)。]

TABLE 4.3.
表 4.3. Properties of Subsets.
子集的属性。

Reflexivity

反身性

A A .

A⊆ A。

Antisymmetry

反对称性

( A B B A ) A = B .

(A⊆B∧B⊆A)⇒A=B。

Transitivity

传递性

( A B B C ) A C .

(A⊆B∧B⊆C)⇒A⊆C。

Lower Bound

下限

A B A and A B B .

A∩B⊆A 和 A∩B⊆B。

Greatest Lower Bound

最大下限

C A B iff ( C A and C B ).

C⊆A∩B 当且仅当(C⊆ A 且 C⊆ B)。

Upper Bound

上限

A A B and B A B .

A⊆A∪B 且 B⊆A∪B。

Least Upper Bound

最小上界

A B C iff ( A C and B C ).

A∪B⊆C 当且仅当(A⊆C 且 B⊆C)。

Least Element

最小元素

A .

∅⊆A。

Difference Property

差额财产

A B A .

A − B⊆ A。

Monotonicity of

∩的单调性

A B A C B C .

A⊆B⇒A∩C⊆B∩C。

Monotonicity of

∪ 的单调性

A B A C B C .

A⊆B⇒A∪C⊆B∪C。

Monotonicity of −

单调性 −

A B A C B C .

A⊆ B ⇒ A − C⊆ B − C。

Inclusion Reversal of −

包含反转 −

A B C B C A .

A⊆ B ⇒ C − B⊆ C − A。

Subset Characterizations

子集特征

A B A B = A A B = B A B = .

A⊆B⇔A∩B=A⇔A∪B=B⇔A−B=∅。

TABLE 4.4.
表 4.4. Set Equality Theorems.
集合等式定理。

Reflexivity

反身性

A = A .

A = A。

Symmetry

对称

A = B B = A .

A = B ⇒ B = A。

Transitivity

传递性

( A = B B = C ) A = C .

(A=B∧B=C)⇒A=C。

Commutativity

交换律

A B = B A , A B = B A .

A∪B=B∪A,A∩B=B∩A。

Associativity

结合律

( A B ) C = A ( B C ) , ( A B ) C = A ( B C ) .

(A∩B)∩C=A∩(B∩C), (A∪B)∪C=A∪(B∪C).

Distributive Law 1

分配律 1

A ( B C ) = ( A B ) ( A C ) .

A∩(B∪C)=(A∩B)∪(A∩C)。

Distributive Law 2

分配律 2

A ( B C ) = ( A B ) ( A C ) .

A∪(B∩C)=(A∪B)∩(A∪C)。

Idempotent Laws

等权律

A A = A , A A = A .

A∩A=A,A∪A=A。

Absorption Laws

吸收定律

A ( A B ) = A , A ( A B ) = A .

A∪(A∩B)=A,A∩(A∪B)=A。

Properties of

∅ 的性质

A = A , A = , A × = , × A = ,

A∪∅=A,A∩∅=∅,A×∅=∅,∅×A=∅,

A = A , A = , A A = .

A−∅=A,∅−A=∅,A−A=∅。

De Morgan Law 1

德摩根定律 1

A ( B C ) = ( A B ) ( A C ) .

A−(B∪C)=(A−B)∩(A−C)。

De Morgan Law 2

德摩根定律2

A ( B C ) = ( A B ) ( A C ) .

A−(B∩C)=(A−B)∪(A−C)。

Set Decomposition

集合分解

A = ( A B ) ( A B ) , ( A B ) ( A B ) = .

A=(A−B)∪(A∩B), (A−B)∩(A∩B)=∅.

Difference Properties

差异属性

A ( A B ) = A B , A ( A B ) = A B .

A−(A∩B)=A−B,A−(A−B)=A∩B。

TABLE 4.5.
表 4.5. Properties of Power Sets, Product Sets, and General Unions and Intersections.
幂集、乘积集、一般并集和交集的性质。

Power Sets:

力量集:

Min/Max Subsets

最小/最大子集

P A , A P A .

∅∈PA,A∈PA。

Monotonicity

单调性

A B iff P A P B .

A⊆B 当且仅当 PA⊆PB。

Intersection Property

交叉属性

P A B = P A P B .

PA∩B=PA∩PB。

Union Property

联合财产

P A P B P A B .

PA∪PB⊆PA∪B。

Difference Property

差额财产

P A B { } P A P B .

PA−B−{∅}⊆PA−PB。

Product Sets:

产品套装:

Monotonicity

单调性

( A C B D ) A × B C × D .

(A⊆C∧B⊆D)⇒A×B⊆C×D。

Distributive Laws

分配法

( A B ) × C = ( A × C ) ( B × C ) ,

(A∪B)×C=(A×C)∪(B×C),

( A B ) × C = ( A × C ) ( B × C ) ,

(A∩B)×C=(A×C)∩(B×C),

( A B ) × C = ( A × C ) − ( B × C ),

(A − B)× C = (A × C) − (B × C),

similarly if C appears in the first factor.

同样地,如果 C 出现在第一个因子中。

Intersection Property

交叉属性

( A × B ) ( C × D ) = ( A C ) × ( B D ) .

(A×B)∩(C×D)=(A∩C)×(B∩D)。

Union Property

联合财产

( A × B ) ( C × D ) ( A C ) × ( B D ) .

(A×B)∪(C×D)⊆(A∪C)×(B∪D)。

Unions and Intersections:

并集与交叉集:

Monotonicity of

⋃的单调性

( i I , A i B i ) i I A i i I B i .

(∀i∈I,Ai⊆Bi)⇒⋃i∈IAi⊆⋃i∈IBi。

Monotonicity of

⋂的单调性

( i I , A i B i ) i I A i i I B i .

(∀i∈I,Ai⊆Bi)⇒⋂i∈IAi⊆⋂i∈IBi。

De Morgan Law 1

德摩根定律 1

X i I A i = i I ( X A i ) .

X−⋃i∈IAi=⋂i∈I(X−Ai).

De Morgan Law 2

德摩根定律2

X i I A i = i I ( X A i ) .

X−⋂i∈IAi=⋃i∈I(X−Ai).

Distributive Laws

分配法

X i I A i = i I ( X A i ) ,

X∩⋃i∈IAi=⋃i∈I(X∩Ai),

X i I A i = i I ( X A i ) ,

X∪⋂i∈IAi=⋂i∈I(X∪Ai),

X × i I B i = i I ( X × B i ) ,

X×⋃i∈IBi=⋃i∈I(X×Bi),

X × i I B i = i I ( X × B i ) ,

X×⋂i∈IBi=⋂i∈I(X×Bi),

similarly if X appears as the second factor.

同样地,如果 X 作为第二个因素出现。

Lower and Upper Bound

下限和上限

k I , i I A i A k i I A i .

∀k∈I,⋂i∈IAi⊆Ak⊆⋃i∈IAi。

Greatest Lower Bound

最大下限

X i I A i iff ( i I , X A i ).

X⊆⋂i∈IAi 当且仅当(∀i∈I,X⊆ A i )。

Least Upper Bound

最小上界

i I A i X iff ( i I , A i X ).

⋃i∈IAi⊆X 当且仅当(∀i∈I,A i ⊆ X)。

Enlarging the Index Set

扩大索引集

I J i I A i j J A j ,

I⊆J⇒⋃i∈IAi⊆⋃j∈JAj,

I J j J A j i I A i .

I⊆J⇒⋂j∈JAj⊆⋂i∈IAi。

Combining Index Sets

合并索引集

( i I A i ) ( j J A j ) = k I J A k ,

(⋃i∈IAi)∪(⋃j∈JAj)=⋃k∈I∪JAk,

( i I A i ) ( j J A j ) = k I J A k .

(⋂i∈IAi)∩(⋂j∈JAj)=⋂k∈I∪JAk。

TABLE 4.6.
表 4.6. Definitions Involving Recursion or Divisibility.
涉及递归或整除性的定义。

Concept

概念

Defined Term

定义术语

Definition Text

定义文本

Sum

k = b n x k

∑k=bnxk

k = b b x k = x b ,

∑k=bbxk=xb,

(for fixed b n )

(对于固定的 b ≤ n)

k = b n + 1 x k = ( k = b n x k ) + x n + 1 .

∑k=bn+1xk=(∑k=bnxk)+xn+1。

Product

产品

k = b n x k

∏k=bnxk

k = b b x k = x b ,

∏k=bbxk=xb,

(for fixed b n )

(对于固定的 b ≤ n)

k = b n + 1 x k = ( k = b n x k ) x n + 1 .

∏k=bn+1xk=(∏k=bnxk)⋅xn+1。

Factorials

阶乘

n ! ( n Z 0 )

n! (n∈Z≥0)

0! = 1, ( n + 1)! = ( n + 1) · n !.

0! = 1,(n + 1)! = (n + 1) · n!。

Exponents

指数

x n ( x R , n Z 0 )

x n (x∈R,n∈Z≥0)

x 0 = 1, x n +1 = x n · x .

x 0 = 1,x n +1 = x n · x。

Fibonacci sequence

斐波那契数列

F n ( n Z 0 )

n (n∈Z≥0)

F 0 = 0, F 1 = 1,

F 0 = 0, F 1 = 1,

F n = F n −1 + F n −2 for all n Z 2 .

对于所有 n∈Z≥2,F n = F n −1 + F n −2

Divisor

除数

a | b ( a divides b )

a|b(a 能整除 b)

u Z , b = a u .

∃u∈Z,b=au。

Common Divisor

公因数

d | a and d | b

d|a 和 d|b

s , t Z , a = d s and b = d t .

∃s,t∈Z,a=ds 且 b=dt。

Greatest Common Divisor

最大公约数

d = gcd( a , b )

d = gcd(a, b)

d | a d | b c Z > 0 , ( c | a c | b c d ) .

d|a∧d|b∧∀c∈Z>0,(c|a∧c|b⇒c≤d).

Least Common Multiple

最小公倍数

= lcm ( a , b )

ℓ=lcm(a,b)

a | b | e Z > 0 , ( a | e b | e e ) .

a|ℓ∧b|ℓ∧∀e∈Z>0,(a|e∧b|e⇒ℓ≤e).

Composite

合成的

n is composite

n 是复合数

n Z > 1 and a , ( 1 < a < n a | n ) .

n∈Z>1 且 ∃a,(1<a<n∧a|n)。

Prime

主要的

n is prime

n 是素数

n Z > 1 and ¬ a , ( 1 < a < n a | n ) .

n∈Z>1 且 ¬∃a,(1<a<n∧a|n)。

Rational

合理的

x Q

x∈Q

m Z , n Z , n 0 x = m / n .

∃m∈Z,∃n∈Z,n≠0∧x=m/n。

Set Theory Pitfalls

集合论的陷阱

(a) Order and repetition do not matter in sets. For example, {1, 2} = {2, 1} = {1, 1, 2, 2}.

(a)集合中元素的顺序和重复并不重要。例如,{1, 2} = {2, 1} = {1, 1, 2, 2}。

(b) Distinguishing ∈ and ⊆. A B means the object A is a member of the set B . A B means every member of the set A is a member of the set B . Do not confuse the set membership relation ( A B ) with the subset relation ( A B ). Note that sets are allowed to be members of other sets.

(b) 区分 ∈ 和 ⊆。A ∈ B 表示对象 A 是集合 B 的成员。A ⊆ B 表示集合 A 中的每个元素都是集合 B 的成员。不要将集合成员关系 (A ∈ B) 与子集关系 (A ⊆ B) 混淆。注意,集合可以属于其他集合。

(c) Types of Brackets. There are three kinds of brackets used in set theory: round parentheses (…), square brackets [· · ·], and curly braces {…}. Each shape of bracket has a different meaning: use round parentheses for ordered pairs or open intervals, use square brackets for closed intervals, and use curly braces for sets. Adding more braces changes the meaning; for example, and { } and { { } } are all different sets. Similarly, S = {1, {2, 3, 4}, 5, 6, {{7}, 8}, [3, 5]} is a set with six members: the integers 1, 5, and 6; the set containing 2 and 3 and 4; another set with members 8 and {7}; and the closed interval [3, 5].

(c) 括号的类型。集合论中常用的括号有三种:圆括号 (…)、方括号 [· · ·] 和花括号 {…}。每种形状的括号都有不同的含义:圆括号表示有序对或开区间,方括号表示闭区间,花括号表示集合。添加更多的括号会改变集合的含义;例如,∅、{∅} 和 {{∅}} 都是不同的集合。类似地,S = {1, {2, 3, 4}, 5, 6, {{7}, 8}, [3, 5]} 是一个包含六个元素的集合:整数 1、5 和 6;包含 2、3 和 4 的集合;包含 8 和 {7} 的另一个集合;以及闭区间 [3, 5]。

Theorems on Divisibility, Primes, and GCDs

关于整除性、素数和最大公约数的定理

  1. Integer Division Theorem. For all integers a and b with b ≠ 0, there exist unique integers q , r with a = bq + r and 0 ≤ r < | b |. We call q the quotient and r the remainder when a is divided by b .
    整数除法定理。对于任意整数 a 和 b(其中 b ≠ 0),存在唯一的整数 q 和 r,使得 a = bq + r 且 0 ≤ r < |b|。我们称 q 为 a 除以 b 的商,r 为 a 除以 b 的余数。
  2. Theorem on Computing GCDs. For all a Z , gcd( a , 0) = | a |. For all a , b , q , r Z , if a = bq + r then gcd( a , b ) = gcd( b , r ).
    计算最大公约数的定理。对于所有 a∈Z,gcd(a, 0) = |a|。对于所有 a,b,q,r∈Z,如果 a = bq + r,则 gcd(a, b) = gcd(b, r)。
  3. Linear Combination Property of GCDs. For all a , b Z , there exist x , y Z such that gcd( a , b ) = ax + by . This result extends to the gcd of any finite list of integers.
    最大公约数的线性组合性质。对于任意整数 a, b∈Z,存在整数 x, y∈Z,使得 gcd(a, b) = ax + by。此结果可推广至任意有限整数列表的最大公约数。
  4. Euclid’s Recursive GCD Algorithm.

    欧几里得递归最大公约数算法。

    Input: a , b Z 0 .

    输入:a,b∈Z≥0。

    Output: d = gcd( a , b ), and x , y Z with d = ax + by .

    输出:d = gcd(a, b),且 x,y∈Z,其中 d = ax + by。

    Procedure:

    程序:

    1. If b = 0, return d = a , x = 1, y = 0.
      如果 b = 0,则返回 d = a,x = 1,y = 0。
    2. If b > 0, divide a by b to get a = bq + r with 0 ≤ r < b .
      如果 b > 0,则将 a 除以 b 得到 a = bq + r,其中 0 ≤ r < b。
    3. Recursively compute d ′ = gcd( b , r ) and s , t Z with d ′ = bs + rt .
      递归计算 d′ = gcd(b, r) 和 s,t∈Z,其中 d′ = bs + rt。
    4. Return d = d ′, x = t , and y = s qt .
      返回 d = d′,x = t,y = s − qt。

    The iterative version of this algorithm uses repeated division to find gcd( a , b ) (the last nonzero remainder), then works backwards through the divisions to find x and y .

    该算法的迭代版本使用重复除法来找到 gcd(a, b)(最后一个非零余数),然后反向进行除法运算来找到 x 和 y。

  5. Theorem on Divisibility by Primes. For all primes p and all a 1 , , a k Z , if p divides a 1 a 2 · · · a k , then p divides some a i .
    素数整除定理。对于所有素数 p 和所有 a1,…,ak∈Z,如果 p 整除 a 1 a 2 · · · a k ,则 p 整除某个 a i
  6. Infinitude of Primes. For every integer n , there exists a prime p > n . So there are infinitely many primes.
    素数的无穷性。对于每个整数 n,都存在一个大于 n 的素数 p。因此,素数有无穷多个。
  7. Unique Prime Factorization of Integers. Let p 1 < p 2 < · · · < p j < · · · be the list of all prime integers. For all nonzero n Z , there exist unique integers s ( n ) ∈ { + 1, − 1} and e j ( n ) ≥ 0 (for j = 1, 2, 3, …) such that all but finitely many e j ( n ) are zero, and n = s ( n ) j 1 p j e j ( n ) . In words, every nonzero integer n can be written as a product of zero or more primes, times a sign factor. The sign and primes occurring in the factorization (including the number of times a given prime appears) are uniquely determined by n .
    整数的唯一素因数分解。设 p 1 < p 2 < · · · < p j < · · · 为所有素数的列表。对于所有非零整数 n∈Z,存在唯一的整数 s(n) ∈ { + 1, − 1} 和 e j (n) ≥ 0(其中 j = 1, 2, 3, …),使得除有限个 e j (n) 外,其余均为零,且 n=s(n)∏j≥1pjej(n)。换句话说,每个非零整数 n 都可以表示为零个或多个素数乘以一个符号因子。分解中出现的符号和素数(包括给定素数出现的次数)由 n 唯一确定。
  8. Unique Prime Factorization of Rational Numbers. Let p 1 < p 2 < · · · < p j < · · · be the list of all prime integers. For all nonzero q Q , there exist unique integers s ( q ) ∈ { + 1, − 1} and e j ( q ) Z (for j = 1, 2, 3, …) such that all but finitely many e j ( q ) are zero, and q = s ( q ) j 1 p j e j ( q ) .
    有理数的唯一素因数分解。设 p 1 < p 2 < · · · < p j < · · · 为所有素数的列表。对于所有非零素数 q∈Q,存在唯一的整数 s(q) ∈ { + 1, − 1} 和 ej(q)∈Z(其中 j = 1, 2, 3, …),使得除有限个 e j (q) 外,其余所有 ej(q) 均为零,且 q=s(q)∏j≥1pjej(q)。
  9. Divisibility Concepts and Prime Factorizations. For all nonzero x , y Q and all j Z 1 , s ( xy ) = s ( x ) s ( y ) and e j ( xy ) = e j ( x ) + e j ( y ). For all nonzero a , b Z , a divides b iff for all j Z 1 , e j ( a ) ≤ e j ( b ). For all nonzero a 1 , , a k Z ,
    gcd ( a 1 , , a k ) = j 1 p j min ( e j ( a 1 ) , , e j ( a k ) ) ,
    1 cm ( a 1 , , a k ) = j 1 p j max ( e j ( a 1 ) , , e j ( a k ) ) .

    For all q Q > 0 and n Z 1 , q n is rational iff for all j Z 1 , n divides e j ( q ); and in this case, q n = j 1 p j e j ( q ) / n .


    整除性概念和质因数分解。对于所有非零 x,y∈Q 和所有 j∈Z≥1,s(xy) = s(x)s(y) 且 e j (xy) = e j (x) + e j (y)。对于所有非零 a,b∈Z,a 整除 b 当且仅当对于所有 j∈Z≥1,e j (a) ≤ e j (b)。对于所有非零 a1,…,ak∈Z, gcd(a1,…,ak)=∏j≥1pjmin(ej(a1),…,ej(ak)),1cm(a1,…,ak)=∏j≥1pjmax(ej(a1),…,ej(ak))。对于所有 q∈Q>0 和 n∈Z≥1,qn 是有理数当且仅当对于所有 j∈Z≥1,n 整除 e j (q);在这种情况下,qn=∏j≥1pjej(q)/n。
  10. Summation Formulas. (These can be proved by induction.)
    • k = b n 1 = n b + 1 .
    • k = 1 n k = n ( n + 1 ) / 2 .
    • k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) / 6 .
    • k = 1 n k 3 = n 2 ( n + 1 ) 2 / 4 .
    • k = 0 n r k = ( r n + 1 1 ) / ( r 1 ) for all real r ≠ 1
    • k = b n ( x k + y k ) = k = b n x k + k = b n y k for all x k , y k R and all n b .
    • k = b n ( a x k ) = a k = b n x k for all a , x k R and all n b .

    求和公式。(这些公式可以用数学归纳法证明。) ∑k=bn1=n−b+1. ∑k=1nk=n(n+1)/2. ∑k=1nk2=n(n+1)(2n+1)/6. ∑k=1nk3=n2(n+1)2/4. ∑k=0nrk=(rn+1−1)/(r−1),对于所有实数r≠1。∑k=bn(xk+yk)=∑k=bnxk+∑k=bnyk,对于所有xk,yk∈R和所有n≥b。∑k=bn(axk)=a∑k=bnxk,对于所有a,xk∈R和所有n≥b。

Review Problems

复习题

  1. Complete the following definitions and theorem statement. (a) x A B iff ... (b) x i I A i iff ... (c) ( y , z ) ∈ C × D iff ... (d) Integer Division Theorem: ...
    完成下列定义和定理陈述。 (a) x∈A∪B 当且仅当…… (b) x∈⋂i∈IAi 当且仅当…… (c) (y, z) ∈ C × D 当且仅当…… (d) 整数除法定理:……
  2. Use Euclid’s algorithm to compute gcd(210, 51), and find integers x and y such that gcd(210, 51) = 210 x + 51 y .
    使用欧几里得算法计算 gcd(210, 51),并找到整数 x 和 y,使得 gcd(210, 51) = 210x + 51y。
  3. Prove this statement by ordinary induction: for all n Z 2 , k = 2 n 1 k ( k 1 ) = n 1 n .
    用普通归纳法证明这个命题:对于所有 n∈Z≥2,∑k=2n1k(k−1)=n−1n。
  4. Let A , B , C be fixed sets. Prove using only the definitions: if B A C and B A C , then B C .
    设 A、B、C 为固定集合。仅使用定义证明:如果 B∩A⊆C 且 B − A⊆ C,则 B⊆ C。
  5. Define a sequence recursively by setting a 0 = 8, a 1 = 13, and a n = 7 a n −1 − 6 a n −2 for all integers n ≥ 2. Prove by strong induction: for all n Z 0 , a n = 6 n + 7.
    递归地定义一个序列,令 a 0 = 8,a 1 = 13,且对于所有整数 n ≥ 2,a n = 7a n −1 − 6a n −2 。用强归纳法证明:对于所有 n∈Z≥0,a n = 6 n + 7。
  6. (a) Let A be a fixed set. Which of the following sets must equal the empty set? A , A , A , A × , P ( ) , { } . (b) Disprove: for all sets B , B P ( B ) = .
    (a) 设 A 为一个固定的集合。下列哪些集合一定等于空集? A∪∅, A∩∅, A−∅, A×∅, P(∅), {∅}. (b) 反证:对于所有集合 B,B∩P(B)=∅。
  7. True or false? Explain briefly. (a) {3, 5, 8} = {5, 8, 3}. (b) {1, 2, 2, 4, 4, 4, 4} = {1, 2, 4}. (c) {2} ∈ {{{2}}}. (d) {7, 8} ∈ {{7, 8}, 9}. (e) {7, 8}⊆{{7, 8}, 9}. (f) For all sets A and B , A B = B A . (g) For all sets A and B , A × B = B × A . (h) For all sets A and B , A × B B × A . (i) For all sets A , A . (j) For all sets A , A .
    对还是错?简要解释。 (a) {3, 5, 8} = {5, 8, 3}. (b) {1, 2, 2, 4, 4, 4, 4} = {1, 2, 4}. (c) {2} ∈ {{{2}}}. (d) {7, 8} ∈ {{7, 8}, 9}. (e) {7, 8}⊆{{7, 8}, 9}. (f) 对于所有集合 A 和 B,A∪B=B∪A. (g) 对于所有集合 A 和 B,A × B = B × A. (h) 对于所有集合 A 和 B,A × B ≠ B × A。 (i) 对于所有集合 A,∅∈A。 (j) 对于所有集合 A,∅⊆A。
  8. For every positive integer n , define a set A n = [ 1 / n , n + 2 ] { n } . Draw pictures of each set (on a number line or in R 2 , as appropriate). (a) A 1 × A 2 (b) n = 1 A n (c) A 2 A 1 (d) A 2 × A 3
    对于每个正整数 n,定义集合 An=[1/n,n+2]∪{−n}。 画出每个集合的图像(在数轴上或在 R² 中,视情况而定)。 (a) A 1 × A 2 (b) ⋃n=1∞An (c) A 2 − A 1 (d) A 2 × A 3
  9. Use Euclid’s Algorithm to compute gcd(528, 209), and find integers x and y such that gcd(528, 209) = 528 x + 209 y .
    使用欧几里得算法计算 gcd(528, 209),并找到整数 x 和 y,使得 gcd(528, 209) = 528x + 209y。
  10. Prove using only the definitions: for all sets A , B , C ,
    A × ( B C ) = ( A × B ) ( A × C ) .

    仅使用定义证明:对于所有集合 A、B、C,A×(B∪C)=(A×B)∪(A×C)。
  11. Define a sequence recursively by a 0 = 0 and a n +1 = 3 a n + 2 for all n Z 0 . Compute a 1 , a 2 , a 3 , and a 4 , guess a closed (non-recursive) formula for a n , and then prove your guess by ordinary induction.
    递归地定义一个序列,对于所有 n∈Z≥0,有 a 0 = 0 且 a n +1 = 3a n + 2。计算 a 1 、a 2 、a 3 和 a 4 ,猜测 a n 的一个封闭(非递归)公式,然后用普通归纳法证明你的猜测。
  12. Prove using only the definitions: for all sets A and B , A ( A B ) = A B .
    仅使用定义证明:对于所有集合 A 和 B,A−(A−B)=A∩B。
  13. Write just the outline of a proof by strong induction of the following statement: “Every odd integer n ≥ 7 can be written as the sum of three primes.”
    请用强归纳法证明以下命题:“每个奇数 n ≥ 7 都可以表示为三个素数之和”。
  14. For each n Z 1 , define a set A n = [1/ n , n + 3) − { n }. Draw each of the following sets on a number line. (a) n = 1 A n (b) n = 1 A n (c) (c) A 2 A 1 (d) A 2 × A 3 .
    对于每个 n∈Z≥1,定义集合 A n = [1/n, n + 3) − {n}。在数轴上画出下列集合。(a) ⋃n=1∞An (b) ⋂n=1∞An (c) (c)A2−A1 (d) A 2 × A 3
  15. Prove: for all sets A , B , C , ( A B ) × C = ( A × C ) − ( B × C ).
    证明:对于所有集合 A、B、C,(A − B) × C = (A × C) − (B × C)。
  16. Prove: for all n Z , the following statements are equivalent: 4 divides n + 2; 4 divides n − 2; n is even and n /2 is odd; 2 divides n but 4 does not divide n .
    证明:对于所有 n∈Z,下列语句等价:4 能整除 n + 2;4 能整除 n − 2;n 是偶数且 n/2 是奇数;2 能整除 n 但 4 不能整除 n。
  17. Prove: for all sets A , B , C , if A C = B C and A C = B C , then A = B .
    证明:对于所有集合 A、B、C,如果 A∩C=B∩C 且 A∪C=B∪C,则 A = B。
  18. Let I and J be fixed, nonempty index sets. Let A i (for each i I ) and A j (for each j J ) be fixed sets. Give two proofs of this fact: ( i I A i ) ( j J A j ) = k I J A k . In the first proof, use only the definitions. In the second proof, try to use previously known theorems wherever possible (except the result being proved).
    设 I 和 J 为固定的非空索引集。设 A i (对于每个 i ∈ I)和 A j (对于每个 j ∈ J)为固定集。给出以下结论的两个证明:(⋂i∈IAi)∩(⋂j∈JAj)=⋂k∈I∪JAk。在第一个证明中,仅使用定义。在第二个证明中,尽可能使用已知的定理(待证明的结果除外)。
  19. Compute d = gcd(1702, 483), and find integers x , y with d = 1702 x + 483 y .
    计算 d = gcd(1702, 483),并找到整数 x, y,使得 d = 1702x + 483y。
  20. Prove: a R , n Z 0 , k = 0 n a 5 k = a ( 5 n + 1 1 ) / 4 .
    证明:∀a∈R,∀n∈Z≥0,∑k=0na5k=a(5n+1−1)/4。
  21. Define c 1 = c 2 = 1, and c n = 2 c n −1 + 3 c n −2 for all n ≥ 3. (a) Prove that c n > 3 n −2 for all n ≥ 3. (b) Prove that c n < 2 · 3 n −2 for all n ≥ 3.
    定义 c 1 = c 2 = 1,且对于所有 n ≥ 3,c n = 2c n −1 + 3c n −2 。(a)证明对于所有 n ≥ 3,c n > 3 n −2 。(b)证明对于所有 n ≥ 3,c n < 2 · 3 n −2
  22. (a) Prove: for all sets A and B , P ( A B ) { } P ( A ) P ( B ) . (b) Disprove: for all sets A and B , P ( A B ) P ( A ) P ( B ) .
    (a)证明:对于所有集合 A 和 B,P(A−B)−{∅}⊆P(A)−P(B)。(b)反证:对于所有集合 A 和 B,P(A−B)−∅⊆P(A)−P(B)。
  23. Recursively define a sequence of sets T n as follows. Define T 0 = . For all integers n ≥ 0, define T n + 1 = T n { T n } . Prove:
    n Z 0 , x , ( x T n x T n ) .

    递归地定义集合序列 T n 如下。定义 T0=∅。对于所有整数 n ≥ 0,定义 Tn+1=Tn∪{Tn}。证明: ∀n∈Z≥0,∀x,(x∈Tn⇒x⊆Tn)。
#

5

Relations and Functions

关系与函数

5.1  Relations

5.1 关系

In this chapter, we study relations and functions, which are two of the most central ideas in mathematics. Informally, a function f is a rule that takes each object x in some set of inputs and transforms it into a new object y = f ( x ) in some set of outputs. A key point is that for every allowable input x , there must exist exactly one output y assigned to this x by the function. One of our main goals is to translate this informal description into a formal definition within set theory. To do so, we first define and study relations, which can be regarded as generalizations of functions in which each input x can be associated with more than one output, or perhaps no outputs at all. Many important mathematical concepts — including functions, partial orders, and equivalence relations — can be developed within the unifying general framework of relation theory.

本章我们将研究关系和函数,它们是数学中最核心的两个概念。通俗地说,函数 f 是一条规则,它将一组输入中的每个对象 x 转换为一组输出中的每个新对象 y = f(x)。关键在于,对于每个允许的输入 x,函数必须恰好为其分配一个输出 y。我们的主要目标之一是将这种通俗描述转化为集合论中的形式化定义。为此,我们首先定义并研究关系,关系可以被视为函数的推广,其中每个输入 x 可以关联多个输出,或者可能根本没有输出。许多重要的数学概念——包括函数、偏序关系和等价关系——都可以在关系理论这一统一的框架内得到发展。

Relations

关系

In set theory, a relation is any set of ordered pairs. If R is a relation and ( x , y ) ∈ R , we can think of x as an input that is associated with the output y . In this setting, it is important to keep track of which objects might appear as potential inputs and outputs. This leads to the following definition.

在集合论中,关系是指任意有序对的集合。如果 R 是一个关系,且 (x, y) ∈ R,我们可以将 x 视为与输出 y 相关联的输入。在这种情况下,重要的是要追踪哪些对象可能作为潜在的输入和输出出现。由此可得出以下定义。

5.1. Definition: Relations. For all sets R , X , Y :

5.1. 定义:关系。对于所有集合 R、X、Y:

R is a relation from X to Y iff R X × Y .

R 是从 X 到 Y 的关系,当且仅当 R⊆X×Y。

This means that every member of R has the form ( x , y ), for some x X and some y Y .

这意味着 R 中的每个成员都具有 (x, y) 的形式,其中 x ∈ X 且 y ∈ Y。

5.2. Example: Arrow Diagram of a Relation. Let X = {1, 2, 3, 4, 5, 6} and Y = { a , b , c , d , e , f }. Then R = {(1, c ), (1, e ), (1, f ), (3, f ), (4, a ), (4, b ), (4, f ), (6, c )} is a relation from X to Y . We can visualize the relation R by the following arrow diagram .

5.2 示例:关系的箭头图。设 X = {1, 2, 3, 4, 5, 6},Y = {a, b, c, d, e, f}。则 R = {(1, c), (1, e), (1, f), (3, f), (4, a), (4, b), (4, f), (6, c)} 是从 X 到 Y 的关系。我们可以用以下箭头图来表示关系 R。

ufig5_1.jpg

The arrow diagram consists of an oval with members of X (potential inputs for R ) on the left side and an oval with members of Y (potential outputs for R ) on the right side. For all x X and y Y , we draw an arrow pointing from x to y iff ( x , y ) ∈ R . When X and Y are finite sets, this arrow diagram contains exactly the same information as the set R .

箭头图由一个椭圆和一个椭圆组成:左侧椭圆包含集合 X(R 的潜在输入)的元素,右侧椭圆包含集合 Y(R 的潜在输出)的元素。对于所有 x ∈ X 和 y ∈ Y,我们绘制一条从 x 指向 y 的箭头,当且仅当 (x, y) ∈ R。当 X 和 Y 为有限集时,该箭头图包含的信息与集合 R 完全相同。

In the current example, we see that the input 1 is related (under R ) to the outputs c and e and f . Input 3 is related to output f only. Inputs 2 and 5 are not related to any outputs under R . Similarly, d is a possible output that does not get associated with any input by the relation R . In particular, R is also a relation from {1, 3, 4, 6} to { a , b , c , e , f }, or a relation from Z to { a , b , c , …, z }, and so on. However, R is not a relation from X ′ = {1, 2, 3, 4} to Y since (6, c ) ∈ R but ( 6 , c ) X × Y .

在当前示例中,我们看到输入 1(在关系 R 下)与输出 c、e 和 f 相关。输入 3 仅与输出 f 相关。输入 2 和 5 在关系 R 下与任何输出均无关联。类似地,d 是一个可能的输出,它不通过关系 R 与任何输入关联。特别地,R 也是从 {1, 3, 4, 6} 到 {a, b, c, e, f} 的关系,或者从 Z 到 {a, b, c, …, z} 的关系,依此类推。然而,R 不是从 X′ = {1, 2, 3, 4} 到 Y 的关系,因为 (6, c) ∈ R 但 (6, c) ∉ X′×Y。

5.3. Example: Graph of a Relation. If X and Y are subsets of R , we can draw the graph of a relation S from X to Y by drawing a dot at each point ( x , y ) in the plane R 2 such that ( x , y ) ∈ S . For example, suppose X = [ − 3, 3], Y = [ − 2, 2], and S is defined by letting ( x , y ) ∈ S iff x R and y R and ( x 2 /4) + y 2 = 1. The graph of the relation S is the ellipse shown below.

5.3. 示例:关系图。如果 X 和 Y 是 R 的子集,我们可以通过在 R² 平面上每个满足 (x, y) ∈ S 的点 (x, y) 处画一个点来绘制从 X 到 Y 的关系 S 的图像。例如,假设 X = [ − 3, 3],Y = [ − 2, 2],并且 S 定义为:(x, y) ∈ S 当且仅当 x∈R 且 y∈R 且 (x 2 /4) + y 2 = 1。关系 S 的图像是下图所示的椭圆。

ufig5_2.jpg

Since this ellipse is completely contained in the solid rectangle X × Y (whose boundary is indicated by dashed lines in the figure), S is a relation from X = [ − 3, 3] to Y = [ − 2, 2]. Since ( − 2, 0) ∈ S , the input x = −2 has associated output y = 0. The input 0 is related to outputs 1 and −1. The input 3 ∈ X is not related to any output, and the possible output 2 ∈ Y is not related to any input. S is also a relation from R to R , and S is a relation from [ − 2, 2] to [ − 1, 1]. But S is not a relation from [0, 1] to [0, 1], because the graph of S does not lie completely within the square [0, 1] × [0, 1].

由于该椭圆完全包含在实心矩形 X × Y 内(其边界在图中用虚线标出),因此 S 是从 X = [ − 3, 3] 到 Y = [ − 2, 2] 的关系。因为 ( − 2, 0) ∈ S,所以输入 x = −2 对应输出 y = 0。输入 0 与输出 1 和 −1 相关。输入 3 ∈ X 与任何输出均不相关,可能的输出 2 ∈ Y 也与任何输入均不相关。S 也是从 R 到 R 的关系,并且 S 是从 [ − 2, 2] 到 [ − 1, 1] 的关系。但是 S 不是从 [0, 1] 到 [0, 1] 的关系,因为 S 的图像并不完全位于正方形 [0, 1] × [0, 1] 内。

5.4. Example. A relation can be any set of ordered pairs, not necessarily a set determined by an explicit equation involving x and y . For example, the next figure shows the graph of a relation T from R to R .

5.4 示例。关系可以是任意有序对的集合,不一定是由包含 x 和 y 的显式方程确定的集合。例如,下图显示了从 R 到 R 的关系 T 的图像。

ufig5_3.jpg

5.5. Example. Suppose X and Y are any sets. Is 0 a relation from X to Y ? Yes, since 0 X × Y by an earlier theorem. When we are thinking of 0 as a relation, we could call it the empty relation . For finite sets X and Y , the arrow diagram of 0 contains no arrows. When X and Y are subsets of R , the graph of 0 has nothing in it.

5.5 示例。假设 X 和 Y 是任意集合。0 是从 X 到 Y 的关系吗?是的,因为根据前面的定理,0⊆X×Y。当我们把 0 看作一个关系时,我们可以称它为空关系。对于有限集合 X 和 Y,0 的箭头图不包含任何箭头。当 X 和 Y 是实数集 R 的子集时,0 的图中没有任何元素。

At the other extreme, the full product set X × Y is a relation from X to Y , since X × Y X × Y by an earlier theorem. In this relation, every possible input x X is related to every possible output y Y . The arrow diagram for X × Y contains every possible arrow leading from a point in X to a point in Y . When X and Y are closed intervals, the graph of X × Y (as we have seen before) is a solid rectangle of points with x -coordinates coming from X and y -coordinates coming from Y .

另一方面,完整的乘积集 X × Y 是一个从 X 到 Y 的关系,因为根据之前的定理,X × Y ⊆ X × Y。在这个关系中,每个可能的输入 x ∈ X 都与每个可能的输出 y ∈ Y 相关。X × Y 的箭头图包含了从 X 中的一点指向 Y 中的一点的所有可能的箭头。当 X 和 Y 是闭区间时,X × Y 的图像(如前所述)是一个实心矩形,其 x 坐标来自 X,y 坐标来自 Y。

Images of Sets under Relations

关系下的集合图像

Suppose R is a relation from X to Y . We have seen that each individual member x of X is related by R to zero or more members y of Y , namely those y for which ( x , y ) ∈ R . More generally, if we are given a set C of possible inputs to R , we could examine the ordered pairs in R to find the set of all possible outputs associated with the inputs coming from C . This new set is called the image of C under the relation R and is denoted R [ C ]. Formally, we make the following definition.

假设 R 是从 X 到 Y 的关系。我们已经看到,X 中的每个元素 x 都通过 R 与 Y 中的零个或多个元素 y 相关,即满足 (x, y) ∈ R 的 y。更一般地,如果给定 R 的可能输入集合 C,我们可以检查 R 中的有序对,找到与来自 C 的输入相关的所有可能输出的集合。这个新集合称为 C 在关系 R 下的像,记为 R[C]。形式上,我们给出以下定义。

5.6. Definition: Image of a Set under a Relation. For all relations R , all sets C , and all objects y , y R [ C ] iff x C , ( x , y ) R .

5.6. 定义:集合在关系下的像。对于所有关系 R、所有集合 C 和所有对象 y,y∈R[C] 当且仅当存在 x∈C,(x,y)∈R。

Note that this definition places no restriction on the set C , although we usually take C to be a subset of the input space X . To get used to this definition, let us see how to compute images of sets for relations given by arrow diagrams and graphs.

请注意,此定义对集合 C 没有限制,尽管我们通常将 C 视为输入空间 X 的子集。为了熟悉此定义,让我们看看如何计算由箭头图和图给出的关系的集合像。

5.7. Example. Consider the relation

5.7 示例。考虑以下关系

R = { ( 1 , c c ) , ( 1 , e e ) , ( 1 , f f ) , ( 3 , f f ) , ( 4 , a 一个 ) , ( 4 , b b ) , ( 4 , f f ) , ( 6 , c c ) }

from Example 5.2 . Given C = {1, 3, 5}, we compute R [ C ] = R [{1, 3, 5}] by scanning the ordered pairs in R and listing the second component of each ordered pair whose first component is in C . This produces the set R [ C ] = { c , e , f , f } = { c , e , f }. Note that R [ C ] = R [{1, 3}] since the input 5 is not related to any output. In fact, R [ C ] = R [{1}] since the output f (related to input 3) is already related to input 1. Image computations can be performed conveniently using the arrow diagram of R (shown on page 211). For example, to compute R [{2, 4, 6}] from the arrow diagram, start at the dots 2 and 4 and 6 on the left side and follow all possible outgoing arrows. The set of outputs that we reach, namely { a , b , f , c } = { a , b , c , f }, is the image R [{2, 4, 6}]. The following generic arrow diagram shows how R [ C ] is obtained from C .

以例 5.2 为例。给定集合 C = {1, 3, 5},我们通过扫描 R 中的有序对,并列出每个有序对的第二个分量(前提是该有序对的第一个分量在 C 中),来计算 R[C] = R[{1, 3, 5}]。这样就得到了集合 R[C] = {c, e, f, f} = {c, e, f}。注意,由于输入 5 与任何输出都无关,因此 R[C] = R[{1, 3}]。实际上,R[C] = R[{1}],因为输出 f(与输入 3 相关)已经与输入 1 相关。使用 R 的箭头图(如图 211 页所示)可以方便地进行图像计算。例如,要从箭头图计算 R[{2, 4, 6}],可以从左侧的点 2、4 和 6 开始,沿着所有可能的出箭头进行计算。我们得到的输出集合,即{a, b, f, c} = {a, b, c, f},是图像R[{2, 4, 6}]。下面的通用箭头图展示了如何从C得到R[C]。

ufig5_4.jpg

If we think of the arrows as rays of light leading from C to a screen represented by the set Y , then R [ C ] is the image cast by the object C on this screen. This picture is the source of the term image .

如果我们把箭头看作是从物体 C 射向由集合 Y 表示的屏幕的光线,那么 R[C] 就是物体 C 在该屏幕上投射的像。这个图像就是“像”这个术语的由来。

5.8. Example. Define a relation S from R to R by letting S consist of all ( x , y ) R 2 with x = y 2 . In the following pictures, we use the graph of S to compute S [{4}], S [(1, 4)], S [ R > 0 ] , and S [[ − 1, 1/4]].

5.8. 示例。定义从 R 到 R 的关系 S,令 S 由所有满足 x = y 2 的 (x,y)∈R2 组成。在下面的图中,我们使用 S 的图来计算 S[{4}]、S[(1, 4)]、S[R>0] 和 S[[ − 1, 1/4]]。

ufig5_5.jpg

First, S [{4}] = {2, − 2} since (4, 2) ∈ S and (4, − 2) ∈ S and no output other than 2 or −2 is related to the input 4 (as 4 = y 2 is solved only by 2 and −2). To obtain this answer from the graph of S , we move up and down from the input 4 on the x -axis until we hit the graph of S . We hit the graph twice, at heights y = 2 and y = −2, so the image is {2, − 2}.

首先,S[{4}] = {2, −2},因为 (4, 2) ∈ S 且 (4, −2) ∈ S,并且除了 2 或 −2 之外没有其他输出与输入 4 相关(因为 4 = y 2 只能通过 2 和 −2 来求解)。为了从 S 的图像中获得这个答案,我们在 x 轴上从输入 4 上下移动,直到与 S 的图像相交。我们在 y = 2 和 y = −2 处两次与图像相交,因此图像为 {2, −2}。

We compute the image of the open interval (1, 4) similarly, but now we imagine moving up and down from all points on the x -axis in the range 1 < x < 4. We hit two arcs of the graph, which occupy heights between 1 and 2 (exclusive) and between −2 and −1 (exclusive). In interval notation, we have S [ ( 1 , 4 ) ] = ( 2 , 1 ) ( 1 , 2 ) .

我们以类似的方式计算开区间 (1, 4) 的图像,但现在我们想象从 x 轴上 1 < x < 4 范围内的所有点上下移动。我们遇到了图像上的两个弧,它们的高度分别介于 1 和 2 之间(不含 2)以及介于 −2 和 −1 之间(不含 −1)。用区间表示法表示,我们有 S[(1,4)]=(−2,−1)∪(1,2)。

ufig5_6.jpg

Next, if we allow all strictly positive real numbers as inputs to S , we see from the graph (or the formula x = y 2 ) that we obtain all nonzero real numbers as the related outputs. Thus, S [ R > 0 ] = ( , 0 ) ( 0 , ) = R 0 . Starting with the closed interval [ − 1, 1/4], we find S [[ − 1, 1/4]] = [ − 1/2, 1/2]. On the other hand, S [ [ 1 , 1 / 4 ] ] = 0 .

接下来,如果我们允许所有严格正实数作为 S 的输入,从图中(或公式 x = y 2 )可以看出,所有非零实数都是对应的输出。因此,S[R>0]=(−∞,0)∪(0,∞)=R≠0。从闭区间 [ − 1, 1/4] 开始,我们发现 S[[ − 1, 1/4]] = [ − 1/2, 1/2]。另一方面,S[[−1,−1/4]]=0。

5.9. Example. Consider the relation T whose graph was drawn in Example 5.4 . Use the graph to find (approximately) T [{0}], T [{1.5}], and T [ R < 0 ] .

5.9. 示例。考虑在示例 5.4 中绘制了图的关系 T。利用该图求出(近似地)T[{0}]、T[{1.5}] 和 T[R<0]。

Solution. To find T [{0}], draw a vertical line through the graph of T at x = 0, and look at the heights of the points on the graph that meet this line. We estimate T [ { 0 } ] = [ 2 , 4 / 3 ] [ 1 / 4 , 1 / 2 ] . Similarly, we estimate

解:为了求出 T[{0}],在 x = 0 处作一条垂直线穿过 T 的图像,并观察图像上与这条线相交的点的高度。我们估计 T[{0}]=[−2,−4/3]∪[−1/4,1/2]。类似地,我们估计

T T [ { 1.5 } ] = [ 1.5 , 1.1 ] { 9 / 8 , 15 / 8 , 17 / 8 } .

Finally, T [ R < 0 ] = ( 2 , 1 ] [ 1 / 4 , 1 / 2 ) [ 9 / 8 , 15 / 8 ] [ 2 , 9 / 4 ] .

最后,T[R<0]=(−2,−1]∪[−1/4,1/2)∪[9/8,15/8]∪[2,9/4].

Properties of Images

图像的属性

The next theorem lists properties of the image construction for relations .

下一个定理列出了关系的像构造的性质。

5.10. Theorem on Images. For all sets X , Y , A , B , I 0 , A i (for i I ) and all relations R and S

5.10. 关于像的定理。对于所有集合 X, Y, A, B, I≠0,A i (对于 i ∈ I)以及所有关系 R 和 S,有

(a) Monotonicity: If A B , then R [ A ]⊆ R [ B ].

(a)单调性:如果 A⊆ B,则 R[A]⊆ R[B]。

(b) Empty Set Images: R [ 0 ] = 0 and 0 [ A ] = 0 .

(b)空集图像:R[0]=0 和 0[A]=0。

(c) Image of Unions: R [ A B ] = R [ A ] R [ B ] and R [ i I A i ] = i I R [ A i ] and ( R S ) [ A ] = R [ A ] S [ A ] .

(c)并集的图像:R[A∪B]=R[A]∪R[B] 且 R[⋃i∈IAi]=⋃i∈IR[Ai] 且 (R∪S)[A]=R[A]∪S[A]。

(d) Image of Intersections: R [ A B ] R [ A ] R [ B ] and R [ i I A i ] i I R [ A i ] and ( R S ) [ A ] R [ A ] S [ A ] . (Equality is not always true.)

(d) 交集的图像:R[A∩B]⊆R[A]∩R[B] 且 R[⋂i∈IAi]⊆⋂i∈IR[Ai] 且 (R∩S)[A]⊆R[A]∩S[A]。(等式并非总是成立。)

(e) Image of Set Difference: R [ A ] − R [ B ] ⊆ R [ A B ]. (Equality is not always true.)

(e)集合差的像:R[A] − R[B] ⊆ R[A − B]。(等式并非总是成立。)

(f) Input/Output Property: If R is a relation from X to Y , then R [ A ] ⊆ Y and R [ A ] = R [ A X ] .

(f)输入/输出性质:如果 R 是从 X 到 Y 的关系,则 R[A] ⊆ Y 且 R[A]=R[A∩X]。

We prove some representative parts of this theorem, asking you to prove other parts in the exercises. Fix arbitrary relations R and S , and sets A , B , etc., as in the theorem statement.

我们证明了该定理的一些代表性部分,其他部分则需要你在练习中证明。固定任意关系 R 和 S,以及集合 A、B 等,如同定理陈述中所述。

5.11. Proof of Monotonicity. Assume A B . Prove R [ A ] ⊆ R [ B ]. [Continue with the subset template.] Fix an arbitrary object y ; assume y R [ A ]; prove y R [ B ]. We have assumed there exists x A with ( x , y ) ∈ R . We must prove z B , ( z , y ) R . Choose z = x , so that ( z , y ) = ( x , y ) ∈ R . Since x A and A B , we have z B , as needed.

5.11. 单调性证明。假设 A ⊆ B。证明 R[A] ⊆ R[B]。[继续使用子集模板。] 固定任意对象 y;假设 y ∈ R[A];证明 y ∈ R[B]。我们假设存在 x ∈ A 使得 (x, y) ∈ R。我们必须证明存在 z ∈ B,(z, y) ∈ R。选择 z = x,使得 (z, y) = (x, y) ∈ R。由于 x ∈ A 且 A ⊆ B,我们有 z ∈ B,满足要求。

5.12. Proof that R [ 0 ] = 0 Assume, to get a contradiction, that R [ 0 ] 0 . This assumption means there exists an object y R [ 0 ] . Expanding the definition of image, we have assumed x 0 , ( x , y ) R . But we know x 0 . The contradiction “ x 0 and x 0 ” proves R [ 0 ] = 0 .

5.12. R[0]=0 的证明 为了得出矛盾,假设 R[0]≠0。这个假设意味着存在一个对象 y∈R[0]。扩展像的定义,我们假设存在 x∈0,(x,y)∈R。但我们知道 x∉0。矛盾“x∈0 且 x∉0”证明了 R[0]=0。

5.13 Proof that R [ i I A i ] = i I R [ A i ] [We try a chain proof.] Fix an arbitrary object y . We know

5.13 证明 R[⋃i∈IAi]=⋃i∈IR[Ai] [我们尝试链式证明。] 固定任意对象 y。我们知道

y R [ i I A 一个 i ] x x i I A 一个 i , ( x x , y ) R (by definition of image) (根据图像的定义) x x , ( x x i I A 一个 i ( x x , y ) R ) (by quantifier conversion rule) (根据量词转换规则) x x , ( ( i I , x x A 一个 i ) ( x x , y ) R ) (by definition of union) (根据工会的定义) x x , i I , ( x x A 一个 i ( x x , y ) R ) (by a quantifier property) (通过量词属性) i I , x x , ( x x A 一个 i ( x x , y ) R ) (by reordering (通过重新排序) quantifiers) 量词) i I , x x A 一个 i , ( x x , y ) R (by quantifier conversion rule) (根据量词转换规则) i I , y R [ A 一个 i ] (by definition of image) (根据图像的定义) y i I R [ A 一个 i ] (by definition of union). (根据工会的定义)。

5.14. Proof of Input/Output Property. Assume R is a relation from X to Y , which means R X × Y .

5.14. 输入/输出性质的证明。假设 R 是从 X 到 Y 的关系,这意味着 R ⊆ X × Y。

Part 1 . Prove R [ A ] ⊆ Y . Fix an object z ; assume z R [ A ]; prove z Y . We have assumed x A , ( x , z ) R . Since R X × Y , we know ( x , z ) ∈ X × Y . Thus x X and z Y , so z Y .

第一部分:证明 R[A] ⊆ Y。固定一个对象 z;假设 z ∈ R[A];证明 z ∈ Y。我们假设存在 x∈A,(x,z)∈R。由于 R ⊆ X × Y,我们知道 (x, z) ∈ X × Y。因此 x ∈ X 且 z ∈ Y,所以 z ∈ Y。

Part 2 . Prove R [ A ] = R [ A X ] . Part 2a. Prove R [ A ] R [ A X ] . Fix z ; assume z R [ A ]; prove z R [ A X ] . We assumed x A , ( x , z ) R . We must prove w A X , ( w , z ) R . Choose w = x , so ( w , z ) = ( x , z ) ∈ R . We must check that w A X , which means w A and w X . On one hand, since w = x and x A , w A . On the other hand, since ( w , z ) ∈ R and R X × Y , we deduce ( w , z ) ∈ X × Y and hence w X . Part 2b. Prove R [ A X ] R [ A ] . Since A X A by an earlier theorem, we know R [ A X ] R [ A ] by the previously proved monotonicity property of images.

第二部分:证明 R[A]=R[A∩X]。第二部分a:证明 R[A] ⊆R[A∩X]。固定 z;假设 z ∈ R[A];证明 z∈R[A∩X]。我们假设 ∃x∈A,(x,z)∈R。我们必须证明 ∃w∈A∩X,(w,z)∈R。选择 w = x,所以 (w, z) = (x, z) ∈ R。我们必须验证 w∈A∩X,这意味着 w ∈ A 且 w ∈ X。一方面,因为 w = x 且 x ∈ A,所以 w ∈ A。另一方面,因为 (w, z) ∈ R 且 R ⊆ X × Y,我们推断 (w, z) ∈ X × Y,因此 w ∈ X。第二部分b:证明 R[A∩X]⊆R[A]。根据先前的定理,由于 A∩X⊆A,我们知道 R[A∩X]⊆R[A],这是由先前证明的像的单调性性质决定的。

Section Summary

章节概要

  1. Relations. R is a relation from X to Y iff R X × Y . To make the arrow diagram of a relation R , draw an arrow from the input x X to the output y Y for each ordered pair ( x , y ) ∈ R . When X and Y are subsets of R , make the graph of R by drawing a dot at each point in the plane with coordinates ( x , y ) ∈ R .
    关系。R 是从 X 到 Y 的关系当且仅当 R ⊆ X × Y。要绘制关系 R 的箭头图,对于每个有序对 (x, y) ∈ R,从输入 x ∈ X 到输出 y ∈ Y 画一条箭头。当 X 和 Y 是 R 的子集时,通过在平面上每个坐标为 (x, y) ∈ R 的点处画一个点来绘制 R 的图形。
  2. Images of Sets under Relations. For any set C , relation R , and object y , y R [ C ] iff x C , ( x , y ) R . In an arrow diagram for R , R [ C ] is the set of all objects reachable by following arrows starting at points in C . In the graph of R , we find R [ C ] by drawing C on the x -axis, moving up and down from C into the graph, and finding the set of all y -coordinates reachable in this way.
    关系下集合的像。对于任意集合 C、关系 R 和对象 y,y ∈ R[C] 当且仅当存在 x∈C,(x,y)∈R。在 R 的箭头图中,R[C] 是所有从 C 中的点出发,沿箭头到达的对象的集合。在 R 的图中,我们可以通过在 x 轴上绘制 C,从 C 向上向下移动,并找到所有沿此方式可达的 y 坐标的集合来找到 R[C]。
  3. Properties of Images. For all relations R and all sets A and B , A B implies R [ A ] ⊆ R [ B ]; R [ A B ] = R [ A ] R [ B ] ; and R [ 0 ] = 0 . So the image construction preserves inclusions, unions, and the empty set. For intersections and differences, we have the weaker properties R [ A B ] R [ A ] R [ B ] and R [ A ] − R [ B ] ⊆ R [ A B ]. Similar results hold for indexed unions and intersections.
    像的性质。对于所有关系 R 和所有集合 A 和 B,A ⊆ B 蕴含 R[A] ⊆ R[B];R[A∪B]=R[A]∪R[B];且 R[0]=0。因此,像的构造保持包含关系、并关系和空集不变。对于交集和差集,我们有较弱的性质 R[A∩B]⊆R[A]∩R[B] 和 R[A] − R[B] ⊆ R[A − B]。类似的结果也适用于索引并集和交集。

Exercises

练习

  1. Let X = {1, 2, 3, 4}, Y = {1, 2, 3, 4, 5}, and R = {(1, 4), (1, 1), (2, 3), (2, 4), (4, 3)}. (a) Draw an arrow diagram for R . (b) Draw the graph of R . (c) Compute R [{1, 2}], R [{3}], R [{2, 4}], and R [ X ]. (d) Find all sets A , B X with R [ A ] = R [ B ].
    设 X = {1, 2, 3, 4},Y = {1, 2, 3, 4, 5},R = {(1, 4), (1, 1), (2, 3), (2, 4), (4, 3)}。 (a) 画出 R 的箭头图。 (b) 画出 R 的图像。 (c) 计算 R[{1, 2}]、R[{3}]、R[{2, 4}] 和 R[X]。 (d) 找出所有满足 R[A] = R[B] 的集合 A, B ⊆ X。
  2. Let R be the relation with the following arrow diagram.

    ufig5_7.jpg

    (a) Describe R as a set of ordered pairs. (b) Find R [{1, 2, 3}], R [{2, 4, 5}], R [{3, 6}], and R [ X ]. (c) Find a relation S such that S [ X ] = Y and R S = 0 .
    设 R 为具有以下箭头图的关系。 ufig5_7.jpg (a) 将 R 描述为有序对的集合。 (b) 求 R[{1, 2, 3}]、R[{2, 4, 5}]、R[{3, 6}] 和 R[X]。 (c) 求关系 S,使得 S[X] = Y 且 R∩S=0。
  3. Let X = { a , b , c , d , e }, Y = { v , w , x , y , z }, R = {( a , z ), ( b , y ), ( c , x ), ( d , y ), ( e , z )}, and S = {( a , v ), ( b , w ), ( c , x ), ( d , y ), ( e , v )}. (a) Draw arrow diagrams for R , S , R S , and R S . (b) Find the image of { a , c , d } under each relation in (a). (c) Which subsets of Y have the form R [ A ] for some set A ?
    设 X = {a, b, c, d, e},Y = {v, w, x, y, z},R = {(a, z), (b, y), (c, x), (d, y), (e, z)},S = {(a, v), (b, w), (c, x), (d, y), (e, v)}。 (a) 画出 R、S、R∩S 和 R∪S 的箭头图。 (b) 求出 {a, c, d} 在 (a) 中每个关系下的像。 (c) Y 的哪些子集具有 R[A] 的形式,其中 A 为某个集合?
  4. Let R = { ( x , y ) R × R : x 2 + y 2 = 25 } . Draw the graph of R . Find R [{1}], R [{ − 3}], R [[6, 9]], R [( − 5, 0)], and R [ Z ] .
    设 R={(x,y)∈R×R:x2+y2=25}。画出 R 的图像。求 R[{1}]、R[{ − 3}]、R[[6, 9]]、R[( − 5, 0)] 和 R[Z]。
  5. Let X = {1, 2, 3, 4, 5}, and let R = { ( x , y ) X × X : x y is even } . Draw an arrow diagram and graph for R . Compute R [{1, 3}] and R [{2, 4}].
    设 X = {1, 2, 3, 4, 5},R = {(x,y)∈X×X:xy 为偶数}。画出 R 的箭头图和图像。计算 R[{1, 3}] 和 R[{2, 4}]。
  6. Let R = { ( x , y ) R : | x y | = 1 } . (a) Draw the graph of R . (b) Find R [{2}], R [{0}], R [(3, 4]], and R [ R < 0 ] . (c) Find R ( Z × Z ) .
    设 R={(x,y)∈R:|xy|=1}。 (a) 画出 R 的图像。 (b) 求 R[{2}]、R[{0}]、R[(3, 4]] 和 R[R<0]。 (c) 求 R∩(Z×Z)。
  7. Define a relation R = { ( x , y ) : x [ 3 , 3 ] and ( y = 2 x or y = x ) } . (a) Draw the graph of R in the xy -plane. (b) Given A = [1, 2] and B = ( − 3, − 1], compute the sets R [ A ], R [ B ], R [ A B ], R [ A ] − R [ B ], R [ A B ] , and R [ A ] R [ B ] .
    定义关系 R={(x,y):x∈[−3,3] 且 (y=2x 或 y=−x)}。 (a) 在 xy 平面上绘制 R 的图像。 (b) 给定 A = [1, 2] 和 B = ( − 3, − 1],计算集合 R[A]、R[B]、R[A − B]、R[A] − R[B]、R[A∩B] 和 R[A]∩R[B]。
  8. Let R = { ( x , y ) R × R : y = sin x } and S = { ( x , y ) R × R : x = sin y } . Compute the image of the following sets under R and S : (a) {0} (b) { π } (c) { − 1/2} (d) [ − 1, 1] (e) R .
    设 R={(x,y)∈R×R:y=sin⁡x} 和 S={(x,y)∈R×R:x=sin⁡y}。计算下列集合在 R 和 S 下的像:(a) {0} (b) {π} (c) { − 1/2} (d) [ − 1, 1] (e) R。
  9. Let X = {1, 2, 3, 4, 5}, Y = { i , j , k , m , n }, R = {(1, n ), (1, i ), (2, m ), (4, i ), (4, m ), (5, j ), (5, k ), (5, n )}, and S = {(1, n ), (2, j ), (3, k ), (3, m ), (3, n ), (4, m ), (5, n )}. (a) Draw arrow diagrams for R , S , R S , and R S , viewed as relations from X to Y . (b) Let A = {2, 3, 5}. Compute the sets R [ A ], S [ A ], ( R S ) [ A ] , R [ A ] S [ A ] , ( R S )[ A ], and R [ A ] − S [ A ].
    设 X = {1, 2, 3, 4, 5},Y = {i, j, k, m, n},R = {(1, n), (1, i), (2, m), (4, i), (4, m), (5, j), (5, k), (5, n)},S = {(1, n), (2, j), (3, k), (3, m), (3, n), (4, m), (5, n)}。(a) 画出 R、S、R∩S 和 R − S 的箭头图,将它们视为从 X 到 Y 的关系。(b) 设 A = {2, 3, 5}。计算集合 R[A]、S[A]、(R∩S)[A]、R[A]∩S[A]、(R − S)[A] 和 R[A] − S[A]。
  10. Let S be the relation from R to R with the graph shown here.

    ufig5_8.jpg

    Find S [{1, 2, 3, 4}], S [{ − 4, − 1}], S [( − 2, 1]], S [[ − 1, 2)], S [(2, 3)], S [ R < 0 ] , and S [ R 0 ] .
    设 S 是从 R 到 R 的关系,其图如下所示。 ufig5_8.jpg 求 S[{1, 2, 3, 4}]、S[{ − 4, − 1}]、S[( − 2, 1]]、S[[ − 1, 2)]、S[(2, 3)]、S[R<0] 和 S[R≥0]。
  11. (a) Prove: for all relations R and S and all sets A , if R S , then R [ A ] ⊆ S [ A ]. (b) Disprove: for all relations R and all sets A and B , if R [ A ] ⊆ R [ B ], then A B . (c) Prove or disprove: for all relations R and S and all sets A , if R [ A ] ⊆ S [ A ], then R S . (d) Prove or disprove: for all relations R and S , if R [ A ] ⊆ S [ A ] for every set A , then R S .
    (a) 证明:对于所有关系 R 和 S 以及所有集合 A,如果 R ⊆ S,则 R[A] ⊆ S[A]。 (b) 反证:对于所有关系 R 以及所有集合 A 和 B,如果 R[A] ⊆ R[B],则 A ⊆ B。 (c) 证明或反证:对于所有关系 R 和 S 以及所有集合 A,如果 R[A] ⊆ S[A],则 R ⊆ S。 (d) 证明或反证:对于所有关系 R 和 S,如果对于每个集合 A,R[A] ⊆ S[A],则 R ⊆ S。
  12. Prove: for all sets A , 0 [ A ] = 0 .
    证明:对于所有集合 A,0[A]=0。
  13. Prove: for all relations R and S and all sets A , ( R S ) [ A ] = R [ A ] S [ A ] . (b) State and prove a generalization of (a) involving the image of an indexed union of relations.
    证明:对于所有关系 R 和 S 以及所有集合 A,(R∪S)[A]=R[A]∪S[A]. (b) 陈述并证明 (a) 的一个推广,涉及关系的索引并集的像。
  14. (a) Prove part (d) of the Theorem on Images. (b) For each set inclusion proved in (a), give a specific example where set equality does not hold.
    (a) 证明像定理的第 (d) 部分。(b) 对于 (a) 中证明的每个集合包含关系,给出集合相等性不成立的具体例子。
  15. Prove part (e) of the Theorem on Images, and give a specific example where equality does not hold.
    证明像定理的(e)部分,并给出一个等式不成立的具体例子。
  16. For any sets A , B , C , find (with proof) the set ( A × B )[ C ].
    对于任意集合 A、B、C,求(并证明)集合 (A × B)[C]。
  17. Suppose R is a relation having the following property: for all w , x , y , if ( w , y ) ∈ R and ( x , y ) ∈ R , then w = x . (a) What does this property say about the arrow diagram of R ? (b) What does this property say about the graph of R ? (c) Prove: for all sets A and B , R [ A B ] = R [ A ] R [ B ] . (d) Prove: for all sets A and B , R [ A ] − R [ B ] = R [ A B ].
    假设关系 R 具有以下性质:对于所有 w、x、y,如果 (w, y) ∈ R 且 (x, y) ∈ R,则 w = x。 (a) 该性质对 R 的箭头图有何说明? (b) 该性质对 R 的图有何说明? (c) 证明:对于所有集合 A 和 B,R[A∩B]=R[A]∩R[B]。 (d) 证明:对于所有集合 A 和 B,R[A] − R[B] = R[A − B]。

5.2 Inverses, Identity, and Composition of Relations

5.2 关系的逆元、恒等元和复合

This section introduces some concepts for relations — inverses, composition, and identity relations — that foreshadow and generalize the corresponding concepts for functions. One advantage of working with relations is that (unlike functions) every relation has an inverse relation.

本节介绍关系的一些概念——逆关系、复合关系和恒等关系——这些概念预示并概括了函数中相应的概念。使用关系的一个优势在于(与函数不同),每个关系都存在逆关系。

Inverse of a Relation

关系的逆

Recall that a relation R from X to Y is a set of ordered pairs ( x , y ), where x is a member of the input set X and y is a member of the output set Y . The relation R associates zero or more outputs y Y with each input x X . We could reverse the action of R by interchanging the roles of inputs and outputs. This produces a new relation R −1 from Y to X , called the inverse of R , which is found by replacing each ( x , y ) in R by ( y , x ). Formally, we have the following definition.

回想一下,从 X 到 Y 的关系 R 是一组有序对 (x, y),其中 x 是输入集 X 的元素,y 是输出集 Y 的元素。关系 R 将零个或多个输出 y ∈ Y 与每个输入 x ∈ X 关联起来。我们可以通过交换输入和输出的角色来逆转 R 的作用。这将产生一个新的从 Y 到 X 的关系 R −1 ,称为 R 的逆关系,它是通过将 R 中的每个 (x, y) 替换为 (y, x) 而得到的。形式上,我们有以下定义。

5.15. Definition: Inverse of a Relation. For all relations R and all objects a and b : R −1 is the relation such that ( a , b ) R 1 iff ( b , a ) R .

5.15. 定义:关系的逆。对于所有关系 R 和所有对象 a 和 b:R −1 是关系,使得 (a,b)∈R−1 当且仅当 (b,a)∈R。

This definition states that R −1 is a relation , so the only possible elements of R −1 are ordered pairs. We could be more explicit by saying: for all objects z , z R −1 iff a , b , z = ( a , b ) ( b , a ) R . A similar comment applies to other relations defined below.

此定义表明 R −1 是一个关系,因此 R −1 的唯一元素是有序对。我们可以更明确地表述:对于所有对象 z,z ∈ R −1 当且仅当 ∃a,∃b,z=(a,b)∧(b,a)∈R。类似的结论也适用于下文定义的其他关系。

5.16. Example: Inverses via Arrow Diagrams. Suppose X = {1, 2, 3, (4, 5), (5, 4)}, Y = { a , b , c , d , 0 } , and

5.16. 示例:利用箭头图求逆矩阵。假设 X = {1, 2, 3, (4, 5), (5, 4)},Y = {a, b, c, d, 0},

R = { ( 1 , a 一个 ) , ( 1 , c c ) , ( 3 , 0 ) , ( ( 4 , 5 ) , a 一个 ) , ( ( 5 , 4 ) , c c ) , ( ( 5 , 4 ) , d d ) } .

By reversing all the ordered pairs, we see that

通过反转所有有序对,我们发现

R 1 = { ( a 一个 , 1 ) , ( c c , 1 ) , ( 0 , 3 ) , ( a 一个 , ( 4 , 5 ) ) , ( c c , ( 5 , 4 ) ) , ( d d , ( 5 , 4 ) ) } .

The arrow diagrams for R and R −1 are shown below. We get the arrow diagram for R −1 by reversing all the arrows in the arrow diagram for R .

R 和 R −1 的箭头图如下所示。R −1 的箭头图是通过将 R 的箭头图中的所有箭头反转而得到的。

ufig5_9.jpg

Let us compute the images of some sets under R and R −1 . First,

让我们计算一些集合在 R 和 R −1 下的像。首先,

R [ { 1 , 2 , 3 } ] = { a 一个 , c c , 0 } , R 1 [ { a 一个 , c c , 0 } ] = { 1 , ( 4 , 5 ) , ( 5 , 4 ) , 3 } ,
so 所以 R 1 [ R [ { 1 , 2 , 3 } ] ] = { 1 , ( 4 , 5 ) , ( 5 , 4 ) , 3 } { 1 , 2 , 3 } .

Similarly, we find

同样地,我们发现

R 1 [ { b b , c c } ] = { 1 , ( 5 , 4 ) } , R [ { 1 , ( 5 , 4 ) } ] = { a 一个 , c c , d d } , so 所以 R [ R 1 [ { b b , c c } ] ] = { a 一个 , c c , d d } { b b , c c } .

Note R 1 [ 0 ] = 0 , but R 1 [ { 0 } ] = { 3 } .

注意 R−1[0]=0,但 R−1[{0}]={3}。

5.17. Example: Inverses via Graphs. The left side of the figure below shows the graph of a relation S from [ − 2, 2] to [ − 2, 2]. The right side of the figure shows the graph of the inverse relation S −1 .

5.17. 示例:通过图表示逆关系。下图左侧显示了从 [ − 2, 2] 到 [ − 2, 2] 的关系 S 的图像。图右侧显示了逆关系 S −1 的图像。

ufig5_10.jpg

To see how the graph of the inverse is found, first consider the isolated point (2, − 1) ∈ S . Switching the coordinates, we see that ( − 1, 2) is a point in S −1 . Next, the part of the graph of S that looks like the letter N consists of line segments from (1/2, 1/2) to (1/2, 2) to (3/2, 1/2) to (3/2, 2). Reversing the coordinates, we obtain three line segments from (1/2, 1/2) to (2, 1/2) to (1/2, 3/2) to (2, 3/2). The counterparts in S −1 of the solid triangle and circular arc in S are computed similarly. Geometrically, the graph of S −1 is obtained from the graph of S by reflecting the picture through the line y = x , since this reflection has the effect of replacing each ( a , b ) in the original graph by ( b , a ).

为了了解如何找到逆图,首先考虑孤立点 (2, −1) ∈ S。交换坐标后,我们发现 (−1, 2) 是 S −1 中的一个点。接下来,S 中形似字母 N 的部分由从 (1/2, 1/2) 到 (1/2, 2) 到 (3/2, 1/2) 到 (3/2, 2) 的线段组成。交换坐标后,我们得到从 (1/2, 1/2) 到 (2, 1/2) 到 (1/2, 3/2) 到 (2, 3/2) 的三条线段。S 中实心三角形和圆弧在 S −1 中的对应点可以用类似的方法计算。从几何角度来看,S −1 的图像是通过对 S 的图像沿直线 y = x 进行反射而得到的,因为这种反射的效果是将原始图像中的每个 (a, b) 替换为 (b, a)。

Note that S [ { 0 , 1 / 2 } ] = { 1 / 2 } [ 1 / 2 , 2 ] , whereas S −1 [{0, 1/2}] = {1/2, 3/2}. The latter image can be found either by looking at vertical lines through x = 0 and x = 1/2 in the graph of S −1 , or by looking at horizontal lines through y = 0 and y = 1/2 in the graph of S . We have S 1 [ [ 0 , 2 ] ] = [ 2 , 1 ] [ 1 / 2 , 3 / 2 ] = S 1 [ [ 1 / 2 , ) ] . Next, S [ S 1 [ R < 0 ] ] = S [ [ 2 , 0 ] { 2 } ] = [ 3 / 2 , 1 / 2 ] [ 1 , 3 / 2 ] .

注意,S[{0,1/2}]={−1/2}∪[1/2,2],而S −1 [{0, 1/2}] = {1/2, 3/2}。后者的图像可以通过观察S −1 图像中过x = 0和x = 1/2的垂直线,或者通过观察S图像中过y = 0和y = 1/2的水平线来找到。我们有S−1[[0,2]]=[−2,−1]∪[1/2,3/2]=S−1[[1/2,∞)]。接下来,S[S−1[R<0]]=S[[−2,0]∪{2}]=[−3/2,−1/2]∪[1,3/2]。

5.18 Example. We know 0 is a relation containing no ordered pairs. It follows that 0 1 = 0 . Drawing a picture, you can check that the inverse of the relation R = [0, 3] × [1, 2] is the relation R −1 = [1, 2] × [0, 3]. More generally, for any sets X and Y , we claim that ( X × Y ) −1 = Y × X . Here is a quick chain proof: for any ordered pair ( a , b ), ( a , b ) ∈ ( X × Y ) −1 iff ( b , a ) ∈ X × Y iff b X and a Y iff a Y and b X iff ( a , b ) ∈ Y × X .

5.18 例。我们知道 0 是一个不包含任何有序对的关系。由此可知 0−1=0。画图可知,关系 R = [0, 3] × [1, 2] 的逆关系是关系 R −1 = [1, 2] × [0, 3]。更一般地,对于任意集合 X 和 Y,我们断言 (X × Y) −1 = Y × X。以下是一个简单的链式证明:对于任意有序对 (a, b),(a, b) ∈ (X × Y) −1 当且仅当 (b, a) ∈ X × Y 当且仅当 b ∈ X 且 a ∈ Y 当且仅当 a ∈ Y 且 b ∈ X 当且仅当 (a, b) ∈ Y × X。

Identity Relation on a Set

集合上的恒等关系

For any set X , we can create an especially simple relation from X to X called the identity relation on X and denoted I X . This relation consists of all ordered pairs ( x , x ) with x X . The name “identity” is used since each x X is related to itself and nothing else. The formal definition follows.

对于任意集合 X,我们可以创建一个从 X 到 X 的特别简单的关系,称为 X 上的恒等关系,记为 I X 。该关系由所有满足 x ∈ X 的有序对 (x, x) 构成。之所以称为“恒等关系”,是因为每个 x ∈ X 都只与自身相关,而与其他任何元素都不相关。其正式定义如下。

5.19. Definition: Identity Relation on a Set. For all sets X and all objects a and b : I X is the relation such that ( a , b ) I X iff a X and a = b .

5.19. 定义:集合上的恒等关系。对于所有集合 X 和所有对象 a 和 b:I X 是关系,使得 (a,b)∈IX 当且仅当 a∈Xanda=b。

5.20 Example: Arrow Diagram of an Identity Relation

5.20 示例:恒等关系的箭头图

Let X be the set {1, 2, 3, (4, 5), (5, 4)}. The identity relation on X is

设 X 为集合 {1, 2, 3, (4, 5), (5, 4)}。X 上的恒等关系是

I X X = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( ( 4 , 5 ) , ( 4 , 5 ) ) , ( ( 5 , 4 ) , ( 5 , 4 ) ) } .

The arrow diagram for I X is shown below. For each x X , there is an arrow pointing from x to itself. We see that I X 1 = I X . For all A X , I X [ A ] = A . These properties hold in general.

下图展示了 I X 的箭头图。对于每个 x ∈ X,都有一条箭头从 x 指向自身。我们可以看到 IX−1=IX。对于所有 A ⊆ X,I X [A] = A。这些性质普遍成立。

ufig5_11.jpg

5.21. Example: Graph of an Identity Relation.

5.21. 示例:恒等关系图。

Let A = [ 2 , 0 ) ( 1 , 2 ] and B = { − 2, − 1, 0, 1/2, 1}. The next figure shows the graphs of the identity relations I A and I B . The graph I A consists of the portion of the line y = x consisting of points ( c , c ) with c A ; similarly for I B . The entire line y = x is the graph of I R . Note I 0 = 0 , so the graph of I 0 is empty.

令 A=[−2,0)∪(1,2],B={−2,−1,0,1/2,1}。下图展示了恒等关系 I A 和 I B 的图像。图像 I A 由直线 y = x 上由点 (c, c) 构成的部分组成,其中 c ∈ A;I B 的情况类似。整条直线 y = x 构成恒等关系 I B 的图像。注意 I0=0,因此 I0 的图像为空集。

ufig5_12.jpg

Composition of Relations

关系构成

Informally, given two real-valued functions f and g , we can form a new function g f (the composition of g and f ) as follows. Starting with an input x R , we feed this number into f to obtain an output y = f ( x ). Now we regard this output y as the input to g , obtaining a new output z = g ( y ). The net effect is that z = g ( y ) = g ( f ( x )) = ( g f )( x ). We can generalize this idea to relations, as follows.

通俗地说,给定两个实值函数 f 和 g,我们可以构造一个新的函数 g ∘ f(g 和 f 的复合函数),如下所示。首先,给定一个输入 x∈R,我们将这个数输入到 f 中,得到输出 y = f(x)。现在,我们将这个输出 y 作为 g 的输入,得到一个新的输出 z = g(y)。最终结果是 z = g(y) = g(f(x)) = (g ∘ f)(x)。我们可以将这个思想推广到关系式,如下所示。

5.22. Definition: Composition of Relations. For all relations R and S and all ordered pairs ( a , b ), S R is the relation such that ( a , b ) S R iff c , ( a , c ) R ( c , b ) S .

5.22. 定义:关系的复合。对于所有关系 R 和 S 以及所有有序对 (a, b),S ∘ R 是关系,使得 (a,b)∈S∘R 当且仅当存在 c,(a,c)∈R∧(c,b)∈S。

The relation S R is the composition of the relations S and R , in this order.

关系 S ∘ R 是关系 S 和 R 按此顺序的复合。

Note carefully the order of the relations in the expression S R , which is motivated by the analogy with the function composition g f . Observe that when computing ( g f )( x ), we start with the input x , apply the function f , and then apply the function g . Analogously, given an ordered pair ( a , b ) ∈ S R , we know there exists c with ( a , c ) ∈ R and ( c , b ) ∈ S . This means that the original input a is related (under R ) to the intermediate output c , and then c (viewed as an input to S ) is related to the final output b . The net effect is that input a is related to output b under S R .

仔细观察表达式 S ∘ R 中关系的顺序,这是基于与函数复合 g ∘ f 的类比。注意,在计算 (g ∘ f)(x) 时,我们首先输入 x,然后应用函数 f,最后应用函数 g。类似地,给定有序对 (a, b) ∈ S ∘ R,我们知道存在 c 使得 (a, c) ∈ R 且 (c, b) ∈ S。这意味着原始输入 a(在 R 下)与中间输出 c 相关,然后 c(作为 S 的输入)与最终输出 b 相关。最终结果是,输入 a 与输出 b 在 S ∘ R 下相关。

5.23. Example: Composition of Relations via Arrow Diagrams.

5.23. 示例:通过箭头图表示关系。

Let X = {1, 2, 3, 4} and Y = { a , b , c , d , e }. Define a relation R from X to Y and a relation S from Y to X by

设 X = {1, 2, 3, 4},Y = {a, b, c, d, e}。定义从 X 到 Y 的关系 R 和从 Y 到 X 的关系 S,如下所示:

R = { ( 1 , d d ) , ( 3 , d d ) , ( 4 , a 一个 ) , ( 4 , e e ) } , S S = { ( a 一个 , 2 ) , ( b b , 1 ) , ( c c , 3 ) , ( c c , 4 ) , ( d d , 2 ) , ( e e , 1 ) } .

The arrow diagrams for R and S are shown below.

R 和 S 的箭头图如下所示。

ufig5_13.jpg

We can compute S R , which is a relation from X to X , by starting at points in X and seeing where we can go by traveling along an R -arrow followed by an S -arrow. By examination of the figure, we arrive at this arrow diagram for S R :

我们可以通过从 X 中的点出发,沿着 R 箭头和 S 箭头依次移动,来计算从 X 到 X 的关系 S ∘ R。通过观察图示,我们可以得到 S ∘ R 的箭头图:

ufig5_14.jpg

Symbolically, S R = {(1, 2), (3, 2), (4, 2), (4, 1)}. For example, (4, 1) ∈ S R because (4, e ) ∈ R and ( e , 1) ∈ S . Similarly, we can compute R S , which is a relation from Y to Y , by combining the arrow diagrams for R and S in the other order, as shown here:

符号上,S ∘ R = {(1, 2), (3, 2), (4, 2), (4, 1)}。例如,(4, 1) ∈ S ∘ R,因为 (4, e) ∈ R 且 (e, 1) ∈ S。类似地,我们可以通过反序组合 R 和 S 的箭头图来计算 R ∘ S,它是从 Y 到 Y 的关系,如下图所示:

ufig5_15.jpg

The arrow diagram for R S looks like this:

R ∘ S 的箭头图如下所示:

ufig5_16.jpg

So R S = {( b , d ), ( c , a ), ( c , d ), ( c , e ), ( e , d )}. Note carefully that S R R S in this example (and in most examples). This means that composition of relations is not commutative . Indeed, in our example, ( R S ) ( S R ) = 0 , which says that these two sets of ordered pairs have no members in common.

因此,R ∘ S = {(b, d), (c, a), (c, d), (c, e), (e, d)}。请注意,在本例(以及大多数例)中,S∘R≠R∘S。这意味着关系的复合运算不满足交换律。事实上,在本例中,(R∘S)∩(S∘R)=0,这表明这两个有序对集合没有共同的元素。

5.24. Example. Let X = {1, 2, 3}, R = {(1, 2), (2, 3), (3, 1)}, and S = {(1, 3), (2, 1), (3, 2)}. You can check that R S = {(1, 1), (2, 2), (3, 3)} = S R = I X . Thus, it does sometimes happen that R S = S R . In this example, it is also true that R −1 = S = R R and S −1 = R = S S .

5.24. 示例。设 X = {1, 2, 3},R = {(1, 2), (2, 3), (3, 1)},S = {(1, 3), (2, 1), (3, 2)}。可以验证 R ∘ S = {(1, 1), (2, 2), (3, 3)} = S ∘ R = I X 。因此,R ∘ S = S ∘ R 的情况确实存在。在本例中,R −1 = S = R ∘ R 和 S −1 = R = S ∘ S 也成立。

5.25. Example. Let R = [0, 2] × [1, 3] and S = I R { ( x , 2 x ) : x [ 0 , 2 ] } . Trying to read off the compositions S R and R S from the graphs of R and S can be tricky, so let us return to the definition. Which ordered pairs ( a , b ) belong to S R ? First, the original input a must belong to [0, 2], and all such inputs are related to all numbers c ∈ [1, 3] (and nothing else). Proceeding to S , a given input c ∈ [1, 3] is always related (under S ) to c ; furthermore, each number c in the range [0, 2] is also related to 2 − c . Thus, the final output b can be any number in [1, 3], and (by considering c ∈ [1, 2]) b can also be any number in [0, 1]. In summary, ( a , b ) ∈ S R iff a ∈ [0, 2] and b ∈ [0, 3], so that S R = [0, 2] × [0, 3].

5.25. 示例。设 R = [0, 2] × [1, 3],S = IR∪{(x,2−x):x∈[0,2]}。从 R 和 S 的图中读取组合 S ∘ R 和 R ∘ S 可能比较棘手,因此让我们回到定义。哪些有序对 (a, b) 属于 S ∘ R?首先,原始输入 a 必须属于 [0, 2],并且所有这样的输入都与 [1, 3] 中的所有数字 c 相关(且仅与 c 相关)。对于 S,给定的输入 c ∈ [1, 3] 总是(在 S 下)与 c 相关;此外,[0, 2] 范围内的每个数字 c 也与 2 − c 相关。因此,最终输出 b 可以是 [1, 3] 中的任何数字,并且(考虑到 c ∈ [1, 2])b 也可以是 [0, 1] 中的任何数字。综上所述,(a, b) ∈ S ∘ R 当且仅当 a ∈ [0, 2] 且 b ∈ [0, 3],因此 S ∘ R = [0, 2] × [0, 3]。

On the other hand, when is ( a , b ) ∈ R S ? Now we need to find c with ( a , c ) ∈ S and ( c , b ) ∈ R . One possibility is that ( a , c ) I R , which means a R and c = a . Then ( c , b ) is in R iff c = a is in [0, 2] and b is in [1, 3]. So every element of [0, 2] × [1, 3] is in R S . The other possibility for ( a , c ) ∈ S is that a ∈ [0, 2] and c = 2 − a ∈ [0, 2]. In this case, ( c , b ) ∈ R holds iff b ∈ [1, 3]. We conclude R S = [0, 2] × [1, 3] = R S R .

另一方面,(a, b) 何时 ∈ R ∘ S?现在我们需要找到满足 (a, c) ∈ S 且 (c, b) ∈ R 的 c。一种可能性是 (a, c) ∈ IR,这意味着 a ∈ R 且 c = a。那么 (c, b) ∈ R 当且仅当 c = a ∈ [0, 2] 且 b ∈ [1, 3]。因此,[0, 2] × [1, 3] 中的每个元素都在 R ∘ S 中。(a, c) ∈ S 的另一种可能性是 a ∈ [0, 2] 且 c = 2 − a ∈ [0, 2]。在这种情况下,(c, b) ∈ R 当且仅当 b ∈ [1, 3]。我们得出结论:R ∘ S = [0, 2] × [1, 3] = R ≠ S ∘ R。

Section Summary

章节概要

  1. Inverse of a Relation. For any relation R , the inverse of R is the relation R −1 such that for all a and b , ( a , b ) ∈ R −1 iff ( b , a ) ∈ R . To go from the arrow diagram of R to the arrow diagram of R −1 , reverse all the arrows. To go from the graph of R to the graph of R −1 , reflect the xy -plane through the line y = x .
    关系的逆。对于任意关系 R,R 的逆关系是关系 R −1 ,使得对于任意 a 和 b,(a, b) ∈ R −1 当且仅当 (b, a) ∈ R。要从关系 R 的箭头图得到关系 R −1 的箭头图,只需反转所有箭头即可。要从关系 R 的图像得到关系 R −1 的图像,只需将 xy 平面关于直线 y = x 反射即可。
  2. Identity Relation. For any set X , the identity relation on X is the relation I X such that for all a and b , ( a , b ) ∈ I X iff a X and a = b . The arrow diagram for I X has an arrow from x to x for each x X . The graph of I X is the part of the line y = x consisting of those points ( c , c ) with c X .
    恒等关系。对于任意集合 X,X 上的恒等关系是关系 I X ,使得对于任意 a 和 b,(a, b) ∈ I X 当且仅当 a ∈ X 且 a = b。I X 的箭头图包含从 x 到 x 的箭头,其中 x ∈ X。I X 的图像是直线 y = x 上由所有点 (c, c) 组成的部分,其中 c ∈ X。
  3. Composition of Relations. For any relations R and S , the composition of R followed by S is the relation S R such that for all a and b , ( a , b ) ∈ S R iff c , ( a , c ) R ( c , b ) S . The arrow diagram for S R is found by drawing an arrow from a to b whenever there is a path from a to b consisting of an arrow for R followed by an arrow for S . In most cases, S R R S . So composition of relations is not commutative.
    关系的复合。对于任意关系 R 和 S,R 后接 S 的复合关系 S ∘ R 满足:对于任意 a 和 b,(a, b) ∈ S ∘ R 当且仅当存在 c,(a, c) ∈ R ∧ (c, b) ∈ S。S ∘ R 的箭头图是通过绘制一条从 a 到 b 的箭头得到的,只要存在一条从 a 到 b 的路径,该路径由一条指向 R 的箭头和一条指向 S 的箭头组成。在大多数情况下,S ∘ R ≠ R ∘ S。因此,关系的复合不满足交换律。

Exercises

练习

  1. Let X = {1, 2, 3, 4, 5, 6}. Define a relation R from X to X as follows:
    R = { ( 1 , 2 ) , ( 1 , 4 ) , ( 1 , 6 ) , ( 2 , 4 ) , ( 4 , 5 ) , ( 4 , 6 ) , ( 6 , 3 ) } .
    (a) Draw an arrow diagram for R and the graph of R in R 2 . (b) Describe R −1 as a set of ordered pairs, as an arrow diagram, and as a graph. (c) Describe R R as a set of ordered pairs, as an arrow diagram, and as a graph. (d) Compute the following images of sets under the relations R and R −1 :
    R [ { 2 , 3 , 4 } ] , R [ { 3 , 5 } ] , R [ { 1 } ] , R 1 [ { 2 , 3 , 4 } ] , R 1 [ { 3 , 5 } ] ,
    R 1 [ { 1 } ] , R [ R 1 [ { 1 , 2 , 3 } ] ] , R 1 [ R [ { 1 , 2 , 3 } ] ] .

    令 X = {1, 2, 3, 4, 5, 6}。定义从 X 到 X 的关系 R 如下: R={(1,2),(1,4),(1,6),(2,4),(4,5),(4,6),(6,3)}。 (a) 为 R 绘制箭头图,并在 R2 中绘制 R 的图像。 (b) 将 R −1 描述为有序对集合、箭头图和图。 (c) 将 R ∘ R 描述为有序对集合、箭头图和图。 (d) 计算以下集合在关系 R 和 R −1 下的像:R[{2,3,4}],R[{3,5}],R[{1}],R−1[{2,3,4}],R−1[{3,5}],R−1[{1}],R[R−1[{1,2,3}]],R−1[R[{1,2,3}]]。
  2. Define X and R as in the previous problem. Define
    S = { ( 2 , 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 3 , 3 ) , ( 4 , 6 ) } .

    Draw arrow diagrams for each relation:

    (a) R S (b) S R (c) ( R S ) −1 (d) R −1 S −1 (e) S −1 R −1 (f) ( R S ) ∘ R .


    定义 X 和 R 与上题相同。定义 S= S={(2,1),(2,2),(2,3),(2,4),(1,5),(1,6),(3,3),(4,6)}。绘制下列关系的箭头图:(a) R ∘ S (b) S ∘ R (c) (R ∘ S) −1 (d) R −1 ∘ S −1 (e) S −1 ∘ R −1 (f) (R ∘ S) ∘ R。
  3. Let X = {1, 2, 3, 4}, R = {( a , b ) ∈ X × X : a < b } and S = {( a , b ) ∈ X × X : a b }. (a) Draw arrow diagrams for R and S . (b) Compute R −1 , S −1 , R S , S R , R R , S S , R S , and R S .
    设 X = {1, 2, 3, 4},R = {(a, b) ∈ X × X : a < b},S = {(a, b) ∈ X × X : a ≤ b}。 (a) 画出 R 和 S 的箭头图。 (b) 计算 R −1 、S −1 、R ∘ S、S ∘ R、R ∘ R、S ∘ S、R∪S 和 R∩S。
  4. Define a relation T from R to R by the following graph.

    ufig5_17.jpg

    (a) Carefully sketch the graph of T −1 . (b) What is the set T I Z ? (c) Compute T [{2}], T −1 [{2}], T [[6, 7]], and T −1 [[0, 1)].
    根据以下图定义从 R 到 R 的关系 T。 ufig5_17.jpg (a) 仔细绘制 T 的图。 −1 (b) 集合 T∩IZ 是什么? (c) 计算 T[{2}]、T[ −1 [{2}]、T[[6, 7]] 和 T[ −1 [[0, 1)]。
  5. Let R be the relation in Example 5.16 . (a) Find all subsets A of X such that R −1 [ R [ A ]] = A . (b) Find all subsets B of Y such that R [ R −1 [ B ]] = B .
    设 R 为例 5.16 中的关系。 (a) 求 X 的所有子集 A,使得 R −1 [R[A]] = A。 (b) 求 Y 的所有子集 B,使得 R[R −1 [B]] = B。
  6. Let S be the relation in Example 5.17 . (a) Find S [{ − 2, − 1}]. (b) Find S −1 [{ − 2, − 1}]. (c) Find S [[1, 2]]. (d) Find S −1 [[1, 2)]. (e) Find S 1 [ S [ R < 0 ] ] .
    设 S 为例 5.17 中的关系。(a) 求 S[{ − 2, − 1}]。(b) 求 S −1 [{ − 2, − 1}]。(c) 求 S[[1, 2]]。(d) 求 S −1 [[1, 2]]。(e) 求 S−1[S[R<0]]。
  7. (a) Draw an arrow diagram and a graph for I {−1,0,2} . (b) Draw the graph of I R Z . (c) Let X = {0, 1} × {1, 2, 3}. Draw an arrow diagram for I X .
    (a)画出 I {−1,0,2} 的箭头图和图像。(b)画出 IR−Z 的图像。(c)设 X = {0, 1} × {1, 2, 3}。画出 I X 的箭头图。
  8. Let R = { ( x , x 2 ) : x R } , S = { ( x , 3 x ) : x R } , and T = { ( x , x 1 ) : x R } . Draw graphs of the following relations: (a) R S (b) S R (c) R T (d) T R (e) S T (f) T R (g) R R −1 (h) R −1 R .
    设 R={(x,x2):x∈R},S={(x,3x):x∈R},T={(x,x−1):x∈R}。绘制下列关系的图像:(a) R ∘ S (b) S ∘ R (c) R ∘ T (d) T ∘ R (e) S ∘ T (f) T ∘ R (g) R ∘ R −1 (h) R −1 ∘ R。
  9. Let R = { ( x , y ) R 2 : x 2 + y 2 = 4 } and S = { ( x , y ) R 2 : y 2 = x 2 } . True or false? Explain each answer. (a) R −1 = R . (b) S −1 = S . (c) R R = S . (d) S S = S . (e) R S = R . (f) S R = R .
    设 R={(x,y)∈R2:x2+y2=4},S={(x,y)∈R2:y2=x2}。判断正误并解释每个答案。(a) R −1 = R。(b) S −1 = S。(c) R ∘ R = S。(d) S ∘ S = S。(e) R ∘ S = R。(f) S ∘ R = R。
  10. Repeat the previous exercise taking R = { ( x , y ) R 2 : x 2 + y = 4 } and S = { ( x , y ) R 2 : y = x } .
    重复前面的练习,取 R={(x,y)∈R2:x2+y=4} 和 S={(x,y)∈R2:y=−x}。
  11. Let A , B , C , and D be arbitrary sets. Describe the following relations as explicitly as possible: (a) ( A × B ) ∘ ( C × D ) (b) ( A × B ) −1 (c) I A I B (d) I A I B (e) ( A × B ) 0 (f) ( B × A ) ∘ R , where R is a relation from A to B .
    设 A、B、C 和 D 为任意集合。请尽可能明确地描述下列关系:(a)(A × B) ∘ (C × D)(b)(A × B) −1 (c)IA∩IB(d)IA∪IB(e)(A×B)∘0(f)(B × A) ∘ R,其中 R 是从 A 到 B 的关系。
  12. Let X = {1, 2, 3}. Give four examples of relations R from X to X satisfying R R = R = R −1 . In one of your examples, find a relation R consisting of five ordered pairs.
    设 X = {1, 2, 3}。给出四个从 X 到 X 的关系 R,满足 R ∘ R = R = R −1 。在你的一个例子中,找到一个由五个有序对组成的关系 R。
  13. Define X = {1, 2, 3, 4}, Y = { a , b , c }, and R = {(1, b ), (4, a ), (3, c )}. (a) Find a relation S from Y to X such that S R = I X , or explain why no such S exists. (b) Find a relation S from Y to X such that R S = I Y , or explain why no such S exists.
    定义 X = {1, 2, 3, 4},Y = {a, b, c},R = {(1, b), (4, a), (3, c)}。(a) 找到一个从 Y 到 X 的关系 S,使得 S ∘ R = I X ,或者解释为什么不存在这样的 S。(b) 找到一个从 Y 到 X 的关系 S,使得 R ∘ S = I Y ,或者解释为什么不存在这样的 S。
  14. Repeat Problem 13 taking X = R , Y = R 0 , and R = { ( x , y ) R 2 : y = x 2 } .
    重复问题 13,取 X=R,Y=R≥0,且 R={(x,y)∈R2:y=x2}。
  15. Repeat Problem 13 taking X = Y = [ − 1, 1] and R = { ( x , y ) R 2 : x 2 + y 2 = 1 } .
    重复问题 13,取 X = Y = [ − 1, 1] 和 R={(x,y)∈R2:x2+y2=1}。
  16. Define a relation R by the following arrow diagram.

    ufig5_18.jpg

    (a) Compute R −1 , R R −1 , and R −1 R . (b) Find three relations S from Y to X such that S R = I {1,4} . (c) Find all subsets B Y for which there exists a relation S from Y to X with R S = I B .
    用以下箭头图定义关系 R。 ufig5_18.jpg (a) 计算 R −1 、R ∘ R −1 和 R −1 ∘ R. (b) 找到从 Y 到 X 的三个关系 S,使得 S ∘ R = I {1,4} . (c) 找到所有子集 B ⊆ Y,使得存在从 Y 到 X 的关系 S,且 R ∘ S = I B
  17. Let R be a relation from X to Y .

    (a) Find a condition on R that guarantees the existence of a relation S from Y to X with S R = X × X . (b) Find a condition on R that guarantees the existence of a relation T from Y to X with R T = Y × Y .
    设 R 是从 X 到 Y 的关系。(a) 求 R 的一个条件,保证存在从 Y 到 X 的关系 S,且 S ∘ R = X × X。(b) 求 R 的一个条件,保证存在从 Y 到 X 的关系 T,且 R ∘ T = Y × Y。

5.3 Properties of Relations

5.3 关系的性质

In this section, we state and prove many properties of inverses, identity relations, and compositions. Before starting, you are encouraged to review the relevant proof templates from set theory and the definitions from the last two sections.

本节将阐述并证明逆元、恒等关系和复合关系的诸多性质。在开始之前,建议您回顾集合论中的相关证明模板以及前两节中的定义。

Closure Theorem for Relations

关系的封闭定理

We begin with a preliminary theorem indicating what kind of relations are produced by various constructions.

我们首先给出一个初步定理,指出各种构造会产生什么样的关系。

5.26. Closure Theorem for Relations. For all sets X , Y , Z , I 0 and all relations R , S , R i (for i I ):

5.26. 关系的封闭性定理。对于所有集合 X, Y, Z, I≠0 和所有关系 R, S, R i (对于 i ∈ I):

(a) If R and S are relations from X to Y , then R S and R S and R S are relations from X to Y .

(a)如果 R 和 S 是从 X 到 Y 的关系,那么 R∪S、R∩S 和 R − S 都是从 X 到 Y 的关系。

(b) If R i is a relation from X to Y for each i I , then i I R i and i I R i are relations from X to Y .

(b)如果对于每个 i ∈ I,R i 是从 X 到 Y 的关系,那么 ⋃i∈IRi 和 ⋂i∈IRi 是从 X 到 Y 的关系。

(c) R is a relation from X to Y iff R −1 is a relation from Y to X .

(c)R 是从 X 到 Y 的关系,当且仅当 R −1 是从 Y 到 X 的关系。

(d) I X is a relation from X to X .

(d)I X 是从 X 到 X 的关系。

(e) If R is a relation from X to Y and S is a relation from Y to Z , then S R is a relation from X to Z .

(e)如果 R 是从 X 到 Y 的关系,S 是从 Y 到 Z 的关系,那么 S ∘ R 是从 X 到 Z 的关系。

We prove (a) and (e) as illustrations. Fix arbitrary sets X , Y , and Z and arbitrary relations R and S . For (a), assume R and S are relations from X to Y , which means R X × Y and S X × Y . We must show R S and R S and R S are relations from X to Y . Part 1 . Prove R S X × Y . Since we assumed R X × Y and S X × Y , this follows from the least upper bound property (part (f) of the Theorem on Subsets; see page 106). Part 2 . Prove R S X × Y . We know R S R by part (c) of the Theorem on Subsets, and we assumed R X × Y , so R S X × Y follows by transitivity of ⊆ (part (b) of the Theorem on Subsets). Part 3 . Prove R S X × Y . As in part 2, we know R S R and R X × Y , so R S X × Y follows.

我们以证明 (a) 和 (e) 为例进行说明。固定任意集合 X、Y 和 Z 以及任意关系 R 和 S。对于 (a),假设 R 和 S 是从 X 到 Y 的关系,这意味着 R ⊆ X × Y 且 S ⊆ X × Y。我们必须证明 R∪S、R∩S 和 R − S 都是从 X 到 Y 的关系。第一部分:证明 R∪S⊆X×Y。由于我们假设 R ⊆ X × Y 且 S ⊆ X × Y,这可由最小上界性质(子集定理的 (f) 部分;参见第 106 页)得出。第二部分:证明 R∩S⊆X×Y。由子集定理的(c)可知R∩S⊆R,并且假设R⊆X×Y,因此由⊆的传递性(子集定理的(b))可知R∩S⊆X×Y。第三部分:证明R−S⊆X×Y。与第二部分类似,我们已知R−S⊆R且R⊆X×Y,因此R−S⊆X×Y成立。

For (e), assume R is a relation from X to Y and S is a relation from Y to Z . Prove S R is a relation from X to Z . We have assumed R X × Y and S Y × Z . We must prove S R X × Z . To prove this, fix an arbitrary ordered pair ( a , b ), assume ( a , b ) ∈ S R , and prove ( a , b ) ∈ X × Z . We have assumed c , ( a , c ) R ( c , b ) S . We must prove a X and b Z . First, since ( a , c ) ∈ R and R X × Y , we see that ( a , c ) ∈ X × Y . So a X and c Y , hence a X . Second, since ( c , b ) ∈ S and S Y × Z , we see that ( c , b ) ∈ Y × Z . So c Y and b Z , hence b Z .

对于 (e),假设 R 是从 X 到 Y 的关系,S 是从 Y 到 Z 的关系。证明 S ∘ R 是从 X 到 Z 的关系。我们假设 R ⊆ X × Y 且 S ⊆ Y × Z。我们需要证明 S ∘ R ⊆ X × Z。为了证明这一点,固定任意有序对 (a, b),假设 (a, b) ∈ S ∘ R,并证明 (a, b) ∈ X × Z。我们假设存在 c,(a, c) ∈ R ∧ (c, b) ∈ S。我们必须证明 a ∈ X 且 b ∈ Z。首先,因为 (a, c) ∈ R 且 R ⊆ X × Y,所以 (a, c) ∈ X × Y。因此 a ∈ X 且 c ∈ Y,故 a ∈ X。其次,因为 (c, b) ∈ S 且 S ⊆ Y × Z,所以 (c, b) ∈ Y × Z。因此 c ∈ Y 且 b ∈ Z,故 b ∈ Z。

Properties of Inverses, Identity Relations, and Composition

逆元、恒等关系和复合的性质

Our next theorem lists some general properties of inverse relations, identity relations, and composition of relations.

我们的下一个定理列出了逆关系、恒等关系和关系复合的一些一般性质。

5.27. Theorem on Relations. For all sets X , Y , C and all relations R , S , T :

5.27. 关于关系的定理。对于所有集合 X、Y、C 和所有关系 R、S、T:

(a) Inverse of Identity: I X 1 = I X .

(a)恒等式的逆元:IX−1=IX。

(b) Double Inverse: ( R −1 ) −1 = R .

(b)双重逆元:(R −1 −1 = R。

(c) Inverse of Union: ( S T ) 1 = S 1 T 1 = T 1 S 1 .

(c)并集的逆运算:(S∪T)−1=S−1∪T−1=T−1∪S−1。

(d) Inverse of Intersection: ( S T ) 1 = S 1 T 1 = T 1 S 1 .

(d)交集的逆运算:(S∩T)−1=S−1∩T−1=T−1∩S−1。

(e) Composition with Identity: If R is a relation from X to Y , then R I X = R and I Y R = R .

(e)带恒等元的组合:如果 R 是从 X 到 Y 的关系,则 R ∘ I X = R 且 I Y ∘ R = R。

(f) Associativity of Composition: T ∘ ( S R ) = ( T S ) ∘ R .

(f)组合的结合律:T ∘ (S ∘ R) = (T ∘ S) ∘ R。

(g) Inverse of a Composition: ( R S ) −1 = S −1 R −1 .

(g)复合的逆:(R ∘ S) −1 = S −1 ∘ R −1

(g) Image of a Composition: ( S R )[ C ] = S [ R [ C ]].

(g)构图的图像:(S ∘ R)[C] = S[R[C]]。

#

(i) Monotonicity Properties: If X Y , then I X I Y . If R S , then R −1 S −1 and T R T S and R T S T .

(i)单调性性质:如果 X ⊆ Y,则 I X ⊆ I Y 。如果 R ⊆ S,则 R −1 ⊆ S −1 且 T ∘ R ⊆ T ∘ S 且 R ∘ T ⊆ S ∘ T。

(j) Empty Set Properties: I 0 = 0 and 0 1 = 0 and R 0 = 0 and 0 R = 0 .

(j)空集性质:I0=0 且 0−1=0 且 R∘0=0 且 0∘R=0。

(k) Distributive Laws for and ∘:

(k)∪ 和 ∘ 的分配律:

( S S T T ) R = ( S S R ) ( T T R ) and T T ( S S R ) = ( T T S S ) ( T T R ) .

(l) Partial Distributive Laws for and ∘:

(l)∩和∘的部分分配律:

( S S T T ) R ( S S R ) ( T T R ) and T T ( S S R ) ( T T S S ) ( T T R ) .

Equality does not always hold.

平等并非总是成立。

Items involving and generalize to indexed unions and intersections. As noted earlier, composition of relations is not commutative : in most cases, R S S R .

涉及 ∪ 和 ∩ 的项可以推广到索引并集和交集。如前所述,关系的复合不满足交换律:在大多数情况下,R ∘ S ≠ S ∘ R。

We prove some representative parts of the theorem, leaving other parts to the exercises. You are strongly encouraged to work out proofs of all parts of this theorem now, before reading further. [ Hints: When proving subset statements or set equality statements involving relations, you should fix arbitrary ordered pairs ( a , b ) rather than fixing arbitrary objects. Chain proofs may work well for sufficiently simple set equalities, but two-part set equality proofs are preferred for more complex statements. When proving a set is empty, contradiction proofs are often the fastest way].

我们证明了定理的一些代表性部分,其余部分留给练习。强烈建议您在继续阅读之前,先完成定理所有部分的证明。[提示:在证明涉及关系的子集命题或集合等式时,您应该固定任意有序对 (a, b),而不是固定任意对象。链式证明对于足够简单的集合等式可能有效,但对于更复杂的命题,两步集合等式证明更佳。证明集合为空时,反证法通常是最快的方法。]

Have you proved the theorem yet? To see our proofs, read on. Fix arbitrary sets X , Y , C and relations R , S , and T .

你已经证明了这个定理吗?要查看我们的证明,请继续阅读。固定任意集合 X、Y、C 和关系 R、S、T。

5.28. Proof that ( R −1 ) −1 = R . We give a chain proof. Let ( a , b ) be an arbitrary ordered pair. Using the definition of inverse relation twice, we see that ( a , b ) ∈ ( R −1 ) −1 iff ( b , a ) ∈ R −1 iff ( a , b ) ∈ R . So ( R −1 ) −1 = R since these two relations contain exactly the same ordered pairs.

5.28. 证明 (R −1 ) −1 = R。我们给出链式证明。设 (a, b) 为任意有序对。两次使用逆关系定义,可知 (a, b) ∈ (R −1 ) −1 当且仅当 (b, a) ∈ R −1 当且仅当 (a, b) ∈ R。因此,(R −1 ) −1 = R,因为这两个关系包含完全相同的有序对。

5.29. Proof of Composition with Identity. Assume R is a relation from X to Y , which means R X × Y . We prove R I X = R (letting you prove I Y R = R ).

5.29. 用恒等式证明复合关系。假设 R 是从 X 到 Y 的关系,这意味着 R ⊆ X × Y。我们证明 R ∘ I X = R(由你来证明 I Y ∘ R = R)。

Part 1 . Prove R I X R . Fix ( a , b ); assume ( a , b ) ∈ R I X ; prove ( a , b ) ∈ R . By definition of composition, we know there exists c with ( a , c ) ∈ I X and ( c , b ) ∈ R . Next, by definition of I X , we know a X and a = c . Since ( c , b ) ∈ R , we deduce ( a , b ) ∈ R , as needed.

第一部分:证明 R ∘ I X ⊆ R。固定 (a, b);假设 (a, b) ∈ R ∘ I X ;证明 (a, b) ∈ R。根据复合的定义,我们知道存在 c 使得 (a, c) ∈ I X 且 (c, b) ∈ R。接下来,根据 I X 的定义,我们知道 a ∈ X 且 a = c。由于 (c, b) ∈ R,我们推导出 (a, b) ∈ R,满足要求。

Part 2 . Prove R R I X . Fix ( s , t ); assume ( s , t ) ∈ R ; prove ( s , t ) ∈ R I X . We must prove w , ( s , w ) I X ( w , t ) R . Choose w = s . On one hand, since ( s , t ) ∈ R and R X × Y , we see that ( s , t ) ∈ X × Y . So s X and t Y , which shows that w = s is a member of X . Then ( s , w ) = ( s , s ) ∈ I X . Next, ( w , t ) = ( s , t ) ∈ R by assumption.

第二部分:证明 R ⊆ R ∘ I X 。固定 (s, t);假设 (s, t) ∈ R;证明 (s, t) ∈ R ∘ I X 。我们需要证明存在 w,(s,w)∈IX∧(w,t)∈R。选择 w = s。一方面,由于 (s, t) ∈ R 且 R ⊆ X × Y,我们看到 (s, t) ∈ X × Y。因此 s ∈ X 且 t ∈ Y,这表明 w = s 是 X 的元素。那么 (s, w) = (s, s) ∈ I X 。接下来,根据假设,(w, t) = (s, t) ∈ R。

5.30. Proof of Distributive Law for and ∘ We give a chain proof of ( S T ) R = ( S R ) ( T R ) , leaving the proof of the companion identity to you. Fix an arbitrary ordered pair ( a , b ). We know

5.30. ∪ 和 ∘ 的分配律证明 我们给出 (S∪T)∘R=(S∘R)∪(T∘R) 的链式证明,伴随恒等式的证明留给您。固定任意有序对 (a, b)。我们知道

( a 一个 , b b ) ( S S T T ) R c c , ( a 一个 , c c ) R ( c c , b b ) S S T T c c , ( a 一个 , c c ) R [ ( c c , b b ) S S ( c c , b b ) T T ] c c , ( [ ( a 一个 , c c ) R ( c c , b b ) S S ] [ ( a 一个 , c c ) R ( c c , b b ) T T ] ) ( c c , [ ( a 一个 , c c ) R ( c c , b b ) S S ] ) ( c c , [ ( a 一个 , c c ) R ( c c , b b ) T T ] ) ( a 一个 , b b ) S S R ( a 一个 , b b ) T T R ( a 一个 , b b ) ( S S R ) ( T T R ) .

The steps follow by definition of composition, definition of union, the distributive law for and ∨, the distributive law for and ∨, the definition of composition, and the definition of union.

接下来的步骤依次是:组合的定义、并集的定义、∧ 和 ∨ 的分配律、∃ 和 ∨ 的分配律、组合的定义和并集的定义。

5.31. Proof of Empty Set Properties. We prove that 0 1 = 0 . Assume, to get a contradiction, that 0 1 0 . Then there exists an object z 0 1 ; in fact, there exists an ordered pair ( a , b ) 0 1 . By definition of inverse, ( b , a ) 0 . But we also know ( b , a ) 0 , which gives us a contradiction.

5.31. 空集性质的证明。我们证明 0−1=0。为了得出矛盾,假设 0−1≠0。那么存在一个对象 z∈0−1;事实上,存在一个有序对 (a,b)∈0−1。根据逆的定义,(b,a)∈0。但我们也知道 (b,a)∉0,这就产生了矛盾。

Next we prove that R 0 = 0 . Assume, to get a contradiction, that R 0 0 . Then there exists an ordered pair ( a , b ) R 0 . By definition of composition, we know there exists c with ( a , c ) 0 and ( c , b ) ∈ R . But ( a , c ) 0 , so we have found a contradiction.

接下来我们证明 R∘0=0。为了得出矛盾,假设 R∘0≠0。那么存在有序对 (a,b)∈R∘0。根据复合的定义,我们知道存在 c 使得 (a,c)∈0 且 (c,b)∈R。但 (a,c)∉0,所以我们得到了矛盾。

5.32. Proof of Partial Distributive Law. We prove T ( S R ) ( T S ) ( T R ) . Fix ( a , b ); assume ( a , b ) T ( S R ) ; prove ( a , b ) ( T S ) ( T R ) . Our assumption means there exists w with ( a , w ) S R and ( w , b ) ∈ T ; so there exists w with ( a , w ) ∈ S and ( a , w ) ∈ R and ( w , b ) ∈ T . We must prove ( a , b ) ∈ T S and ( a , b ) ∈ T R . First, to prove ( a , b ) ∈ T S , we must prove x , ( a , x ) S ( x , b ) T . Choose x = w ; note that ( a , w ) ∈ S and ( w , b ) ∈ T have been assumed. Second, to prove ( a , b ) ∈ T R , we must prove y , ( a , y ) R ( y , b ) T . Choose y = w ; note that ( a , w ) ∈ R and ( w , b ) ∈ T have been assumed.

5.32. 部分分配律的证明。我们证明 T∘(S∩R)⊆(T∘S)∩(T∘R)。固定 (a, b);假设 (a,b)∈T∘(S∩R);证明 (a,b)∈(T∘S)∩(T∘R)。我们的假设意味着存在 w 使得 (a,w)∈S∩R 且 (w, b) ∈ T;因此存在 w 使得 (a, w) ∈ S 且 (a, w) ∈ R 且 (w, b) ∈ T。我们必须证明 (a, b) ∈ T ∘ S 且 (a, b) ∈ T ∘ R。首先,为了证明 (a, b) ∈ T ∘ S,我们必须证明存在 x,(a,x)∈S∧(x,b)∈T。选择 x = w;注意,这里假设了 (a, w) ∈ S 且 (w, b) ∈ T。其次,为了证明 (a, b) ∈ T ∘ R,我们必须证明存在 y,(a, y) ∈ R ∧ (y, b) ∈ T。取 y = w;注意,这里假设了 (a, w) ∈ R 且 (w, b) ∈ T。

We give an example to show that set equality may not hold. Let R = {(1, 2)}, S = {(1, 3)}, and T = {(2, 4), (3, 4)}. By drawing arrow diagrams or looking at the ordered pairs, we find that T S = {(1, 4)}, T R = {(1, 4)}, and so ( T S ) ( T R ) = { ( 1 , 4 ) } . But S R = 0 , so T ( S R ) = 0 . Thus T ( S R ) is a proper subset of ( T S ) ( T R ) in this example.

我们给出一个例子来说明集合相等性可能不成立。设 R = {(1, 2)},S = {(1, 3)},T = {(2, 4), (3, 4)}。通过绘制箭头图或观察有序对,我们发现 T ∘ S = {(1, 4)},T ∘ R = {(1, 4)},因此 (T∘S)∩(T∘R) = {(1, 4)}。但 S∩R = 0,所以 T∘(S∩R) = 0。因此,在本例中,T∘(S∩R) 是 (T∘S)∩(T∘R) 的真子集。

5.33. Partial Proof of Associativity of Composition. We prove T ∘ ( S R ) ⊆ ( T S ) ∘ R ; we let you prove ( T S ) ∘ R T ∘ ( S R ). Fix ( a , b ); assume ( a , b ) ∈ T ∘ ( S R ); prove ( a , b ) ∈ ( T S ) ∘ R . We have assumed there exists w with ( a , w ) ∈ S R and ( w , b ) ∈ T . Expanding this again, we have assumed there exists w such that there exists x with ( a , x ) ∈ R and ( x , w ) ∈ S and ( w , b ) ∈ T . Now, since ( x , w ) ∈ S and ( w , b ) ∈ T , we see that ( x , b ) ∈ T S . Since ( a , x ) ∈ R and ( x , b ) ∈ T S , it follows that ( a , b ) ∈ ( T S ) ∘ R , as needed.

5.33. 组合结合律的部分证明。我们证明 T ∘ (S ∘ R) ⊆ (T ∘ S) ∘ R;我们留给你们证明 (T ∘ S) ∘ R ⊆ T ∘ (S ∘ R)。固定 (a, b);假设 (a, b) ∈ T ∘ (S ∘ R);证明 (a, b) ∈ (T ∘ S) ∘ R。我们假设存在 w,使得 (a, w) ∈ S ∘ R 且 (w, b) ∈ T。进一步扩展,我们假设存在 w,使得存在 x,使得 (a, x) ∈ R 且 (x, w) ∈ S 且 (w, b) ∈ T。现在,由于 (x, w) ∈ S 且 (w, b) ∈ T,我们看到 (x, b) ∈ T ∘ S。由于 (a, x) ∈ R 且 (x, b) ∈ T ∘ S,因此 (a, b) ∈ (T ∘ S) ∘ R,证毕。

5.34. Proof of ( R S ) −1 = S −1 R −1 . We try a chain proof. Fix an ordered pair ( a , b ). We know

5.34. 证明 (R ∘ S) −1 = S −1 ∘ R −1 。我们尝试链式证明。固定一个有序对 (a, b)。我们知道

( a 一个 , b b ) ( R S S ) 1 ( b b , a 一个 ) R S S (by definition of inverse) (根据逆的定义) c c , ( b b , c c ) S S ( c c , a 一个 ) R (by definition of composition) (根据组成原理) c c , ( c c , a 一个 ) R ( b b , c c ) S S (by commutativity of AND) (根据“与”的交换律) c c , ( a 一个 , c c ) R 1 ( c c , b b ) S S 1 (by definition of inverse) (根据逆的定义) ( a 一个 , b b ) S S 1 R 1 (by definition of composition). (根据组成原理)

Section Summary

章节概要

  1. Closure Properties of Relations. The union, intersection, and set difference of relations from X to Y are also relations from X to Y . The inverse of a relation from X to Y is a relation from Y to X . The identity relation I X is a relation from X to X . If R is a relation from X to Y and S is a relation from Y to Z , then S R is a relation from X to Z .
    关系的封闭性质。从 X 到 Y 的关系的并集、交集和差集仍然是从 X 到 Y 的关系。从 X 到 Y 的关系的逆关系是从 Y 到 X 的关系。恒等关系 I X 是从 X 到 X 的关系。如果 R 是从 X 到 Y 的关系,S 是从 Y 到 Z 的关系,那么 S ∘ R 是从 X 到 Z 的关系。
  2. Properties of Inverse Relations. For all sets X and relations R and S , I X 1 = I X , ( R −1 ) −1 = R , ( R S ) 1 = R 1 S 1 , ( R S ) 1 = R 1 S 1 , ( R S ) −1 = S −1 R −1 , R S implies R −1 S −1 , and 0 1 = 0 .
    逆关系的性质。对于所有集合 X 和关系 R 和 S,IX−1=IX,(R −1 ) −1 = R,(R∪S)−1=R−1∪S−1,(R∩S)−1=R−1∩S−1,(R ∘ S) −1 = S −1 ∘ R −1 ,R ⊆ S 蕴含 R −1 ⊆ S −1 ,且 0−1=0。
  3. Properties of Composition of Relations. For all relations R from X to Y , R I X = R = I Y R . Composition of relations is associative but not commutative: T ∘ ( S R ) = ( T S ) ∘ R always holds, but S R = R S does not always hold. We have distributive laws such as ( S T ) R = ( S R ) ( T R ) and ( S T ) R ( S R ) ( T R ) .
    关系组合的性质。对于从 X 到 Y 的所有关系 R,R ∘ I X = R = I Y ∘ R。关系组合满足结合律但不满足交换律:T ∘ (S ∘ R) = (T ∘ S) ∘ R 总是成立,但 S ∘ R = R ∘ S 并非总是成立。我们有分配律,例如 (S∪T)∘R=(S∘R)∪(T∘R) 和 (S∩T)∘R⊆(S∘R)∩(T∘R)。

Exercises

练习

  1. Prove parts (b), (c), and (d) of the Closure Theorem for Relations.
    证明关系封闭定理的(b)、(c)和(d)部分。
  2. Prove part (a) of the Theorem on Relations.
    证明关系定理的(a)部分。
  3. Prove parts (c) and (d) of the Theorem on Relations.
    证明关系定理的(c)和(d)部分。
  4. Finish the proof of part (e) of the Theorem on Relations.
    完成关系定理第(e)部分的证明。
  5. Finish the proof of part (f) of the Theorem on Relations.
    完成关系定理(f)部分的证明。
  6. Prove part (h) of the Theorem on Relations.
    证明关系定理的第(h)部分。
  7. Prove part (i) of the Theorem on Relations.
    证明关系定理的第(i)部分。
  8. Finish the proof of part (j) of the Theorem on Relations.
    完成关系定理第(j)部分的证明。
  9. Finish the proof of part (k) of the Theorem on Relations.
    完成关系定理第(k)部分的证明。
  10. Finish the proof of part (l) of the Theorem on Relations.
    完成关系定理第(l)部分的证明。
  11. Let A and B be arbitrary sets. (a) Prove I A B = I A I B . (b) Prove I A B = I A I B = I A I B .
    设 A 和 B 为任意集合。 (a) 证明 IA∪B=IA∪IB。 (b) 证明 IA∩B=IA∩IB=IA∘IB。
  12. Prove: for all relations R and S , R S = S R iff R −1 S −1 = S −1 R −1 .
    证明:对于所有关系 R 和 S,R ∘ S = S ∘ R 当且仅当 R −1 ∘ S −1 = S −1 ∘ R −1
  13. (a) Prove: for all sets X and all relations R , S , T on X , if R −1 R = I X and R S = R T , then S = T . (b) Show that the result in (a) need not hold if we drop the hypothesis R −1 R = I X .
    (a)证明:对于所有集合 X 和 X 上的所有关系 R、S、T,如果 R −1 ∘ R = I X 且 R ∘ S = R ∘ T,则 S = T. 。(b)证明,如果我们放弃假设 R −1 ∘ R = I X ,则 (a) 中的结果不一定成立。
  14. Prove: for all relations R , if R R −1 , then R = R −1 .
    证明:对于所有关系 R,如果 R ⊆ R −1 ,则 R = R −1
  15. (a) Prove: for all sets X and all relations R on X , if I X R , then R R R . (b) Prove or disprove the converse of part (a).
    (a)证明:对于所有集合 X 和 X 上的所有关系 R,如果 I X ⊆ R,则 R ⊆ R ∘ R. (b)证明或反证 (a) 部分的逆命题。
  16. Prove or disprove: for all sets X and all relations R on X , if R = R −1 , then R R = I X .
    证明或反证:对于所有集合 X 和 X 上的所有关系 R,如果 R = R −1 ,则 R ∘ R = I X
  17. Prove or disprove: for all sets X and all relations R on X , if R R = I X , then R = R −1 .
    证明或反证:对于所有集合 X 和 X 上的所有关系 R,如果 R ∘ R = I X ,则 R = R −1
  18. Prove: for all relations R , S , and T , ( S R ) T = 0 iff ( T R 1 ) S = 0 .
    证明:对于所有关系 R、S 和 T,(S∘R)∩T=0 当且仅当 (T∘R−1)∩S=0。
  19. Prove: for all relations R , S , and T , T ( S R ) T ( R 1 R ) . Give an example where equality does not hold.
    证明:对于任意关系 R、S 和 T,T∩(S∘R)⊆T∘(R−1∘R)。举例说明等式不成立的情况。
  20. Prove: for all relations R , S , and T , R S T ( T S 1 ) R . Give an example where equality does not hold.
    证明:对于任意关系 R、S 和 T,R∩S∩T⊆(T∘S−1)∘R。举例说明等式不成立的情况。
  21. Prove: for all relations R , S , and T , if T S S and T −1 S S , then S ( T R ) = T ( S R ) .
    证明:对于所有关系 R、S 和 T,如果 T ∘ S ⊆ S 且 T −1 ∘ S ⊆ S,则 S∩(T∘R)=T∘(S∩R)。
  22. Prove: for all relations R , S , T , U , ( S R ) ( U T ) [ U ( ( U 1 S ) ( T R 1 ) ) ] R . Does equality always hold?
    证明:对于所有关系 R、S、T、U,(S∘R)∩(U∘T)⊆[U∘((U−1∘S)∩(T∘R−1))]∘R。等式总是成立吗?

5.4 Definition of Functions

5.4 函数的定义

One of the most fundamental and pervasive concepts in mathematics is the notion of a function. In calculus, we study real-valued functions of a real variable such as the exponential function, the natural logarithm function, polynomial functions, and trigonometric functions. In advanced calculus and other parts of mathematics, we need to consider more general kinds of functions. In this section, we give a formal definition of functions within the framework of set theory. We illustrate the logical nuances of this definition with many examples of functions and non-functions, presented using arrow diagrams, graphs, and formulas.

函数是数学中最基本、最普遍的概念之一。在微积分中,我们研究实变量的实值函数,例如指数函数、自然对数函数、多项式函数和三角函数。在高等微积分和数学的其他分支中,我们需要考虑更一般的函数。本节将在集合论的框架下给出函数的正式定义。我们将通过大量的函数和非函数示例来阐释该定义的逻辑细节,这些示例将使用箭头图、图形和公式来呈现。

Arrow Diagrams of Functions and Non-Functions

函数和非函数的箭头图

We begin with an informal discussion of functions, which motivates the formal definition below. Informally, a function g from a set A into a set B is a rule that assigns to each object x in A a unique object g ( x ) in B . The set A is called the domain of g ; the set B is called the codomain of g ; and g ( x ) is called the value of g at the input x . The crucial point is that each input x in A has exactly one output g ( x ) associated to it, and this output is required to belong to the codomain B . We use the notation g : A B to mean that g is a function with domain A and codomain B .

我们首先对函数进行非正式的讨论,这为下文的正式定义提供了依据。通俗地说,从集合 A 到集合 B 的函数 g 是一条规则,它将 A 中的每个对象 x 映射到 B 中唯一的对象 g(x)。集合 A 称为 g 的定义域;集合 B 称为 g 的值域;g(x) 称为 g 在输入 x 处的值。关键在于,A 中的每个输入 x 都恰好对应一个输出 g(x),并且该输出必须属于值域 B。我们使用符号 g:A→B 来表示 g 是一个定义域为 A、值域为 B 的函数。

These requirements can be conveniently visualized using arrow diagrams. Consider the six arrow diagrams in Figure 5.1 . The three arrow diagrams in the top row all represent valid functions from the set A = { v , w , x , y } to the set B = {1, 2, 3, 4}. In each case, there is exactly one arrow emanating from each point in the domain A , and each such arrow leads into the codomain B . We have, for example, g 1 ( w ) = 4, g 2 ( y ) = 2 = g 2 ( v ), and g 3 ( a ) = 4 for all a A . Note that it is permissible for several arrows to hit the same element of B . For example, in the diagram for g 3 , four different arrows hit the same point 4 ∈ B . Similarly, there may exist points in B that are not hit by any arrows. For example, in the diagram for g 2 , no arrow leads to 1 ∈ B , and yet g 2 is still considered to be a function mapping A into B .

这些要求可以用箭头图方便地可视化。考虑图 5.1 中的六个箭头图。顶行的三个箭头图都表示从集合 A = {v, w, x, y} 到集合 B = {1, 2, 3, 4} 的有效函数。在每种情况下,定义域 A 中的每个点都恰好有一条箭头指向值域 B,且每条箭头都指向值域 B。例如,对于所有 a ∈ A,我们有 g 1 (w) = 4,g 2 (y) = 2 = g 2 (v),以及 g 3 (a) = 4。注意,允许多条箭头指向 B 中的同一个元素。例如,在 g 3 的图中,四条不同的箭头指向同一个点 4 ∈ B。类似地,B 中可能存在一些点没有被任何箭头指向。例如,在 g 2 的图中,没有箭头指向 1 ∈ B,但 g 2 仍然被认为是将 A 映射到 B 的函数。

fig5_1

Figure 5.1

图 5.1

Arrow diagrams illustrating functions and non-functions.

用箭头图表示功能和非功能。

On the other hand, g 4 is not a function from A to B because there is no arrow leaving the input point w A . The arrow diagram g 5 is not a function because there are multiple arrows leaving v (and also y ). Finally, g 6 is not a function from A to B because g ( x ) = 0, which is not an element of the codomain B . However, g 6 can be regarded as a function from A to the set {0, 1, 2, 3, 4}.

另一方面,g 4 不是从 A 到 B 的函数,因为没有箭头从输入点 w ∈ A 出发。箭头图 g 5 也不是函数,因为有多条箭头从 v(以及 y)出发。最后,g 6 不是从 A 到 B 的函数,因为 g(x) = 0,而 0 不是值域 B 的元素。然而,g 6 可以被视为从 A 到集合 {0, 1, 2, 3, 4} 的函数。

Graphs of Functions and Non-Functions

函数和非函数的图像

We continue to build intuition for the function concept by looking at the graphs of some functions and non-functions with domain R and codomain R . Informally, given a function f : R R , the graph of f is the set of ordered pairs G f = { ( x , f ( x ) ) : x R } , which is a certain relation from R to R . The graph G f encodes exactly the information needed to understand how the function f transforms inputs to outputs. Specifically, for each input x R , there must exist a unique y R such that ( x , y ) belongs to the graph G f , and we then know that y = f ( x ) is the output associated with the input x by the function f .

我们继续通过观察一些定义域为 R 且值域为 R 的函数和非函数的图像来构建对函数概念的直觉。通俗地说,给定一个函数 f:R→R,f 的图像是有序对的集合 Gf={(x,f(x)):x∈R},它表示从 R 到 R 的一种特定关系。图 G f 精确地编码了理解函数 f 如何将输入转换为输出所需的信息。具体来说,对于每个输入 x∈R,必然存在唯一的 y∈R,使得 (x, y) 属于图 G f ,因此我们知道 y = f(x) 是函数 f 与输入 x 对应的输出。

For example, consider the graphs shown in Figure 5.2 . Each graph illustrates a relation on R × R consisting of certain ordered pairs ( x , y ). The first three pictures are graphs of functions from R to R . Specifically, G 1 is the graph of the function g 1 : R R defined by g 1 ( x ) = −| x | for each real x . G 2 is the graph of the function g 2 : R R defined by g 2 ( x ) = ( x 2 /4) + 1 for each real x . G 3 is the graph of the constant function g 3 : R R defined by g 3 ( x ) = 2 for all real x . In each case, every particular real input x has exactly one associated output y . Some of the potential outputs y R do not appear as actual outputs of the functions, but this does not matter. Similarly, some outputs have more than one associated input, for instance 2 = g 3 (0) = g 3 (1) = g 3 (3), but this does not matter either.

例如,考虑图 5.2 所示的图形。每个图形都展示了 R×R 上由若干有序对 (x, y) 构成的一个关系。前三个图形是从 R 到 R 的函数图像。具体来说,G 1 是函数 g1:R→R 的图像,其中 g 1 (x) = −|x|,对于每个实数 x。G 2 是函数 g2:R→R 的图像,其中 g 2 (x) = (x 2 /4) + 1,对于每个实数 x。G 3 是常数函数 g3:R→R 的图像,其中 g 3 (x) = 2,对于所有实数 x。在每种情况下,每个特定的实数输入 x 都恰好对应一个输出 y。某些潜在输出 y∈R 并未作为函数的实际输出出现,但这并不重要。类似地,某些输出有多个关联的输入,例如 2 = g 3 (0) = g 3 (1) = g 3 (3),但这同样不重要。

fig5_2

Figure 5.2

图 5.2

Graphs of functions and non-functions.

函数和非函数的图像。

On the other hand, G 4 = { ( x , y ) R × R : x 2 + y 2 = 4 } is not the graph of any function from R to R . One reason is that some inputs x in the domain R do not have any associated output y in the codomain. For example, there is no y R such that (3, y ) ∈ G 4 . We could try to circumvent this difficulty by shrinking the domain of the function from all of R to the closed interval [ − 2, 2]. But we still do not get a function, because there are some inputs that have more than one associated output based on the graph. For example, (0, 2) ∈ G 4 and (0, − 2) ∈ G 4 . In terms of formulas, the condition defining the relation G 4 can be rewritten y = ± 4 x 2 for −2 ≤ x ≤ 2. This is a double-valued expression, but we insist that all functions be single-valued. One possible way to extract a function from this example would be to define g 4 : [ 2 , 2 ] R by g 4 ( x ) = 4 x 2 for −2 ≤ x ≤ 2. The graph of this function would be just the lower half of the circle of radius 2 centered at the origin.

另一方面,G4={(x,y)∈R×R:x²+y²=4} 不是任何从 R 到 R 的函数的图像。原因之一是定义域 R 中的某些输入 x 在值域中没有对应的输出 y。例如,不存在 y∈R 使得 (3, y) ∈ G 4 。我们可以尝试通过将函数的定义域从整个 R 缩小到闭区间 [-2, 2] 来规避这个问题。但是我们仍然无法得到一个函数,因为根据图像,某些输入有多个对应的输出。例如,(0, 2) ∈ G 4 且 (0, -2) ∈ G 4 。就公式而言,定义关系 G 4 的条件可以重写为 y=±4−x2,其中 −2 ≤ x ≤ 2。这是一个双值表达式,但我们坚持所有函数都是单值的。从这个例子中提取函数的一种方法是定义 g4:[−2,2]→R,其中 g4(x)=−4−x2,其中 −2 ≤ x ≤ 2。该函数的图像将是原点为圆心、半径为 2 的圆的下半部分。

Next, G 5 is not the graph of a function with domain R , since each negative real input x has no associated output using this graph. If we shrink the domain to R 0 , we would obtain a function g 5 : R 0 R given by g 5 ( x ) = x for all x ≥ 0. Here we could also shrink the codomain (the set of allowable outputs) to get a function g 5 : R 0 R 0 given by g 5 ( x ) = x for all x ≥ 0. Are g 5 and g 5 * the same function? Based on the formal definition below, the answer turns out to be no, because g 5 and g 5 * have different codomains. Here, g 5 and g 5 * are different functions with the same graph and the same domain.

其次,G 5 不是定义域为 R 的函数的图像,因为每个负实数输入 x 在该图像上都没有对应的输出。如果我们缩小定义域至 R≥0,则会得到一个函数 g5:R≥0→R,其中 g5(x)=x,对于所有 x ≥ 0。这里我们也可以缩小值域(允许输出的集合),得到一个函数 g5∗:R≥0→R≥0,其中 g5∗(x)=x,对于所有 x ≥ 0。g 5 和 g 5 * 是同一个函数吗?根据下面的正式定义,答案是否定的,因为 g 5 和 g 5 * 的值域不同。这里,g 5 和 g 5 * 是不同的函数,但它们具有相同的图像和相同的定义域。

Finally, G 6 is not the graph of a function because of the ordered pairs (1, 1) ∈ G 6 and (1, − 1) ∈ G 6 . If exactly one of these ordered pairs were removed from G 6 , we would get the graph of a function g 6 : R R . We could make additional functions with the same domain and graph by changing the codomain to be any set containing the image G 6 [ R ] of the relation G 6 .

最后,由于有序对 (1, 1) ∈ G 6 和 (1, −1) ∈ G 6 ,G 6 不是函数的图像。如果从 G 6 中移除其中一个有序对,我们将得到函数 g6:R→R 的图像。通过将值域更改为包含关系 G 6 的像 G6[R] 的任意集合,我们可以构造具有相同定义域和图像的其他函数。

Recall the key condition in the informal definition of a function: g is a function from the domain A R to the codomain R iff for each x A there exists exactly one y with y = g ( x ), and moreover y must belong to R . A given relation G R 2 will be the graph of such a function iff the following geometric property holds: for every x 0 A , the vertical line x = x 0 intersects G in exactly one point; and for every x 0 R A , the vertical line x = x 0 does not intersect G at all. In beginning calculus, this is often called the vertical line test for determining whether a given graph is a function. One point not often stressed in elementary treatments is that we must know the intended domain A in advance in order to see if a graph passes this test. Often A is tacitly assumed to be the largest possible domain that might work for the given graph, namely A = { x R : y R , ( x , y ) G } .

回顾函数非正式定义中的关键条件:g 是从定义域 A⊆R 到值域 R 的函数,当且仅当对于每个 x ∈ A,存在唯一一个 y 使得 y = g(x),并且 y 必须属于 R。给定关系 G⊆R² 是此类函数的图像,当且仅当以下几何性质成立:对于每个 x 0 ∈ A,垂直线 x = x 0 与 G 恰好相交于一点;并且对于每个 x0∈R−A,垂直线 x = x 0 与 G 不相交。在初级微积分中,这通常被称为垂直线检验,用于判断给定的图像是否为函数。初等教材中经常忽略的一点是,我们必须预先知道目标定义域 A 才能判断图像是否通过此检验。通常默认 A 是给定图可能适用的最大域,即 A={x∈R:∃y∈R,(x,y)∈G}。

Formal Definition of a Function

函数的正式定义

We are now ready to present the formal definition of a function. This definition needs to capture all the information contained in the arrow diagrams and graphs we have been discussing: the input set A , the output set B , the rule by which inputs x A are associated with outputs y B , and the key requirement that each input have exactly one associated output. The crucial insight needed to formalize these ideas is that the vague idea of a “rule” for transforming inputs to outputs can be captured precisely by forming the relation consisting of all ordered pairs ( x , y ) such that input x is mapped to output y by the function. By analogy with the case where A = B = R , this relation is called the graph of the function. For instance, the function g 1 shown in Figure 5.1 has graph G 1 = {( v , 2), ( w , 4), ( x , 1), ( y , 3)}. These considerations motivate the following formal definition.

现在我们可以给出函数的正式定义。这个定义需要涵盖我们之前讨论过的箭头图和图形中包含的所有信息:输入集 A、输出集 B、输入 x ∈ A 与输出 y ∈ B 之间的关联规则,以及每个输入恰好对应一个输出这一关键要求。将这些概念形式化的关键在于,我们可以精确地理解“规则”这一模糊的输入到输出转换的概念,即通过构建一个关系来精确描述所有满足以下条件的有序对 (x, y):输入 x 被函数映射到输出 y。类比 A=B=R 的情况,这个关系被称为函数的图形。例如,图 5.1 中所示的函数 g 1 的图形为 G 1 = {(v, 2), (w, 4), (x, 1), (y, 3)}。基于这些考虑,我们提出了以下正式定义。

5.35. Definition: Functions. A function g is an ordered triple g = ( A , B , G ) such that: A and B are sets, G is a relation from A to B , and x A , ! y , ( x , y ) G .

5.35. 定义:函数。函数 g 是一个有序三元组 g = (A, B, G),其中:A 和 B 是集合,G 是从 A 到 B 的关系,并且 ∀x∈A,∃!y,(x,y)∈G。

A is called the domain of g ; B is called the codomain of g ; and G is called the graph of g . The notation g : A B means that g is a function with domain A and codomain B .

称 A 为函数 g 的定义域;称 B 为函数 g 的值域;称 G 为函数 g 的图像。记号 g:A→B 表示 g 是一个定义域为 A、值域为 B 的函数。

For all objects x , y , we write y = g ( x ) i f f ( x , y ) G .

对于所有对象 x、y,我们记 y=g(x) 当且仅当 (x,y)∈G。

So we can describe the graph of g as G = { ( x , g ( x ) ) : x A } .

因此,我们可以将 g 的图像描述为 G={(x,g(x)):x∈A}。

We must emphasize at the outset that the formal ordered triple notation g = ( A , B , G ) is almost never used ; instead, we write g : A B to introduce a new function with domain A and codomain B . Similarly, only in rare cases do we need to mention the graph by writing ( x , y ) ∈ G ; instead, we use the equivalent function notation y = g ( x ). If we say “let g : A B be defined by g ( x ) = (expr),” where (expr) is some expression involving x , it is understood that the graph of g is the relation G consisting of all ordered pairs ( x , y ) where x A and y is the value of the expression for this choice of x .

首先必须强调的是,形式化的有序三元组符号 g = (A, B, G) 几乎从不使用;相反,我们用 g : A → B 来表示定义域为 A、值域为 B 的新函数。类似地,只有在极少数情况下才需要用 (x, y) ∈ G 来表示图;相反,我们使用等价的函数符号 y = g(x)。如果我们说“令 g : A → B 由 g(x) = (expr) 定义”,其中 (expr) 是包含 x 的某个表达式,则 g 的图是关系 G,它由所有满足 x ∈ A 且 y 为该表达式在 x 取此值时的值的有序对 (x, y) 组成。

Let us spell out the three conditions that the relation G must satisfy in order to be the graph of a function g : A B .

让我们明确指出关系 G 必须满足的三个条件,才能成为函数 g : A → B 的图像。

(a) Valid Inputs and Outputs: Every member of G is an ordered pair ( x , y ) with x A and y B .

(a)有效输入和输出:G 的每个成员都是一个有序对 (x, y),其中 x ∈ A 且 y ∈ B。

(b) Existence of Outputs: For all x A , there exists y with ( x , y ) ∈ G .

(b)输出的存在性:对于所有 x ∈ A,存在 y 使得 (x, y) ∈ G。

(c) Uniqueness of Outputs: For all x , y , z , if ( x , y ) ∈ G and ( x , z ) ∈ G , then y = z .

(c)输出的唯一性:对于所有 x、y、z,如果(x、y)∈ G 且(x、z)∈ G,则 y = z。

Here are the same three conditions for g : A B to be a function, written in function notation.

以下是函数 g : A → B 成为函数的三个相同条件,用函数符号表示。

(a) Valid Inputs and Outputs: For all x , y , if y = g ( x ), then x A and y B .

(a)有效输入和输出:对于所有 x、y,如果 y = g(x),则 x ∈ A 且 y ∈ B。

(b) Existence of Outputs: For all x A , there exists y such that y = g ( x ).

(b)输出的存在性:对于所有 x ∈ A,存在 y 使得 y = g(x)。

(c) Uniqueness of Outputs: For all x , y , z , if y = g ( x ) and z = g ( x ), then y = z .

(c)输出的唯一性:对于所有 x、y、z,如果 y = g(x) 且 z = g(x),则 y = z。

The uniqueness condition can be stated in a way that does not explicitly mention y and z . (c′) Uniqueness of Outputs: For all x 1 , x 2 A , if x 1 = x 2 , then g ( x 1 ) = g ( x 2 ).

唯一性条件可以用不显式提及 y 和 z 的方式来表述。(c′) 输出的唯一性:对于所有 x 1 , x 2 ∈ A,如果 x 1 = x 2 ,则 g(x 1 ) = g(x 2 )。

The main technical drawback of the function notation g ( x ) is that the very use of the notation suggests that the existence and uniqueness of the output g ( x ) associated with input x are already known. But in fact, when introducing a new function for the first time, we really need to prove these assertions before the function notation can be legitimately used. Some of the dangers of using this notation prematurely are illustrated in the following subsection.

函数符号 g(x) 的主要技术缺陷在于,使用该符号本身就暗示着与输入 x 相关的输出 g(x) 的存在性和唯一性已经已知。但实际上,在首次引入一个新函数时,我们确实需要先证明这些断言,才能合法地使用该函数符号。以下小节将阐述过早使用该符号的一些风险。

5.36. Example. Consider the arrow diagram labeled g 2 in Figure 5.2 . We could formally define the function g 2 by setting

5.36. 示例。考虑图 5.2 中标记为 g 2 的箭头图。我们可以通过设置来正式定义函数 g 2

g 2 = ( { v v , w 西 , x x , y } , { 1 , 2 , 3 , 4 } , { ( v v , 2 ) , ( w 西 , 3 ) , ( x x , 4 ) , ( y , 2 ) } ) .

But we would usually introduce this function by saying “define g 2 :{ v , w , x , y } → {1, 2, 3, 4} by setting g ( v ) = 2, g ( w ) = 3, g ( x ) = 4, and g ( y ) = 2.”

但我们通常会这样引入这个函数:“定义 g 2 :{v, w, x, y} → {1, 2, 3, 4},其中 g(v) = 2,g(w) = 3,g(x) = 4,g(y) = 2。”

5.37. Example. Consider the graph G 2 in Figure 5.2 . One possible function having this graph could be formally defined as

5.37. 示例。考虑图 5.2 中的图 G 2 。具有该图的一个可能函数可以正式定义为

g = ( R , R 0 , G G 2 ) = ( R , R 0 , { ( x x , x x 2 / 4 + 1 ) : x x R } ) .

But we would usually introduce this function by saying “define g : R R 0 by setting g ( x ) = x 2 /4 + 1 for all x R .”

但我们通常会这样引入这个函数:“定义 g:R→R≥0,令 g(x) = x 2 /4 + 1,对于所有 x∈R。”

Formulas for Functions and Non-Functions

函数和非函数的公式

Here are more examples of function notation using some familiar functions from calculus. We defer to the next section the task of proving that these formulas really do define functions.

以下是一些使用微积分中常见函数的函数符号示例。我们将在下一节中证明这些公式确实定义了函数。

5.38. Example. Define the cube function C : R R by C ( x ) = x 3 for x R . Define the absolute value function A : R R by A ( x ) = x for real x ≥ 0, and A ( x ) = − x for x < 0; we often write A ( x ) = | x |. Define the sine function S : R R by S ( x ) = sin x for all real x . If we shrink the domain and codomain of S , we obtain a related function S 1 :[ − π /2, π /2] → [ − 1, 1] given by S 1 ( x ) = sin x for − π /2 ≤ x π /2. This function is not the same as the function S . Finally, define the natural logarithm function L : ( 0 , ) R by L ( x ) = ln x for all real x > 0. We obtain a related function with a larger domain by defining L 1 : R 0 R by L 1 ( x ) = ln| x | for all real x ≠ 0. Note L 1 ( x ) = L ( A ( x )) for all x ≠ 0.

5.38. 示例。定义立方函数 C:R→R,其中 C(x) = x 3 ,x∈R。定义绝对值函数 A:R→R,其中 A(x) = x,x ≥ 0,A(x) = −x,x < 0;我们通常记作 A(x) = |x|。定义正弦函数 S:R→R,其中 S(x) = sin x,x 为所有实数。如果我们缩小 S 的定义域和值域,则得到一个相关函数 S 1 :[ − π/2, π/2] → [ − 1, 1],其中 S 1 (x) = sin x,−π/2 ≤ x ≤ π/2。此函数与函数 S 不同。最后,定义自然对数函数 L:(0,∞)→R 为 L(x) = lnx,其中 x > 0 为实数。通过定义 L1:R≠0→R 为 L 1 (x) = ln|x|,其中 x ≠ 0 为实数,我们得到一个定义域更大的相关函数。注意,对于所有 x ≠ 0,L 1 (x) = L(A(x))。

The next examples illustrate the pitfalls that can occur when we use function notation before checking the three defining properties above.

接下来的例子说明了在检查上述三个定义属性之前使用函数表示法可能会出现的陷阱。

5.39. Example. Suppose we try to define f : R R by f ( x ) = 1/ x for all real x . This definition is incorrect, because the input x = 0 has no associated output in R . We could remedy this by defining f ( x ) = 1/ x for all real nonzero x , and setting f (0) = 0 (say). Alternatively, we could shrink the domain, defining f 0 : R 0 R by f 0 ( x ) = 1/ x for all real x ≠ 0.

5.39. 示例。假设我们尝试定义函数 f:R→R,使得对于所有实数 x,f(x) = 1/x。这个定义是不正确的,因为输入 x = 0 在 R 中没有对应的输出。我们可以通过定义 f(x) = 1/x 对所有非零实数 x 成立,并令 f(0) = 0(例如)来解决这个问题。或者,我们可以缩小定义域,定义 f0:R≠0→R,使得对于所有实数 x ≠ 0,f 0 (x) = 1/x。

5.40. Example. Suppose we try to define g : Z Z by g ( m ) = m /2 for all m Z . The graph of g is { ( m , m / 2 ) : m Z } , but this is not a subset of Z × Z . In other words, the proposed function does not map the given domain Z into the given codomain Z . We could fix this by enlarging the codomain to Q (say), defining g : Z Q by g ( m ) = m /2 for all m Z .

5.40. 示例。假设我们尝试定义函数 g:Z→Z,使得对于所有 m∈Z,g(m) = m/2。g 的图像是 {(m,m/2):m∈Z},但这并非 Z×Z 的子集。换句话说,所提出的函数并没有将给定的定义域 Z 映射到给定的值域 Z。我们可以通过将值域扩展到 Q(例如)来解决这个问题,从而定义函数 g:Z→Q,使得对于所有 m∈Z,g(m) = m/2。

5.41. Example. Suppose we try to define h : Q Z by h ( m / n ) = m for all rational numbers m / n . More precisely, the graph of h is H = { ( m / n , m ) : m Z , n Z 0 } . In fact, h is not a function because its graph fails the Uniqueness of Outputs condition. For instance, choosing x = 1/2 = 2/4, we find that (1/2, 1) ∈ H and (2/4, 2) = (1/2, 2) ∈ H , yet 1 ≠ 2. Rephrasing this example using the (technically forbidden) function notation, we have 1/2 = 2/4 and yet h (1/2) = 1 ≠ 2 = h (2/4). We often say that “ h is not a well-defined function” to indicate that the relation H assigns multiple outputs to a given input. In this situation, it is clearer and safer to discuss the graph H instead of using the function notation h ( m / n ) for the non-function h .

5.41. 示例。假设我们尝试定义函数 h:Q→Z,使得对于所有有理数 m/n,h(m/n) = m。更准确地说,h 的图像是 H={(m/n,m):m∈Z,n∈Z≠0}。事实上,h 不是一个函数,因为它的图像不满足输出唯一性条件。例如,选择 x = 1/2 = 2/4,我们发现 (1/2, 1) ∈ H 且 (2/4, 2) = (1/2, 2) ∈ H,但 1 ≠ 2。用(严格来说禁止的)函数符号重新表述这个例子,我们有 1/2 = 2/4,但 h(1/2) = 1 ≠ 2 = h(2/4)。我们通常说“h 不是一个定义良好的函数”,以表明关系 H 为给定的输入分配了多个输出。在这种情况下,讨论图 H 比使用函数符号 h(m/n) 来表示非函数 h 更清晰、更安全。

5.42. Example. Finally, let us try to define f : Q > 0 Q > 0 by f ( m / n ) = n / m for all positive rationals m / n . More precisely, we are setting f = ( Q > 0 , Q > 0 , F ) , where the graph F is the set { ( m / n , n / m ) : m , n Z > 0 } . We might at first think that f is not single-valued, as in the previous example. But in fact, the f considered here is a legitimate, well-defined function! We verify Uniqueness of Outputs as follows. Suppose x , y , z satisfy ( x , y ) ∈ F and ( x , z ) ∈ F ; we must prove y = z . By definition of F , there exist m , n , p , q Z > 0 with ( x , y ) = ( m / n , n / m ) and ( x , z ) = ( p / q , q / p ). Comparing first coordinates, m / n = x = p / q , so mq = pn . Using this and comparing second coordinates, we see that y = n / m = q / p = z , as needed. In function notation, we have just shown: for all m / n and p / q in Q > 0 , if m / n = p / q then f ( m / n ) = f ( p / q ), i.e., n / m = q / p . (To finish the proof that f is a function, we must also check Existence of Outputs and that F Q > 0 × Q > 0 .)

5.42. 示例。最后,我们尝试定义函数 f:Q>0→Q>0,使得对于所有正有理数 m/n,有 f(m/n) = n/m。更准确地说,我们令 f=(Q>0,Q>0,F),其中图 F 是集合 {(m/n,n/m):m,n∈Z>0}。我们最初可能会认为 f 不是单值的,就像前面的例子一样。但实际上,这里考虑的 f 是一个合法的、定义良好的函数!我们按如下方式验证输出的唯一性。假设 x, y, z 满足 (x, y) ∈ F 且 (x, z) ∈ F;我们必须证明 y = z。根据 F 的定义,存在 m,n,p,q∈Z>0,使得 (x, y) = (m/n, n/m) 且 (x, z) = (p/q, q/p)。比较第一个坐标,m/n = x = p/q,所以 mq = pn。利用这一点并比较第二个坐标,我们发现 y = n/m = q/p = z,符合要求。用函数表示法,我们刚刚证明了:对于 Q>0 中的所有 m/n 和 p/q,如果 m/n = p/q,则 f(m/n) = f(p/q),即 n/m = q/p。(为了完成 f 是函数的证明,我们还必须检查输出的存在性以及 F⊆Q>0×Q>0。)

5.43. Remark. Some texts adopt a different definition of functions that essentially identifies a function g = ( A , B , G ) with its graph G . On one hand, the domain A of g can be reconstructed from G , since it is the set of all first coordinates of the ordered pairs in G . On the other hand, the codomain B of g cannot be inferred from G alone. For example, for the three functions shown in the top row of Figure 5.1 , knowing where all the arrows go does not give us enough information to deduce what the codomain B is. Similarly, we cannot determine the codomain of a real-valued function from inspection of its graph, although we often assume an intended codomain of R . In many areas of mathematics (including abstract algebra, topology, and differential geometry), it is crucial to distinguish functions with the same domain and the same graph but different codomains. Some of the reasons for this appear below when we discuss surjections, bijections, and invertible functions. Thus we must use the more elaborate definition of functions that explicitly includes the codomain.

5.43. 备注。有些教材采用不同的函数定义,本质上是将函数 g = (A, B, G) 与其图像 G 等同起来。一方面,g 的定义域 A 可以从 G 中重构出来,因为它是 G 中所有有序对的第一个坐标的集合。另一方面,g 的值域 B 不能仅从 G 推断出来。例如,对于图 5.1 顶行所示的三个函数,知道所有箭头指向的位置并不足以推断出它们的值域 B。类似地,我们无法仅通过观察实值函数的图像来确定其值域,尽管我们通常假设其值域为 R。在数学的许多领域(包括抽象代数、拓扑学和微分几何),区分具有相同定义域和相同图像但值域不同的函数至关重要。下文在讨论满射、双射和可逆函数时会阐述其中的一些原因。因此,我们必须使用更详细的函数定义,明确地包含值域。

Section Summary

章节概要

  1. Definition of a Function. A function is an ordered triple g = ( A , B , G ) where A and B are sets, G A × B , and x A , ! y , ( x , y ) G . A is the domain of g , B is the codomain of g , and G is the graph of g . The notation g : A B means g is a function with domain A and codomain B . For all x and y , y = g ( x ) means ( x , y ) ∈ G . We have G = {( x , g ( x )) : x A }.
    函数的定义。函数是一个有序三元组 g = (A, B, G),其中 A 和 B 是集合,G ⊆ A × B,且对于任意 x∈A,存在 y,(x,y)∈G。A 是 g 的定义域,B 是 g 的值域,G 是 g 的图像。记号 g : A → B 表示 g 是一个定义域为 A、值域为 B 的函数。对于任意 x 和 y,y = g(x) 表示 (x, y) ∈ G。因此,G = {(x, g(x)) : x ∈ A}。
  2. Arrow Diagrams of Functions. In an arrow diagram for a function g : A B , every input x A has exactly one arrow leaving it, and every arrow arrives at some output y B . It is allowed for a value in the codomain to have no arrows hitting it, or multiple arrows hitting it.
    函数的箭头图。对于函数 g : A → B,其箭头图中每个输入 x ∈ A 都恰好有一条箭头从它出发,并且每条箭头都指向某个输出 y ∈ B。允许值域中的某个值没有箭头指向它,或者有多条箭头指向它。
  3. Graphs of Real-Valued Functions. A relation G R 2 drawn in the xy -plane is the graph of a function g : A R iff for each x 0 A , the vertical line x = x 0 intersects G in exactly one point, and for each x 0 A , the vertical line x = x 0 does not intersect G . Each horizontal line y = y 0 may intersect G once, never, or multiple times.
    实值函数的图像。在 xy 平面上绘制的关系 G⊆R² 是函数 g:A→R 的图像,当且仅当对于每个 x 0 ∈ A,垂直线 x = x 0 与 G 恰好相交于一点,并且对于每个 x0∉A,垂直线 x = x 0 与 G 不相交。每条水平线 y = y 0 可能与 G 相交一次、不相交或相交多次。
  4. Three Conditions to be a Function. We often define functions g : A B by giving a formula specifying g ( x ) for each x A . To be sure we do have a well-defined function, we must check three things: g ( x ) is in the codomain B for all x A ; for each x A there exists at least one associated output g ( x ); and for all x 1 , x 2 A , if x 1 = x 2 , then g ( x 1 ) = g ( x 2 ). The last condition can be stated more precisely using the graph G of g : we must check for all x , y , z , if ( x , y ) ∈ G and ( x , z ) ∈ G then y = z .
    函数的三个条件。我们通常通过给出一个公式来定义函数 g : A → B,该公式指定了每个 x ∈ A 的 g(x)。为了确保我们得到的是一个定义良好的函数,我们必须检查三件事:对于所有 x ∈ A,g(x) 都在值域 B 中;对于每个 x ∈ A,至少存在一个对应的输出 g(x);并且对于所有 x 1 , x 2 ∈ A,如果 x 1 = x 2 ,则 g(x 1 ) = g(x 2 )。最后一个条件可以使用 g 的图像 G 更精确地表述:我们必须检查对于所有 x, y, z,如果 (x, y) ∈ G 且 (x, z) ∈ G,则 y = z。

Exercises

练习

  1. Formally describe the functions g 1 and g 3 pictured in Figure 5.1 as ordered triples. Also give informal definitions of these functions using function notation.
    请用有序三元组的形式正式描述图 5.1 中所示的函数 g 1 和 g 3 。此外,请使用函数符号给出这些函数的非正式定义。
  2. Suppose we reverse all the arrows in Figure 5.1 . Which of the resulting diagrams represents functions from B into A ? Which of these diagrams represents functions from some domain C into A ?
    假设我们将图 5.1 中的所有箭头反转。得到的图中,哪个表示从 B 到 A 的函数?哪个表示从某个区域 C 到 A 的函数?
  3. Which of the arrow diagrams in Figure 5.3 represent functions from A to B ? Explain.
    图 5.3 中的哪些箭头图表示从 A 到 B 的函数?请解释。
  4. Let A = {1, 2, 3, 4, 5, 6, 7}. Which relations are graphs of functions from A to A ? Explain. (a) R 1 = {( i , i + 1) : i = 1, 2, 3, …, 6}. (b) R 2 = R 1 { ( 7 , 1 ) } . (c) R 3 = A × {5}. (d) R 4 = {5} × A . (e) R 5 = {( a , 8 − a ) : a A }. (f) R 6 = I A .
    设 A = {1, 2, 3, 4, 5, 6, 7}。哪些关系是从 A 到 A 的函数的图像?请解释。(a) R 1 = {(i, i + 1) : i = 1, 2, 3, …, 6}。(b) R2=R1∪{(7,1)}。(c) R 3 = A × {5}。(d) R 4 = {5} × A。(e) R 5 = {(a, 8 − a) : a ∈ A}。(f) R 6 = I A
  5. For each relation G i shown in Figure 5.2 , find G i [ R ] and G i 1 [ R ] .
    对于图 5.2 中所示的每个关系 G i ,求 Gi[R] 和 Gi−1[R]。
  6. (a) Which of the graphs in Figure 5.4 are graphs of functions from R to R ? Explain. (b) Repeat (a), replacing each graph in the figure by its inverse.
    (a) 图 5.4 中的哪些图是从 R 到 R 的函数图像?请解释。(b) 重复 (a) 的步骤,将图中的每个图替换为其反函数。
  7. Let A = {1, 2, 3} and B = {4, 5, 6, 7}. For which of the following choices of the graph G is ( A , B , G ) a function? Explain briefly. (a) {(2, 4), (3, 7)}. (b) {(1, 6), (2, 4), (3, 4)}. (c) {( x , x + 4) : x A }. (d) {( x , 2 x ) : x A }. (e) A × B .
    设 A = {1, 2, 3},B = {4, 5, 6, 7}。对于下列图 G 的哪些选择,(A, B, G) 是函数?简要解释。(a) {(2, 4), (3, 7)}。(b) {(1, 6), (2, 4), (3, 4)}。(c) {(x, x + 4) : x ∈ A}。(d) {(x, 2x) : x ∈ A}。(e) A × B。
  8. (a) Draw arrow diagrams for all functions f : {1, 2, 3} → { a , b }. (b) For fixed n Z > 0 , how many functions f : {1, 2, …, n } → { a , b } are there?
    (a)画出所有函数 f : {1, 2, 3} → {a, b} 的箭头图。(b)对于固定的 n∈Z>0,有多少个函数 f : {1, 2, …, n} → {a, b}?
  9. (a) Draw graphs for all functions g : {1, 4} → {2, 3, 5}. (b) For fixed m Z > 0 , how many functions g : {1, 4} → {1, 2, …, m } are there?
    (a)画出所有函数 g : {1, 4} → {2, 3, 5} 的图像。(b)对于固定的 m∈Z>0,有多少个函数 g : {1, 4} → {1, 2, …, m}?
  10. Which of the following relations are graphs of functions? For those relations that are graphs of functions, determine the domain and smallest possible codomain for a function with that graph. (a) { ( x , y ) R × R : x = y } . (b) { ( x , y ) R × R : x 2 = y 2 } . (c) { ( x , y ) R × R : x 3 = y 3 } . (d) { ( x , y ) R × R : x = y 2 } . (e) { ( x , y ) R × R : x 2 = y } . (f) { ( x , y ) R × R : x 2 = y 3 } . (g) { ( x , y ) R × R : x 3 = y 2 } .
    下列哪些关系是函数的图像?对于构成函数图的关系,确定具有该图的函数的定义域和最小可能值域。 (a) {(x,y)∈R×R:x=y}. (b) {(x,y)∈R×R:x2=y2}. (c) {(x,y)∈R×R:x3=y3}. (d) {(x,y)∈R×R:x=y2}. (e) {(x,y)∈R×R:x2=y}. (f) {(x,y)∈R×R:x2=y3}. (g) {(x,y)∈R×R:x3=y2}.
  11. Which of the following relations are graphs of functions? For those relations that are graphs of functions, determine the domain and smallest possible codomain for a function with that graph. (a) { ( x , y ) R × R : x y = 5 } . (b) { ( x , y ) R × R : y = x 2 + 5 x 2 6 x + 8 } . (c) { ( x , y ) R × R : y = e tan x } . (d) R × { 4 } . (e) { 0 } × R . (f) Q × Z . (g) ( Q × { 1 } ) ( ( R Q ) × { 0 } ) . (h) 0 . (i) { ( ( x , y ) , z ) ( R × R ) × R : z = x y } . (j) { ( ( x , y ) , z ) ( R × R ) × R : z y = x } .
    下列哪些关系是函数的图像?对于构成函数图像的关系,确定具有该图像的函数的定义域和最小可能值域。 (a) {(x,y)∈R×R:xy=5}. (b) {(x,y)∈R×R:y=x2+5x2−6x+8}. (c) {(x,y)∈R×R:y=etan⁡x}. (d) R×{4}. (e) {0}×R. (f) Q×Z. (g) (Q×{1})∪((R−Q)×{0}). (h) 0. (i) {((x,y),z)∈(R×R)×R:z=xy}. (j) {((x,y),z)∈(R×R)×R:zy=x}.
  12. Let X = {1, 2, 3} and define h : ( P ( X ) { 0 } ) X by letting h ( A ) be the least element of A for every nonempty A X . Give a formal description of the function h as an ordered triple.
    设 X = {1, 2, 3},定义 h:(P(X)−{0})→X,其中 h(A) 为任意非空集合 A ⊆ X 中 A 的最小元素。给出函数 h 的有序三元组形式化描述。
  13. Let Y = {2, {3}, {2, {3}}} and define f : Y P ( Y ) by f ( y ) = { y } for all y Y . Describe the graph of f as a set of ordered pairs.
    设 Y = {2, {3}, {2, {3}}},定义函数 f:Y→P(Y) 为 f(y) = {y},其中 y ∈ Y。将函数 f 的图像描述为有序对的集合。
  14. Which of the following are well-defined functions? Explain. (Assume m , n , a , b , c , d Z with n , b , d > 0.) (a) f : Q Z given by f ( m / n ) = m + n . (b) g : Q Q given by g ( m / n ) = ( m + 2 n )/ n . (c) h : R { 0 , 1 , , 9 } given by h ( x ) = the second digit after the decimal point in the base-10 representation of x . (d) m : Q × Q Q given by m (( a / b , c / d )) = ( ac )/( bd ). (e) p : Q × Q Q given by p (( a / b , c / d )) = ( a + c )/( b + d ). (f) s : Q × Q Q given by s (( a / b , c / d )) = ( ad + bc )/( bd ).
    下列哪些是定义明确的函数?请解释。 (假设 m,n,a,b,c,d∈Z,且 n,b,d > 0。) (a) f:Q→Z 由 f(m/n) = m + n 给出。 (b) g:Q→Q 由 g(m/n) = (m + 2n)/n 给出。 (c) h:R→{0,1,…,9} 由 h(x) = x 的十进制表示中小数点后的第二位数字 给出。 (d) m:Q×Q→Q 由 m((a/b, c/d)) = (ac)/(bd) 给出。 (e) p:Q×Q→Q 由 p((a/b, c/d)) = (a + c)/(b + d). (f) s:Q×Q→Q 由 s((a/b, c/d)) = (ad + bc)/(bd) 给出。
  15. Prove that for all sets A and B , ( A , B , 0 ) is a function iff A = 0 .
    证明对于所有集合 A 和 B,(A,B,0) 是一个函数当且仅当 A=0。
  16. Enlarging the Codomain. Prove that for all sets A , B , C , if B C and ( A , B , F ) is a function, then ( A , C , F ) is a function.
    扩大值域。证明对于所有集合 A、B、C,如果 B ⊆ C 且 (A, B, F) 是一个函数,则 (A, C, F) 也是一个函数。
  17. Prove or disprove: for all A , B , G , if ( A , B , G ) is a function, then ( B , A , G −1 ) is a function.
    证明或反证:对于所有 A、B、G,如果 (A、B、G) 是一个函数,那么 (B、A、G −1 ) 也是一个函数。

fig5_3

Figure 5.3

图 5.3

More arrow diagrams.

更多箭头图。

fig5_4

Figure 5.4

图 5.4

More graphs in R 2 .

R2 中更多图表。

5.5 Examples of Functions and Function Equality

5.5 函数和函数等式的例子

This section defines some basic examples of functions, such as constant functions, identity functions, inclusion functions, characteristic functions, and arithmetic functions. We also introduce operations (called pointwise sum, difference, product, and quotient) that can be used to construct new functions from old functions. We then develop a proof template for showing that two functions are equal. We use this template to prove some algebraic properties of the pointwise operations on functions, analogous to corresponding properties for addition and multiplication of real numbers.

本节定义了一些基本函数示例,例如常函数、恒等函数、包含函数、特征函数和算术函数。我们还介绍了一些运算(称为逐点和、逐点差、逐点积和逐点商),这些运算可用于从现有函数构造新函数。然后,我们开发了一个证明两个函数相等的证明模板。我们利用该模板证明了函数逐点运算的一些代数性质,这些性质类似于实数加法和乘法的相应性质。

Constant Functions, Inclusion Functions, and Identity Functions

常函数、包含函数和恒等函数

5.44. Definition: Constant Functions. For any sets A and B and any fixed object c B , the function f : A B defined by f ( x ) = c for all x A is called a constant function with value c .

5.44. 定义:常函数。对于任意集合 A 和 B 以及任意固定对象 c ∈ B,由 f(x) = c 对所有 x ∈ A 定义的函数 f : A → B 称为值为 c 的常函数。

The graph of the constant function defined above is G = {( x , c ) : x A } = A × { c }. Let us prove that f = ( A , B , G ) really is a function by checking the three conditions from §5.4. First, every member of G is an ordered pair ( x , c ) with x A and c B , so G A × B . Second, for a fixed x A , there exists y with ( x , y ) ∈ G , namely y = c . Third, fix x , y , z , assume ( x , y ) ∈ G and ( x , z ) ∈ G , and prove y = z . By definition of G , we must have y = c and z = c , so y = z follows.

上述定义的常数函数的图像为 G = {(x, c) : x ∈ A} = A × {c}。我们通过验证 §5.4 中的三个条件来证明 f = (A, B, G) 确实是一个函数。首先,G 中的每个元素都是一个有序对 (x, c),其中 x ∈ A 且 c ∈ B,因此 G ⊆ A × B。其次,对于固定的 x ∈ A,存在 y 使得 (x, y) ∈ G,即 y = c。第三,固定 x、y、z,假设 (x, y) ∈ G 且 (x, z) ∈ G,并证明 y = z。根据 G 的定义,我们必须有 y = c 且 z = c,因此 y = z 成立。

5.45. Definition: Inclusion Functions. Let A and B be two sets such that A B . Define a function j = j A , B : A B by setting j A , B ( x ) = x for all x A . We call j A , B the inclusion function mapping A into B .

5.45. 定义:包含函数。设 A 和 B 是两个集合,且 A ⊆ B。定义函数 j = j A , B : A → B,使得对于所有 x ∈ A,j A , B (x) = x。我们称 j A , B 为将 A 映射到 B 的包含函数。

The graph of the inclusion function defined above is G = {( x , x ) : x A } = I A , the identity relation on A . Let us prove that j A , B = ( A , B , G ) really is a function. First, every member of G is an ordered pair ( x , x ) with x A and x B (since A B ), so that G A × B . Second, for a fixed x A , there exists y with ( x , y ) ∈ G , namely y = x . Third, fix x , y , z , assume ( x , y ) ∈ G and ( x , z ) ∈ G , and prove y = z . By definition of G , we must have y = x and z = x , so y = z follows.

上述定义的包含函数的图像是 G = {(x, x) : x ∈ A} = I A ,它是 A 上的恒等关系。我们来证明 j A , B = (A, B, G) 确实是一个函数。首先,G 中的每个元素都是一个有序对 (x, x),其中 x ∈ A 且 x ∈ B(因为 A ⊆ B),所以 G ⊆ A × B。其次,对于固定的 x ∈ A,存在 y 使得 (x, y) ∈ G,即 y = x。第三,固定 x、y、z,假设 (x, y) ∈ G 且 (x, z) ∈ G,并证明 y = z。根据 G 的定义,我们必须有 y = x 且 z = x,因此 y = z 成立。

5.46. Definition: Identity Functions. Given any set A , define a function Id A : A A by setting Id A ( x ) = x for all x A . We call Id A the identity function on the set A .

5.46. 定义:恒等函数。给定任意集合 A,定义函数 Id A : A → A,使得对于所有 x ∈ A,Id A (x) = x。我们称 Id A 为集合 A 上的恒等函数。

Note that the identity function Id A is the same as the inclusion function j A , A , so Id A is a function. On the other hand, if A is a proper subset of B , then Id A is not the same function as the inclusion j A , B . To check this carefully, note that Id A is the ordered triple ( A , A , G ), where G = {( x , x ) : x A }, whereas j A , B is the ordered triple ( A , B , G ). Since A B , these ordered triples are not equal. Less formally, Id A and j A , B are unequal functions since their codomains are different (even though their domains and graphs agree). We return to this point later when we discuss function equality.

注意,恒等函数 Id A 与包含函数 j A , A 相同,因此 Id A 是一个函数。另一方面,如果 A 是 B 的真子集,则 Id A 与包含函数 j A , B 不是同一个函数。为了仔细验证这一点,请注意 Id A 是有序三元组 (A, A, G),其中 G = {(x, x) : x ∈ A},而 j A , B 是有序三元组 (A, B, G)。由于 A ≠ B,这两个有序三元组不相等。更通俗地说,Id A 和 j A , B 是不相等的函数,因为它们的值域不同(即使它们的定义域和图相同)。我们稍后在讨论函数相等性时会再回到这一点。

5.47. Example. Figure 5.5 gives arrow diagrams for Id A and j A , B when A = {2, 4, 5} and B = {1, 2, 3, 4, 5}.

5.47. 示例。图 5.5 给出了当 A = {2, 4, 5} 和 B = {1, 2, 3, 4, 5} 时,Id A 和 j A , B 的箭头图。

fig5_5

Figure 5.5

图 5.5

Contrasting identity functions and inclusion functions.

对比恒等函数和包含函数。

Arithmetic Functions and Pointwise Operations on Functions

算术函数和函数上的逐点运算

We are all familiar with the arithmetic operations on real numbers: addition, subtraction, multiplication, and division. We do not attempt to define these operations, instead viewing them as undefined concepts that obey certain axioms (see §2.1 and §8.1). However, now that the language of functions is available, we can at least describe more precisely what kind of objects these undefined operations are .

我们都熟悉实数的算术运算:加法、减法、乘法和除法。我们并不试图定义这些运算,而是将它们视为遵循某些公理的未定义概念(参见第2.1节和第8.1节)。然而,现在有了函数语言,我们至少可以更精确地描述这些未定义运算的对象类型。

For example, consider the undefined operation of adding two real numbers. For any two real numbers x and y , we can apply the addition operation to these two numbers to obtain a unique output x + y , called the sum of x and y . More formally, addition transforms the ordered pair of inputs ( x , y ) into the real output x + y . So we could describe addition as a function ADD : R × R R , where add ( ( x , y ) ) is defined to be the sum of x and y . We would almost always write x + y as a shorthand for ADD ( ( x , y ) ) . Similarly, multiplication is a function MULT : R × R R , where MULT ( ( x , y ) ) = x y is the product of x and y . Subtraction and division could be defined similarly, provided we use R × R 0 as the domain of the division function (since we cannot divide by zero). Alternatively, we could first introduce an additive inverse function AINV : R R , denoted AINV ( z ) = z for z R , and then define subtraction via x y = x + ( − y ) for x , y R . Similarly, we could introduce a multiplicative inverse function MINV : R 0 R , denoted MINV ( z ) = z 1 for z R 0 , and then define division via x / y = x · ( y −1 ) for x , y R with y ≠ 0.

例如,考虑两个实数相加这个未定义运算。对于任意两个实数 x 和 y,我们可以对这两个数进行加法运算,得到唯一的输出 x + y,称为 x 和 y 的和。更正式地说,加法将有序输入对 (x, y) 转换为实数输出 x + y。因此,我们可以将加法描述为一个函数 ADD:R×R→R,其中 add((x,y)) 定义为 x 和 y 的和。我们几乎总是将 x + y 简写为 ADD((x,y))。类似地,乘法是一个函数 MULT:R×R→R,其中 MULT((x,y))=x⋅y 表示 x 和 y 的乘积。减法和除法也可以类似地定义,前提是我们使用 R×R≠0 作为除法函数的定义域(因为除数不能为零)。或者,我们可以先引入一个加法逆函数 AINV:R→R,记为 AINV(z)=−z,其中 z∈R,然后定义减法为 x − y = x + ( − y),其中 x,y∈R。类似地,我们可以引入一个乘法逆函数 MINV:R≠0→R,记为 MINV(z)=z−1,其中 z∈R≠0,然后定义除法为 x/y = x · (y −1 ),其中 x,y∈R,y ≠ 0。

We can use the addition operation on real numbers to create an addition operation acting on functions , as follows.

我们可以使用实数的加法运算来创建作用于函数的加法运算,如下所示。

5.48. Definition: Pointwise Sum of Functions. Suppose A is a set and f : A R , g : A R are two real-valued functions with domain A . Define the pointwise sum f + g : A R by letting ( f + g ) ( x ) = f ( x ) + g ( x ) for all x A .

5.48. 定义:函数的逐点和。假设 A 是一个集合,f:A→R,g:A→R 是定义域为 A 的两个实值函数。定义逐点和 f+g:A→R,令对于所有 x ∈ A,(f+g)(x)=f(x)+g(x)。

The graph of f + g is the set {( x , f ( x ) + g ( x ) : x A )}. You can check that f + g is a function, using the fact that f and g are known to be functions.

函数 f + g 的图像是集合 {(x, f(x) + g(x) : x ∈ A)}。你可以利用已知 f 和 g 是函数这一事实来验证 f + g 也是一个函数。

5.49. Example. If f : R R is the exponential function and g : R R is the cosine function, then f + g : R R is given by ( f + g )( x ) = e x + cos x for all real x . For another example, the function h : R R defined by h ( x ) = x + 3 is the pointwise sum of the identity function on R and the constant function with value 3.

5.49. 示例。如果 f:R→R 是指数函数,g:R→R 是余弦函数,则 f+g:R→R 由 (f + g)(x) = e x + cos x 给出,其中 x 为实数。再举一例,函数 h:R→R 由 h(x) = x + 3 定义,它是 R 上的恒等函数与值为 3 的常数函数的逐点和。

Note that the domain A need not be a subset of R to form pointwise sums. It is required that the functions being combined share the same domain. We can define more pointwise operations as follows.

注意,定义域 A 不必是 R 的子集才能构成逐点求和。但要求被组合的函数具有相同的定义域。我们可以定义更多逐点运算如下。

5.50. Definition: More Pointwise Operations. Given functions f , g : A R , we define the following operations (called pointwise product, pointwise additive inverse, pointwise difference, pointwise reciprocal, and pointwise quotient). For parts (d) and (e), we assume g ( x ) ≠ 0 for all x A .

5.50. 定义:更多逐点运算。给定函数 f,g:A→R,我们定义以下运算(分别称为逐点乘积、逐点加逆、逐点差、逐点倒数和逐点商)。对于 (d) 和 (e) 部分,我们假设对于所有 x ∈ A,g(x) ≠ 0。

(a) Define f g : A R by setting ( f g ) ( x ) = f ( x ) g ( x ) for x in A .

(a)定义 f⋅g:A→R,令 (f⋅g)(x)=f(x)⋅g(x),其中 x ∈ A。

(b) Define g : A R by setting ( g ) ( x ) = ( g ( x ) ) for x in A .

(b)定义 −g:A→R,令 (−g)(x)=−(g(x)),其中 x ∈ A。

(c) Define f g : A R by setting ( f g ) ( x ) = f ( x ) g ( x ) for x in A .

(c)定义 f−g:A→R,令 (f−g)(x)=f(x)−g(x),其中 x ∈ A。

(d) Define 1 / g : A R by setting ( 1 / g ) ( x ) = 1 / g ( x ) for x in A .

(d)定义 1/g:A→R,令 (1/g)(x)=1/g(x),其中 x ∈ A。

(e) Define f / g : A R by setting ( f / g ) ( x ) = f ( x ) / g ( x ) for x in A .

(e)定义 f/g:A→R,令 (f/g)(x)=f(x)/g(x),其中 x ∈ A。

Take care not to confuse the pointwise product f · g (also denoted fg ) with the composition of functions f g (to be introduced later). Similarly, we write 1/ g for the pointwise reciprocal to avoid confusion with the inverse function g −1 (to be studied later). Also, be very careful not to confuse a function such as f + g with ( f + g )( x ), which is the output of that function at input x . In particular, ( − g )( x ) is the output of the function − g at input x ; this output is defined to be −( g ( x )), the additive inverse of the real number g ( x ).

注意不要将逐点乘积 f · g(也记作 fg)与函数复合 f ∘ g(稍后介绍)混淆。类似地,我们用 1/g 表示逐点倒数,以避免与反函数 g(稍后学习)混淆。此外,务必注意不要将函数 f + g 与 (f + g)(x) 混淆,后者是该函数在输入 x 处的输出。特别地,( − g)(x) 是函数 −g 在输入 x 处的输出;该输出定义为 −(g(x)),即实数 g(x) 的加法逆元。

5.51. Example.

5.51. 示例。

The function q : R R given by q ( x ) = 5 x is the pointwise product of the constant function with value 5 and the identity function on R . The squaring function s : R R given by s ( x ) = x 2 is the pointwise product s = Id R Id R , since ( Id R Id R ) ( x ) = ( Id R ( x ) ) ( Id R ( x ) ) = x x = x 2 = s ( x ) for all x R .

函数 q:R→R 由 q(x) = 5x 给出,它是值为 5 的常数函数与 R 上的恒等函数的逐点乘积。平方函数 s:R→R 由 s(x) = x 2 给出,它是逐点乘积 s=IdRIdR,因为对于所有 x∈R,(IdRIdR)(x)=(IdR(x))⋅(IdR(x))=x⋅x=x2=s(x)。

By starting with the identity function and constant functions on R and repeatedly iterating the pointwise sum and product constructions, we can build the polynomial functions on R . For example, Id R Id R Id R + 2 Id R 5 is the polynomial p : R R given by p ( x ) = x 3 + 2 x − 5 for x R . The general definition is as follows.

从 R 上的恒等函数和常数函数出发,反复迭代逐点求和与求积的构造,我们可以构造 R 上的多项式函数。例如,IdRIdRIdR+2IdR−5 是多项式 p:R→R,定义为 p(x) = x 3 + 2x − 5,其中 x∈R。一般定义如下。

5.52. Definition: Polynomials. A function p : R R is a polynomial (in one real variable) iff there exist an integer n ≥ 0 and constants a 0 , , a n R such that for all x R ,

5.52. 定义:多项式。函数 p:R→R 是多项式(关于一个实变量)当且仅当存在整数 n ≥ 0 和常数 a0,…,an∈R,使得对于所有 x∈R,

p p ( x x ) = a 一个 0 + a 一个 1 x x + a 一个 2 x x 2 + + a 一个 n n x x n n = i = 0 n n a 一个 i x x i .

Equality of Functions

函数相等

Recall that a function f : A B was formally defined as an ordered triple f = ( A , B , G ), where G = {( x , f ( x )) : x A } is the graph of f . Given another function h = ( A ′, B ′, G ′), when are the functions f and h equal ? By the fundamental property of ordered triples, we see that f = h iff A = A ′ and B = B ′ and G = G ′. In other words, two functions are equal iff they have the same domain, the same codomain, and the same graph . Suppose we know that A = A ′ and B = B ′. You can check that the graph G = {( x , f ( x )) : x A } equals the graph G ′ = {( x , h ( x )) : x A } iff for all x A , f ( x ) = h ( x ). Thus we can say that two functions are equal iff they have the same domain, the same codomain, and the same values on their common domain . This leads to the following proof template for proving the equality of two functions. (Compare this to earlier proof templates for proving equality of sets or equality of numbers.)

回想一下,函数 f : A → B 的形式定义为一个有序三元组 f = (A, B, G),其中 G = {(x, f(x)) : x ∈ A} 是 f 的图像。给定另一个函数 h = (A′, B′, G′),函数 f 和 h 何时相等?根据有序三元组的基本性质,我们知道 f = h 当且仅当 A = A′ 且 B = B′ 且 G = G′。换句话说,两个函数相等当且仅当它们具有相同的定义域、相同的值域和相同的图像。假设我们知道 A = A′ 且 B = B′。你可以验证,图像 G = {(x, f(x)) : x ∈ A} 等于图像 G′ = {(x, h(x)) : x ∈ A} 当且仅当对于所有 x ∈ A,都有 f(x) = h(x)。因此,我们可以说两个函数相等当且仅当它们具有相同的定义域、相同的值域,并且它们的公共定义域上的值也相同。由此可得出以下证明两个函数相等的证明模板。(可将其与之前证明集合相等或数相等的证明模板进行比较。)

5.53. Proof Template for Proving Function Equality. Let functions f : A B and g : A B have the same domain A and the same codomain B . To prove f = g :

5.53. 证明函数相等性的证明模板。设函数 f : A → B 和 g : A → B 具有相同的定义域 A 和值域 B。要证明 f = g:

Fix an arbitrary object x A . Use the definitions of f and g to prove f ( x ) = g ( x ).

固定任意对象 x ∈ A。利用 f 和 g 的定义证明 f(x) = g(x)。

The last step (proving equality of f ( x ) and g ( x ), which are objects in the codomain B ) is often achieved by a chain proof, writing down a chain of known equalities linking the left side f ( x ) to the right side g ( x ). We now illustrate this proof template by proving some algebraic laws satisfied by pointwise operations on functions.

最后一步(证明 f(x) 和 g(x) 相等,它们都是值域 B 中的对象)通常通过链式证明来实现,即写出一系列已知的等式,将左侧的 f(x) 与右侧的 g(x) 联系起来。现在,我们通过证明一些由函数逐点运算满足的代数定律来说明这种证明模板。

5.54. Theorem on Pointwise Operations. Suppose A is any set and f , g , h : A R are functions. Let 0 A and 1 A denote the constant functions from A to R with values 0 and 1 (respectively). The following function equalities hold:

5.54. 逐点运算定理。设 A 为任意集合,f, g, h: A → R 为函数。令 0 和 1 分别表示从 A 到 R 的取值为 0 和 1 的常值函数。下列函数等式成立:

(a) Commutativity: f + g = g + f and f · g = g · f .

(a)交换律:f + g = g + f 且 f · g = g · f。

(b) Associativity: ( f + g ) + h = f + ( g + h ) and ( f · g ) · h = f · ( g · h ).

(b)结合律:(f + g)+ h = f +(g + h)且(f · g)· h = f ·(g · h)。

(c) Identity: f + 0 A = f = 0 A + f and f · 1 A = f = 1 A · f .

(c)恒等式:f + 0 A = f = 0 A + f 且 f · 1 A = f = 1 A · f。

(d) Inverses: f + ( − f ) = 0 A = ( − f ) + f .

(d) 逆:f + ( − f) = 0 A = ( − f) + f。

If g ( x ) ≠ 0 for all x A , then g · (1/ g ) = 1 A = (1/ g ) · g .

如果对于所有 x ∈ A,g(x) ≠ 0,则 g · (1/g) = 1 A = (1/g) · g。

(e) Distributive Laws: f · ( g + h ) = ( f · g ) + ( f · h ) and ( f + g ) · h = ( f · h ) + ( g · h ).

(e)分配律:f · (g + h) = (f · g) + (f · h) 和 (f + g) · h = (f · h) + (g · h)。

Proof. We prove a few identities here, leaving the others as exercises. To prove the first commutative law in (a), first note that f + g and g + f are both functions with domain A and codomain R (by definition of pointwise sum). Following the proof template for function equality, fix an arbitrary object x A , and prove ( f + g )( x ) = ( g + f )( x ). We compute ( f + g )( x ) = f ( x ) + g ( x ) = g ( x ) + f ( x ) = ( g + f )( x ), where the first step uses the definition of f + g , the second step uses the known commutativity of addition of real numbers, and the third step uses the definition of g + f .

证明。这里我们证明几个恒等式,其余的留作练习。为了证明 (a) 中的第一个交换律,首先注意到 f + g 和 g + f 都是定义域为 A、值域为 R 的函数(根据逐点求和的定义)。按照函数相等性的证明模板,固定任意对象 x ∈ A,并证明 (f + g)(x) = (g + f)(x)。我们计算 (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x),其中第一步利用了 f + g 的定义,第二步利用了实数加法的交换律,第三步利用了 g + f 的定义。

Next we prove f · 1 A = f . Both f · 1 A and f have domain A and codomain R . Fix x A . We know ( f · 1 A )( x ) = f ( x ) · 1 A ( x ) = f ( x ) · 1 = f ( x ), using the definition of pointwise product, then the definition of the constant function 1 A , and then the multiplicative identity axiom for R .

接下来我们证明 f · 1 A = f。f · 1 A 和 f 的定义域均为 A,值域均为 R。固定 x ∈ A。我们知道 (f · 1 A )(x) = f(x) · 1 A (x) = f(x) · 1 = f(x),这是利用逐点乘积的定义,然后利用常数函数 1 A 的定义,最后利用 R 的乘法恒等式公理。

Finally, we prove f · ( g + h ) = ( f · g ) + ( f · h ). Repeated use of the definitions shows that g + h , f · ( g + h ), f · g , f · h , and ( f · g ) + ( f · h ) all have domain A and codomain R . Fix x A . Compute

最后,我们证明 f · (g + h) = (f · g) + (f · h)。反复使用定义可知,g + h、f · (g + h)、f · g、f · h 和 (f · g) + (f · h) 的定义域均为 A,值域均为 R。固定 x ∈ A。计算

[ f f ( g + h h ) ] ( x x ) = f f ( x x ) ( g + h h ) ( x x ) (by def. of pointwise product) (根据逐点乘积的定义) = f f ( x x ) [ g ( x x ) + h h ( x x ) ] (by def. of pointwise sum) (根据逐点求和的定义) = ( f f ( x x ) g ( x x ) ) + ( f f ( x x ) h h ( x x ) ) (by the distributive law in (根据分配法) R ) = ( f f g ) ( x x ) + ( f f h h ) ( x x ) (by def. of pointwise product) (根据逐点乘积的定义) = [ ( f f g ) + ( f f h h ) ] ( x x ) (by def. of pointwise sum). (根据逐点求和的定义)。
#

So f · ( g + h ) = ( f · g ) + ( f · h ), as needed.

因此,f · (g + h) = (f · g) + (f · h),符合要求。

In the previous computation, be sure you understand why the input x does or does not appear at various places in each expression. For example, the initial expression [ f · ( g + h )]( x ) is the output of the function f · ( g + h ) at input x , while the next expression f ( x ) · ( g + h )( x ) is the product of the two real numbers f ( x ) and ( g + h )( x ). These two quantities are equal by the definition of the pointwise product of the functions f and g + h .

在之前的计算中,请务必理解输入 x 在每个表达式中出现或不出现的原因。例如,初始表达式 [f · (g + h)](x) 是函数 f · (g + h) 在输入 x 处的输出,而下一个表达式 f(x) · (g + h)(x) 是两个实数 f(x) 和 (g + h)(x) 的乘积。根据函数 f 和 g + h 的逐点乘积的定义,这两个量相等。

5.55. Example: Changing the Codomain. Suppose f : A B is a function with graph G = {( x , f ( x )) : x A }. For any set C such that B C , we can create a new function f * : A C by setting f *( x ) = f ( x ) for all x A . Formally, f = ( A , B , G ) and f * = ( A , C , G ). We have constructed f * from f by enlarging the codomain . More generally, the new codomain C can be any set such that G A × C . This means that for all x A , we must have f ( x ) ∈ C . This condition holds iff G [ A ] (the image of A under the relation G ) is a subset of C . We must remember that when B C , f and f * are not the same function, although some texts use the same letter f to denote both functions.

5.55. 示例:改变值域。假设函数 f : A → B 的图像为 G = {(x, f(x)) : x ∈ A}。对于任意集合 C,使得 B ⊆ C,我们可以构造一个新的函数 f* : A → C,令 f*(x) = f(x) 对所有 x ∈ A 成立。形式上,f = (A, B, G) 且 f* = (A, C, G)。我们通过扩大值域,从 f 构造了 f*。更一般地,新的值域 C 可以是任意满足 G ⊆ A × C 的集合。这意味着对于所有 x ∈ A,必须有 f(x) ∈ C。此条件成立当且仅当 G[A](A 在关系 G 下的像)是 C 的子集。我们必须记住,当 B ≠ C 时,f 和 f* 不是同一个函数,尽管有些教材使用相同的字母 f 来表示这两个函数。

Characteristic Functions (Optional)

特征函数(可选)

We end this section with another class of examples of functions that are often used in probability.

本节最后,我们将给出概率论中常用的另一类函数示例。

5.56. Definition: Characteristic Functions. Let X be any set, and let A be a fixed subset of X . Define χ A : X R by setting χ A ( x ) = 1 for all x A , and χ A ( x ) = 0 for all x A . We call χ A the characteristic function of A in X .

5.56. 定义:特征函数。设 X 为任意集合,A 为 X 的一个固定子集。定义 χA:X→R,使得对于所有 x ∈ A,χ A (x) = 1,对于所有 x∉A,χ A (x) = 0。我们称 χ A 为 A 在 X 中的特征函数。

The graph of χ A is ( A × { 1 } ) ( ( X A ) × { 0 } ) . You can prove that χ A is a function. Intuitively, χ A is an indicator function that tells us which inputs x belong to A and which do not. For example, taking X = R , we have

χ A 的图像是 (A×{1})∪((X−A)×{0})。你可以证明 χ A 是一个函数。直观地说,χ A 是一个指示函数,它告诉我们哪些输入 x 属于 A,哪些不属于 A。例如,取 X=R,我们有

χ x Q ( 0.73 ) = χ x Q ( 0 ) = χ x Q ( 11 / 3 ) = 1 ; χ x Q ( π p ) = χ x Q ( e e ) = χ x Q ( 3 ) = 0.

When using characteristic functions, it is critical to specify what the intended domain X is, since this is not part of the notation χ A .

在使用特征函数时,必须明确指定预期域 X 是什么,因为这不是符号 χ A 的一部分。

We can combine characteristic functions with the same domain using the pointwise operations introduced earlier. We can then verify various identities that connect operations on sets to pointwise operations on functions. The next theorem gives a sample of such identities.

我们可以利用前面介绍的逐点运算,将具有相同定义域的特征函数组合起来。然后,我们可以验证各种将集合运算与函数逐点运算联系起来的恒等式。下一个定理给出了一些这样的恒等式示例。

5.57. Theorem on Characteristic Functions. For all sets X and all A , B X , the following function equalities hold.

5.57. 特征函数定理。对于所有集合 X 和所有 A, B ⊆ X,下列函数等式成立。

(a) χ A B = χ A χ B .

(a)χA∩B=χA⋅χB。

(b) χ X A = 1 − χ A .

(b)x X A = 1 − x A

(c) χ 0 = 0 X (where 0 X : X R is the constant function with value 0).

(c)χ0=0X(其中 0X:X→R 是值为 0 的常数函数)。

(d) χ X = 1 X , where 1 X : X R is the constant function with value 1.

(d)χ X = 1 X ,其中 1X:X→R 是值为 1 的常数函数。

(e) χ A B = χ A + χ B χ A B .

(e) χA∪B=χA+χB−χA∩B。

(f) If A B = 0 , then χ A B = χ A + χ B .

(f)如果 A∩B=0,则 χA∪B=χA+χB。

Proof. We prove part (a) and leave the others as exercises. First note that χ A B and χ A χ B are both functions with domain X and codomain R . Fix x X , and consider four cases. Case 1 . Assume x A and x B . Then χ A B ( x ) = 1 = 1 1 = χ A ( x ) χ B ( x ) = ( χ A χ B ) ( x ) . Case 2 . Assume x A and x B . Then χ A B ( x ) = 0 = 1 0 = χ A ( x ) χ B ( x ) = ( χ A χ B ) ( x ) . Case 3 . Assume x A and x B . Then χ A B ( x ) = 0 = 0 1 = χ A ( x ) χ B ( x ) = ( χ A χ B ) ( x ) . Case 4 . Assume x A and x B . Then χ A B ( x ) = 0 = 0 0 = χ A ( x ) χ B ( x ) = ( χ A χ B ) ( x ) . Thus, χ A B ( x ) = ( χ A χ B ) ( x ) in all cases. So χ A B = χ A χ B .       □

证明。我们证明 (a) 部分,其余部分留作练习。首先注意到 χA∩B 和 χ A χ B 都是定义域为 X、值域为 R 的函数。固定 x ∈ X,考虑以下四种情况。情况 1:假设 x ∈ A 且 x ∈ B。则 χA∩B(x)=1=1⋅1=χA(x)χB(x)=(χAχB)(x)。情况 2:假设 x ∈ A 且 x∉B。则 χA∩B(x)=0=1⋅0=χA(x)χB(x)=(χAχB)(x)。情况 3:假设 x∉A 且 x ∈ B。则 χA∩B(x)=0=0⋅1=χA(x)χB(x)=(χAχB)(x)。情况 4. 假设 x∉A 且 x∉B。则 χA∩B(x)=0=0⋅0=χA(x)χB(x)=(χAχB)(x)。因此,在所有情况下,χA∩B(x)=(χAχB)(x)。所以 χA∩B=χA⋅χB。□

Section Summary

章节概要

  1. Constants, Inclusions, and Identity Functions. For fixed c B , the function f : A B given by f ( x ) = c for all x A is the constant function on A with value c . For sets A B , the function j A , B : A B given by j A , B ( x ) = x for all x A is the inclusion function mapping A into B . For any set A , the function Id A : A A given by Id A ( x ) = x for all x A is the identity function on A .
    常量、包含关系和恒等函数。对于固定的 c ∈ B,函数 f : A → B,即对于所有 x ∈ A,f(x) = c,是 A 上的常量函数,其值为 c。对于集合 A ⊆ B,函数 j A , B : A → B,即对于所有 x ∈ A,j A , B (x) = x,是将 A 映射到 B 的包含函数。对于任意集合 A,函数 Id A : A → A,即对于所有 x ∈ A,Id A (x) = x,是 A 上的恒等函数。
  2. Pointwise Operations on Real-Valued Functions. For functions f , g : A R , we can define new functions f + g , f · g , − g , f g , and (when g ( x ) ≠ 0 for all x A ) f / g by performing the indicated arithmetic operation one input at a time. For instance, ( f + g )( x ) = f ( x ) + g ( x ) for all x A , and ( f · g )( x ) = f ( x ) g ( x ) for all x A . Pointwise operations on functions satisfy algebraic laws such as commutativity, associativity, and the distributive laws.
    实值函数的逐点运算。对于函数 f,g:A→R,我们可以通过逐个输入执行指定的算术运算,定义新的函数 f + g、f · g、−g、f − g 以及(当所有 x ∈ A 的 g(x) ≠ 0 时)f/g。例如,对于所有 x ∈ A,(f + g)(x) = f(x) + g(x),并且对于所有 x ∈ A,(f · g)(x) = f(x)g(x)。函数的逐点运算满足交换律、结合律和分配律等代数定律。
  3. Function Equality. Two functions are equal iff they have the same domain, the same codomain, and the same graph (i.e., the same value at each point in the common domain). To prove that f : A B and g : A B are equal functions, fix x A and use known facts to prove f ( x ) = g ( x ). Changing the codomain of a function produces a different function with the same domain and graph.
    函数相等。两个函数相等当且仅当它们具有相同的定义域、相同的值域和相同的图像(即,在公共定义域内的每个点的值都相同)。要证明 f : A → B 和 g : A → B 是相等的函数,固定 x ∈ A,并利用已知事实证明 f(x) = g(x)。改变函数的值域会得到一个定义域和图像相同的新函数。
  4. Characteristic Functions. For A X , define χ A : X R by χ A ( x ) = 1 for x A and χ A ( x ) = 0 for x X A . For A , B X , we have χ A B = χ A χ B , χ X A = 1 − χ A , and χ A B = χ A + χ B χ A B .
    特征函数。对于 A ⊆ X,定义 χA:X→R,其中 χ A (x) = 1,x ∈ A;χ A (x) = 0,x ∈ X − A。对于 A, B ⊆ X,我们有 χA∩B=χA⋅χB,χ X A = 1 − χ A ,以及 χA∪B=χA+χB−χA∩B。

Exercises

练习

  1. Draw arrow diagrams for the following functions. (a) Id [ 2 , 2 ] Z (b) j {0,3},{0,1,2,3,4} (c) j Z , Q (d) χ A : X R , where X = {1, 2, …, 10} and A consists of the prime numbers in X .
    为下列函数绘制箭头图。(a) Id[−2,2]∩Z (b) j {0,3},{0,1,2,3,4} (c) jZ,Q (d) χA:X→R,其中 X = {1, 2, …, 10},A 由 X 中的素数组成。
  2. Draw graphs in the xy -plane for the following functions. Mark the domain on the x -axis and the codomain on the y -axis. (a) Id [ , 3 ] Z (b) Id [ , 3 ] Z (c) j Z , R (d) χ ( 3 , 1 ] ( 1 , 3 ] : R R (e ) χ { 0 } : R R (f) χ { x : x 3 + 2 x 2 13 x + 10 = 0 } : R R .
    在 xy 平面上绘制下列函数的图像。在 x 轴上标出定义域,在 y 轴上标出值域。(a) Id[−∞,3]∪Z (b) Id[−∞,3]∩Z (c) jZ,R (d) χ(−3,−1]∪(1,3]:R→R (e)χ{0}:R→R (f) χ{x:x3+2x2−13x+10=0}:R→R。
  3. Consider the following six functions:
    f 1 : R R given by f 1 ( x ) = | x | ; f 2 : R [ 0 , ) given by f 2 ( x ) = | x | ; f 3 : [ 0 , ) [ 0 , ) given by f 3 ( x ) = | x | ; f 4 : R R given by f 4 ( x ) = x 2 ; f 5 : [ 0 , ) [ 0 , ) given by f 5 ( x ) = x ; f 6 : R [ 0 , ) given by f 6 ( x ) = x .
    (a) Which pairs of functions in this list are equal? (b) Which pairs of functions have the same graph? (c) Sketch the graph of each function, indicating the domain and codomain of each.
    考虑以下六个函数: f1:R→R,其中 f1(x)=|x|;f2:R→[0,∞),其中 f2(x)=|x|;f3:[0,∞)→[0,∞),其中 f3(x)=|x|;f4:R→R,其中 f4(x)=x²;f5:[0,∞)→[0,∞),其中 f5(x)=x;f6:R→[0,∞),其中 f6(x)=x。 (a) 此列表中哪些函数对相等? (b) 哪些函数对具有相同的图像? (c) 绘制每个函数的图像,并标明每个函数的定义域和值域。
  4. Define f , g , h , k : R R by setting f ( x ) = x 3 , g ( x ) = x − 4, h ( x ) = 7 x , and k ( x ) = 2 x for x R . Find simplified formulas for the following functions constructed using pointwise operations. (a) f + g (b) fh (c) −2 g (d) kk (e) fk ( h g ) (f) ggh − 7 f (g) f / k .
    定义函数 f,g,h,k:R→R,令 f(x) = x 3 ,g(x) = x − 4,h(x) = 7x,k(x) = 2 x ,其中 x∈R。求下列函数的简化表达式,这些函数是通过逐点运算构造的。(a) f + g (b) fh (c) −2g (d) kk (e) fk(h − g) (f) ggh − 7f (g) f/k。
  5. Suppose f , g : R R satisfy f (1) = 5, g (1) = 2, f (3) = −4, g (3) = −3, and g ( x ) ≠ 0 for all x R . Compute: (a) ( f + g )(1) (b) ( f g )(3) (c) ( fg )(1) (d) ( f / g )(3) (e) ( f Id R + g g ) ( 1 ) .
    假设 f,g:R→R 满足 f(1) = 5, g(1) = 2, f(3) = −4, g(3) = −3,且对于所有 x∈R,g(x) ≠ 0。计算:(a) (f + g)(1) (b) (f − g)(3) (c) (fg)(1) (d) (f/g)(3) (e) (fIdR+gg)(1)。
  6. Let X = [0, 4], A = [0, 2], B = (1, 3), C = (2, 4]. Draw the graphs of the following functions from X to R . (a) χ A (b) 3 χ B (c) 2 χ A + 5 χ C (d) 3 χ A χ B + 2 χ C .
    设 X = [0, 4],A = [0, 2],B = (1, 3),C = (2, 4)。绘制下列从 X 到 R 的函数图像。(a) χ A (b) 3χ B (c) 2χ A + 5χ C (d) 3χ A − χ B + 2χ C
  7. Let f , g , h : A R be functions. Prove the following identities from Theorem 5.54 , explaining every step with a definition or an algebraic fact about real numbers. (a) ( f + g ) + h = f + ( g + h ). (b) f + ( − f ) = 0 A = ( − f ) + f . (c) f · g = g · f . (d) ( f · g ) · h = f · ( g · h ). (e) f + 0 A = f = 0 A + f . (f) ( f + g ) · h = ( f · h ) + ( g · h ).
    设 f, g, h: A→R 为函数。证明定理 5.54 中的下列恒等式,并用定义或关于实数的代数事实解释每一步。(a) (f + g) + h = f + (g + h)。(b) f + ( − f) = 0 A = ( − f) + f。(c) f · g = g · f。(d) (f · g) · h = f · (g · h)。(e) f + 0 A = f = 0 A + f。(f) (f + g) · h = (f · h) + (g · h)。
  8. (a) Suppose f , g : A R . Prove carefully that f + g , f · g , and − g are indeed functions. (b) Let Z = { x A : g ( x ) = 0}. Show that the generalized pointwise quotient f / g : ( A Z ) R , defined by ( f / g )( x ) = f ( x )/ g ( x ) for all x A Z , is a function.
    (a) 设 f,g:A→R。仔细证明 f + g、f · g 和 −g 确实是函数。(b) 令 Z = {x ∈ A : g(x) = 0}。证明广义逐点商 f/g:(A−Z)→R,定义为对于所有 x ∈ A − Z,有 (f/g)(x) = f(x)/g(x),是一个函数。
  9. Suppose X is a set and A X . Prove that the characteristic function χ A is indeed a function.
    假设 X 是一个集合,A ⊆ X。证明特征函数 χ A 确实是一个函数。
  10. Compute the following quantities. All characteristic functions mentioned have domain and codomain R . (a) χ Q ( 1.107 ) (b) χ R Q ( 2 ) (c) χ Z ( 111111 / 37 ) (d) χ (0,∞) ( e π π e ) (e) i = 1 10 χ { 1 , 2 , , i } ( 7 ) (f) i = 1 5 χ { 1 , 2 , , i } ( 2 )
    计算下列各量。所有特征函数的定义域和值域均为 R。(a) χQ(1.107) (b) χR−Q(2) (c) χZ(1111/37) (d) χ (0,∞) (e π − π e ) (e) ∑i=110χ{1,2,…,i}(7) (f) ∏i=15χ{1,2,…,i}(2)
  11. Let X be a set and let A , B be subsets of X . Prove parts (b) through (f) of Theorem 5.57 .
    设 X 为一个集合,A、B 为 X 的子集。证明定理 5.57 的 (b) 至 (f) 部分。
  12. Prove: for all sets X and all A , B , C X ,
    χ A B C = χ A + χ B + χ C χ A B χ A C χ B C + χ A B C .

    证明:对于所有集合 X 和所有 A、B、C ⊆ X,χA∪B∪C=χA+χB+χC−χA∩B−χA∩C−χB∩C+χA∩B∩C。
  13. Suppose n Z > 0 and f 1 , , f n : A R are functions. Give recursive definitions of the pointwise sum i = 1 n f i and the pointwise product i = 1 n f i of these n functions.
    假设 n∈Z>0 且 f1,…,fn:A→R 是函数。给出这 n 个函数的逐点和 ∑i=1nfi 和逐点积 ∏i=1nfi 的递归定义。
  14. Suppose n Z > 0 and A 1 , …, A n are subsets of X . (a) Prove by induction that χ A 1 A n = i = 1 n χ A i . (b) Prove: if A i A j = 0 for all i j , then χ A 1 A n = i = 1 n χ A i .
    假设 n∈Z>0 且 A 1 , …, A n 是 X 的子集。 (a) 用归纳法证明 χA1∩⋯∩An=∏i=1nχAi。 (b) 证明:如果对于所有 i ≠ j,Ai∩Aj=0,则 χA1∪⋯∪An=∑i=1nχAi。
  15. Suppose X is a set, I 0 is an index set, and for each i I , A i is a subset of X . (a) Prove: for all x X , χ i I A i ( x ) = min { χ A i ( x ) : i I } . (b) Prove: for all x X , χ i I A i ( x ) = max { χ A i ( x ) : i I } .
    假设 X 是一个集合,I≠0 是一个索引集,对于每个 i ∈ I,A i 是 X 的一个子集。 (a) 证明:对于所有 x ∈ X,χ⋂i∈IAi(x)=min{χAi(x):i∈I}。 (b) 证明:对于所有 x ∈ X,χ⋃i∈IAi(x)=max{χAi(x):i∈I}。
  16. Suppose G is a relation satisfying the uniqueness condition
    x , y , z , ( ( x , y ) G ( x , z ) G ) ( y = z ) .
    Prove that for all sets A and B , ( A , B , G ) is a function iff A = { x : y , ( x , y ) G } and G [ A ] ⊆ B .
    假设 G 是一个满足唯一性条件 ∀x,∀y,∀z,((x,y)∈G∧(x,z)∈G)⇒(y=z) 的关系。证明对于任意集合 A 和 B,(A, B, G) 是一个函数当且仅当 A={x:∃y,(x,y)∈G} 且 G[A] ⊆ B。
  17. (a) Prove: for all f , g : R R , if f and g are polynomials, then f + g is a polynomial. (b) Prove: for all f , g : R R , if f and g are polynomials, then f · g is a polynomial.
    (a) 证明:对于任意函数 f,g:R→R,如果 f 和 g 都是多项式,则 f + g 也是多项式。(b) 证明:对于任意函数 f,g:R→R,如果 f 和 g 都是多项式,则 f · g 也是多项式。
  18. Even Functions. A function f : R R is even iff for all x R , f ( − x ) = f ( x ). (a) Prove: for all f , g : R R , if f and g are even, then f + g is even. (b) Prove: for all f , g : R R , if f and g are even, then f · g is even. (c) Prove: for all f : R R , the function g : R R , given by g ( x ) = [ f ( x ) + f ( − x )]/2 for x R , is even.
    偶函数。函数 f:R→R 是偶函数当且仅当对于所有 x∈R,f( − x) = f(x)。 (a) 证明:对于所有 f,g:R→R,如果 f 和 g 都是偶函数,则 f + g 也是偶函数。 (b) 证明:对于所有 f,g:R→R,如果 f 和 g 都是偶函数,则 f · g 也是偶函数。 (c) 证明:对于所有 f:R→R,函数 g:R→R,由 g(x) = [f(x) + f( − x)]/2 (x∈R) 给出,是偶函数。
  19. Odd Functions. A function f : R R is odd iff for all x R , f ( − x ) = − f ( x ). (a) Prove: for all f , g : R R , if f and g are odd, then f + g is odd. (b) Prove: for all f , g : R R , if f and g are odd, then f · g is even. (c) If f is odd and g is even, what can you say about f + g and f · g ?
    奇函数。函数 f:R→R 是奇函数当且仅当对于所有 x∈R,f( − x) = −f(x)。 (a) 证明:对于所有 f,g:R→R,如果 f 和 g 都是奇函数,则 f + g 也是奇函数。 (b) 证明:对于所有 f,g:R→R,如果 f 和 g 都是奇函数,则 f · g 也是偶函数。 (c) 如果 f 是奇函数且 g 是偶函数,关于 f + g 和 f · g,你能得出什么结论?
  20. Prove: for all f : R R , there exist g , h : R R such that g is even, h is odd, and f = g + h .
    证明:对于所有 f:R→R,存在 g,h:R→R,使得 g 为偶数,h 为奇数,且 f = g + h。
  21. Linear Functions. A function f : R R is linear iff for all c , x , y R , f ( x + y ) = f ( x ) + f ( y ) and f ( cx ) = cf ( x ). (a) Prove: for all f , g : R R , if f and g are linear, then f + g is linear. (b) Prove: for all f : R R and all b R , if f is linear, then bf is linear. (c) Prove or disprove: for all f , g : R R , if f and g are linear, then f · g is linear.
    线性函数。函数 f:R→R 是线性函数当且仅当对于所有 c,x,y∈R,f(x + y) = f(x) + f(y) 且 f(cx) = cf(x)。 (a) 证明:对于所有 f,g:R→R,如果 f 和 g 是线性函数,则 f + g 也是线性函数。 (b) 证明:对于所有 f:R→R 和所有 b∈R,如果 f 是线性函数,则 bf 也是线性函数。 (c) 证明或反证:对于所有 f,g:R→R,如果 f 和 g 是线性函数,则 f · g 也是线性函数。

5.6 Composition, Restriction, and Gluing

5.6 组成、限制和粘合

This section investigates three more operations for building new functions from given functions: composition, restriction, and gluing. The composition operation combines functions f : A B and g : B C to get a new function g f : A C that acts by doing f and then doing g . The restriction operation takes one function f : A B and builds a new function by shrinking the domain of f to some subset of A . The gluing operation allows us to combine two or more functions (under certain conditions) to build a new function with a larger domain. For instance, we need gluing to define the absolute value function and other functions defined by cases.

本节探讨了从给定函数构造新函数的另外三种操作:复合、限制和粘合。复合操作将函数 f : A → B 和 g : B → C 组合起来,得到一个新函数 g ∘ f : A → C,其作用方式为先执行 f,再执行 g。限制操作接受一个函数 f : A → B,并通过将其定义域缩小到 A 的某个子集来构造一个新函数。粘合操作允许我们(在特定条件下)组合两个或多个函数,从而构造一个具有更大定义域的新函数。例如,我们需要使用粘合操作来定义绝对值函数以及其他由条件定义的函数。

Composition of Functions

函数复合

5.58. Definition: Function Composition. Given functions f : A B and g : B C , define the composition g f : A C by setting ( g f )( x ) = g ( f ( x )) for all x A .

5.58. 定义:函数复合。给定函数 f : A → B 和 g : B → C,定义复合函数 g ∘ f : A → C,令 (g ∘ f)(x) = g(f(x)) 对所有 x ∈ A 成立。

Note that the composition g f is only defined when the codomain of f equals the domain of g . In this case, the output of the composite function on an input x in A is found by first performing f on x to obtain an intermediate output f ( x ) in B , and then performing g on f ( x ) to obtain the final output g ( f ( x )) in C . The graph of g f is {( x , g ( f ( x ))) : x A }. It is routine to check that g f really is a function, using the assumption that f and g are functions. You can also verify that if F is the graph of f and G is the graph of g , then the composition of relations G F is the graph of g f .

注意,复合函数 g ∘ f 仅在 f 的值域等于 g 的定义域时才有定义。在这种情况下,复合函数对输入 x(属于集合 A)的输出可以通过以下方式找到:首先对 x 执行 f,得到中间输出 f(x)(属于集合 B);然后对 f(x) 执行 g,得到最终输出 g(f(x))(属于集合 C)。g ∘ f 的图像为 {(x, g(f(x))) : x ∈ A}。假设 f 和 g 都是函数,则很容易验证 g ∘ f 是否确实是一个函数。你还可以验证,如果 F 是 f 的图像,G 是 g 的图像,那么复合函数 G ∘ F 就是 g ∘ f 的图像。

5.59. Example: Composition via Arrow Diagrams. Let A = {1, 2, 3, 4, 5} and B = { a , b , c , d , e }. Figure 5.6 shows the arrow diagrams for functions f : A B and g : B B . For each x A , we obtain ( g f )( x ) = g ( f ( x )) by following an f -arrow and then a g -arrow in succession. The resulting arrow diagram is shown at the bottom of the figure. The graph of g f : A B is {(1, c ), (2, e ), (3, d ), (4, d ), (5, a )}. In this example, the composite function f g is not defined, since the codomain of g (the set B ) does not equal the domain of f (the set A ).

5.59. 示例:利用箭头图进行复合。设 A = {1, 2, 3, 4, 5},B = {a, b, c, d, e}。图 5.6 显示了函数 f : A → B 和 g : B → B 的箭头图。对于每个 x ∈ A,我们依次沿着 f 箭头和 g 箭头,得到 (g ∘ f)(x) = g(f(x))。所得箭头图显示在图的底部。g ∘ f : A → B 的图像为 {(1, c), (2, e), (3, d), (4, d), (5, a)}。在本例中,复合函数 f ∘ g 未定义,因为 g 的值域(集合 B)与 f 的定义域(集合 A)不相等。

fig5_6

Figure 5.6

图 5.6

Arrow diagrams illustrating composition.

用箭头图表示组成成分。

5.60. Example: Composition via Formulas. Let f : R R be given by f ( x ) = x 2 , and let g : R R be given by g ( x ) = x + 2. In this case, both f g and g f are functions with domain R and codomain R . For each real x , we compute ( g f )( x ) = g ( f ( x )) = g ( x 2 ) = x 2 + 2, whereas ( f g )( x ) = f ( g ( x )) = f ( x + 2) = ( x + 2) 2 = x 2 + 4 x + 4. In particular, ( g f )(1) = 3 ≠ 9 = ( f g )(1), so we conclude that g f f g . This shows that function composition is not commutative in general .

5.60. 示例:通过公式进行组合。设函数 f:R→R 由 f(x) = x 2 给出,函数 g:R→R 由 g(x) = x + 2 给出。在这种情况下,f ∘ g 和 g ∘ f 都是定义域为 R 且值域为 R 的函数。对于每个实数 x,我们计算 (g ∘ f)(x) = g(f(x)) = g(x 2 ) = x 2 + 2,而 (f ∘ g)(x) = f(g(x)) = f(x + 2) = (x + 2) 2 = x 2 + 4x + 4。特别地,(g ∘ f)(1) = 3 ≠ 9 = (f ∘ g)(1),因此我们得出结论:g ∘ f ≠ f ∘ g. 这表明函数复合一般不满足交换律。

We can also combine function composition with pointwise operations. For instance, g f + fg is the function sending x R to g ( f ( x )) + f ( x ) g ( x ) = x 2 + 2 + x 2 ( x + 2). We have f f = f · f , since for each x R , ( f f )( x ) = f ( f ( x )) = ( x 2 ) 2 = x 4 and ( f · f )( x ) = f ( x ) f ( x ) = x 2 · x 2 = x 4 . On the other hand, g g g · g , since ( g g )( x ) = g ( g ( x )) = ( x + 2) + 2 = x + 4 and ( g · g )( x ) = g ( x ) g ( x ) = ( x + 2) 2 , and these functions disagree at x = 1.

我们还可以将函数复合与逐点运算结合起来。例如,g ∘ f + fg 是将 x∈R 映射到 g(f(x)) + f(x)g(x) = x 2 + 2 + x 2 (x + 2) 的函数。我们有 f ∘ f = f · f,因为对于每个 x∈R,(f ∘ f)(x) = f(f(x)) = (x 2 ) 2 = x 4 且 (f · f)(x) = f(x)f(x) = x 2 · x 2 = x 4 。另一方面,g ∘ g ≠ g · g,因为 (g ∘ g)(x) = g(g(x)) = (x + 2) + 2 = x + 4 且 (g · g)(x) = g(x)g(x) = (x + 2) 2 ,并且这些函数在 x = 1 处不一致。

5.61. Example. Let A = {1, 2, 3, 4}, and let functions f , g , h : A A have graphs

5.61. 示例。设 A = {1, 2, 3, 4},函数 f, g, h : A → A 的图像如下:

G G f f = { ( 1 , 3 ) , ( 2 , 4 ) , ( 3 , 1 ) , ( 4 , 2 ) } , G G g = { ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 ) , ( 4 , 4 ) } , G G h h = { ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) , ( 4 , 1 ) } .

You can check that g f and f g both have graph {(1, 1), (2, 4), (3, 3), (4, 2)}, so that g f = f g . But g h h g , since (for instance) ( g h )(1) = g ( h (1)) = g (2) = 2 and ( h g )(1) = h ( g (1)) = h (3) = 4.

你可以验证 g ∘ f 和 f ∘ g 的图像均为 {(1, 1), (2, 4), (3, 3), (4, 2)},因此 g ∘ f = f ∘ g。但是 g ∘ h ≠ h ∘ g,因为(例如)(g ∘ h)(1) = g(h(1)) = g(2) = 2 且 (h ∘ g)(1) = h(g(1)) = h(3) = 4。

Although function composition is not commutative, the following algebraic laws do hold.

虽然函数复合不满足交换律,但以下代数定律仍然成立。

5.62. Theorem on Function Composition.

5.62. 函数复合定理。

(a) Associativity: For all f : A B , g : B C , and h : C D , h ∘ ( g f ) = ( h g ) ∘ f .

(a)结合律:对于所有 f : A → B,g : B → C,和 h : C → D,h ∘ (g ∘ f) = (h ∘ g) ∘ f。

(b) Identity: For all f : A B , f ∘Id A = f = Id B f .

(b)恒等式:对于所有 f : A → B,f ∘Id A = f = Id B ∘ f。

(c) Right Distributive Laws: For all f : A R and all g , h : R R , ( g + h ) ∘ f = ( g f ) + ( h f ) and ( g · h ) ∘ f = ( g f ) · ( h f ).

(c)右分配律:对于所有 f:A→R 和所有 g,h:R→R,(g + h)∘ f =(g ∘ f)+(h ∘ f)且(g · h)∘ f =(g ∘ f)·(h ∘ f)。

Proof. We use the proof template for proving equality of functions (see page 241). For part (a), fix f : A B , g : B C , and h : C D . We have functions ( g f ) : A C and ( h g ) : B D , so that the functions h ∘ ( g f ) and ( h g ) ∘ f are both defined with domain A and codomain D . Now, fix x A . Repeatedly using the definition of function composition, we find that

证明。我们使用证明函数相等性的证明模板(参见第 241 页)。对于 (a) 部分,固定 f : A → B,g : B → C 和 h : C → D。我们有函数 (g ∘ f) : A → C 和 (h ∘ g) : B → D,因此函数 h ∘ (g ∘ f) 和 (h ∘ g) ∘ f 的定义域均为 A,值域均为 D。现在,固定 x ∈ A。反复使用函数复合的定义,我们发现

[ h h ( g f f ) ] ( x x ) = h h ( ( g f f ) ( x x ) ) = h h ( g ( f f ( x x ) ) ) ;
[ ( h h g ) f f ] ( x x ) = ( h h g ) ( f f ( x x ) ) = h h ( g ( f f ( x x ) ) ) .

Thus, h ∘ ( g f ) = ( h g ) ∘ f , as needed.

因此,h ∘ (g ∘ f) = (h ∘ g) ∘ f,符合要求。

For part (b), fix f : A B . Recall Id A : A A is defined by Id A ( x ) = x for all x A . We now see that f ∘Id A : A B is defined and has the same domain and codomain as f : A B . Fix x A , and compute ( f ∘Id A )( x ) = f (Id A ( x )) = f ( x ). So f ∘Id A = f . The equality f = Id B f is proved similarly.

对于 (b) 部分,固定 f : A → B。回想一下,Id A : A → A 定义为对于所有 x ∈ A,Id A (x) = x。现在我们看到 f ∘Id A : A → B 已定义,并且与 f : A → B 具有相同的定义域和值域。固定 x ∈ A,并计算 (f ∘Id A )(x) = f(Id A (x)) = f(x)。因此 f ∘Id A = f。类似地证明等式 f = Id B ∘ f。

Finally, we prove the first equality in part (c). Fix f : A R and g , h : R R . Both ( g + h ) ∘ f and ( g f ) + ( h f ) are functions from A to R . Fix x A , and compute

最后,我们证明 (c) 部分中的第一个等式。固定 f:A→R 和 g,h:R→R。(g + h) ∘ f 和 (g ∘ f) + (h ∘ f) 都是从 A 到 R 的函数。固定 x ∈ A,并计算

( ( g + h h ) f f ) ( x x ) = ( g + h h ) ( f f ( x x ) ) = g ( f f ( x x ) ) + h h ( f f ( x x ) )

(5.1)

= ( g f f ) ( x x ) + ( h h f f ) ( x x ) = [ ( g f f ) + ( h h f f ) ] ( x x ) .

(5.2)

So the functions ( g + h ) ∘ f and ( g f ) + ( h f ) are equal.      □

因此,函数 (g + h) ∘ f 和 (g ∘ f) + (h ∘ f) 相等。□

In the exercises, we see that the left distributive law f ∘ ( g + h ) = ( f g ) + ( f h ) is not always true, though it does hold when f has a special property called linearity .

在练习中,我们看到左分配律 f ∘ (g + h) = (f ∘ g) + (f ∘ h) 并不总是成立的,尽管当 f 具有称为线性的特殊性质时,它确实成立。

Restriction of Functions

功能限制

Our next operation on functions, called restriction, shrinks the domain of a given function to a smaller set of inputs.

我们对函数进行的下一个操作称为限制,它将给定函数的定义域缩小到一组较小的输入。

5.63. Definition: Restriction. For any function f : A B and any set S A , the restriction of f to S is the function f | S : S B defined by f | S ( x ) = f ( x ) for all x S .

5.63. 定义:限制。对于任意函数 f : A → B 和任意集合 S ⊆ A,f 在 S 上的限制是函数 f| S : S → B,定义为 f| S (x) = f(x) 对所有 x ∈ S 成立。

The graph of f | S is {( x , f ( x )) : x S }, which can be obtained by intersecting the graph of f with S × B . You can check that f | S really is a function. Note that f | S and f are different functions (except when S = A ), since these functions have different domains.

函数 f| S 的图像是 {(x, f(x)) : x ∈ S},它可以通过 f 的图像与 S × B 的交集得到。你可以验证 f| S 确实是一个函数。注意,f| S 和 f 是不同的函数(当 S = A 时除外),因为它们的定义域不同。

5.64. Example: Arrow Diagrams of Restricted Functions. Let A = {1, 2, 3, 4, 5}, B = { a , b , c , d , e }, and define f : A B by f (1) = c , f (2) = c , f (3) = e , f (4) = b , and f (5) = e . Restricting f to the set S = {1, 3, 5} gives the function f | S : S B with f | S (1) = c , f | S (3) = e , and f | S (5) = e . If we restrict f to T = {2, 4}, we get a function f | T : T B with graph {(2, c ), (4, b )}. Note f | A = f , whereas f | 0 is the empty function mapping 0 into the codomain B . Arrow diagrams for f , f | S , f | T , and f | 0 are shown in Figure 5.7 . Note that all of these restricted functions have the same codomain as the original function. We could obtain more new functions from these by enlarging or shrinking the codomain, as discussed in Example 5.55 .

5.64. 示例:受限函数的箭头图。设 A = {1, 2, 3, 4, 5},B = {a, b, c, d, e},定义函数 f : A → B,其中 f(1) = c,f(2) = c,f(3) = e,f(4) = b,f(5) = e。将 f 限制到集合 S = {1, 3, 5},得到函数 f| S : S → B,其中 f| S (1) = c,f| S (3) = e,f| S (5) = e。如果将 f 限制到 T = {2, 4},则得到函数 f| T : T → B,其箭头图为 {(2, c), (4, b)}。注意,f| A = f,而 f|0 是将 0 映射到值域 B 的空函数。图 5.7 显示了 f、f| S 、f| T 和 f|0 的箭头图。请注意,所有这些受限函数都与原函数具有相同的值域。我们可以通过扩大或缩小值域来获得更多新函数,如例 5.55 所述。

fig5_7

Figure 5.7

图 5.7

Arrow diagrams for restricted functions.

受限函数的箭头图。

5.65. Example: Graphs of Restricted Functions. Suppose g : R R is the squaring function g ( x ) = x 2 for x R . Figure 5.8 displays the graphs of g , g | [0,∞) , g | (−∞,0] , g | (−2,2] , g | {−1,0,1} , and g | 0 . For each S R , we obtain the graph of g | S from the graph of g by intersecting the full graph with the product set S × R . This means that we only keep those points ( x , g ( x )) in the original graph where the first coordinate x is in the set S .

5.65. 示例:受限函数的图像。假设 g:R→R 是平方函数 g(x) = x 2 ,其中 x∈R。图 5.8 展示了 g、g| [0,∞) 、g| (−∞,0] 、g| (−2,2] 、g| {−1,0,1} 和 g|0 的图像。对于每个 S ⊆R,我们可以通过将 g 的图像与乘积集 S×R 相交,得到 g| S 的图像。这意味着我们只保留原图中第一个坐标 x 属于集合 S 的点 (x, g(x))。

fig5_8

Figure 5.8

图 5.8

Graphs for restrictions of the squaring function.

平方函数的限制条件的图像。

Gluing Functions Together (Optional)

将功能粘合在一起(可选)

We now discuss the gluing operation, which is used to combine two or more functions to obtain a single new function with a larger domain. Before presenting the general definition, we consider two examples.

现在我们讨论粘合操作,它用于将两个或多个函数合并,从而得到一个定义域更大的新函数。在给出一般定义之前,我们先考虑两个例子。

5.66. Example. Consider the three functions f , g , h defined as follows:

5.66. 示例。考虑定义如下的三个函数 f、g、h:

f f : { 1 , 2 , 3 , 4 } { a 一个 , b b , c c , d d } with graph 带图表 { ( 1 , a 一个 ) , ( 2 , a 一个 ) , ( 3 , d d ) , ( 4 , b b ) } ; g : { 1 , 3 , 5 , 7 } { a 一个 , b b , c c , d d } with graph 带图表 { ( 1 , a 一个 ) , ( 3 , d d ) , ( 5 , b b ) , ( 7 , a 一个 ) } ; h h : { 0 , 1 , 5 } { a 一个 , b b , c c , d d } with graph 带图表 { ( 0 , d d ) , ( 1 , a 一个 ) , ( 5 , d d ) } .

Arrow diagrams for these functions appear in the top half of Figure 5.9 . Consider the effect of merging (gluing together) the diagrams for f and g . The result is shown in the bottom-left panel of the figure. In this case, a new function results, namely

图 5.9 的上半部分展示了这些函数的箭头图。考虑将 f 和 g 的箭头图合并(粘合)在一起的效果。结果如图左下角所示。在这种情况下,会得到一个新的函数,即

glue 胶水 ( f f , g ) : { 1 , 2 , 3 , 4 , 5 , 7 } { a 一个 , b b , c c , d d } with graph 带图表 { ( 1 , a 一个 ) , ( 2 , a 一个 ) , ( 3 , d d ) , ( 4 , b b ) , ( 5 , b b ) , ( 7 , a 一个 ) } .

Similarly, combining the diagrams for f and h produces the new function

类似地,将 f 和 h 的图结合起来,即可得到新的函数。

glue 胶水 ( f f , h h ) : { 0 , 1 , 2 , 3 , 4 , 5 } { a 一个 , b b , c c , d d } with graph 带图表 { ( 0 , d d ) , ( 1 , a 一个 ) , ( 2 , a 一个 ) , ( 3 , d d ) , ( 4 , b b ) , ( 5 , d d ) } .

On the other hand, if we try to combine the diagrams for g and h , the new graph is

另一方面,如果我们尝试将 g 和 h 的图合并,则新的图是

{ ( 0 , d d ) , ( 1 , a 一个 ) , ( 3 , d d ) , ( 5 , b b ) , ( 5 , d d ) , ( 7 , a 一个 ) } ,
#

which is not the graph of a function. The trouble is that 5 is in the domain of both g and h , but g (5) = b d = h (5). Thus, when we combine g and h , the input 5 is associated with two distinct outputs.

这并非函数图像。问题在于,5 同时在函数 g 和 h 的定义域内,但 g(5) = b ≠ d = h(5)。因此,当我们结合 g 和 h 时,输入 5 对应两个不同的输出。

fig5_9

Figure 5.9

图 5.9

Arrow diagrams illustrating gluing.

用箭头图示说明粘合过程。

5.67. Example. Consider the function f : ( , 0 ] R defined by f ( x ) = − x for x ≤ 0 and the function g : [ 0 , ) R defined by g ( x ) = x for x ≥ 0. Combining the graphs of f and g produces a new function h = glue ( f , g ) : R R such that h ( x ) = − x for x ≤ 0, and h ( x ) = x for x ≥ 0. This function h is the familiar absolute value function from calculus.

5.67. 示例。考虑函数 f:[−∞,0]→R,定义为当 x ≤ 0 时 f(x) = −x,以及函数 g:[0,∞)→R,定义为当 x ≥ 0 时 g(x) = x。将 f 和 g 的图像合并,得到一个新的函数 h=glue(f,g):R→R,使得当 x ≤ 0 时 h(x) = −x,当 x ≥ 0 时 h(x) = x。这个函数 h 就是微积分中我们熟悉的绝对值函数。

On the other hand, suppose we try to glue together the functions f 1 : R R and g 1 : R R defined by f 1 ( x ) = − x and g 1 ( x ) = x for all x R . In this case, the union of the graphs of f 1 and g 1 is not the graph of a function. The trouble is that there exist elements x in the common domain of f 1 and g 1 such that f 1 ( x ) ≠ g 1 ( x ). Indeed, this inequality holds for all x except for x = 0. Note that this difficulty did not arise when we glued together f and g , because these two functions agree on the common part of their domains (which is the one-point set {0}).

另一方面,假设我们尝试将函数 f1:R→R 和 g1:R→R 连接起来,这两个函数分别定义为 f 1 (x) = −x 和 g 1 (x) = x,其中 x∈R。在这种情况下,f 1 和 g 1 的图像的并集并不是一个函数的图像。问题在于,在 f 1 和 g 1 的公共定义域中存在元素 x,使得 f 1 (x) ≠ g 1 (x)。事实上,除了 x = 0 之外,这个不等式对所有 x 都成立。注意,当我们连接 f 和 g 时,并没有出现这个问题,因为这两个函数在它们的公共定义域(即单点集 {0})上是相同的。

Now we define the gluing operation for a family of functions.

现在我们定义一族函数的粘合操作。

5.68. Definition: Gluing. Suppose I is an index set, and for each i I we have a function f i : A i B i with graph G i . We call { f i : i I } a gluable family iff for all i , j I and all x A i A j , f i ( x ) = f j ( x ). When this condition holds, we define a new function f : i I A i i I B i with graph G = i I G i . We use the notation f = glue( f i : i I ).

5.68. 定义:粘合。假设 I 是一个索引集,对于每个 i ∈ I,我们有一个函数 f i : A i → B i ,其图为 G i 。我们称 {f i : i ∈ I} 为可粘合族,当且仅当对于所有 i, j ∈ I 和所有 x∈Ai∩Aj,f i (x) = f j (x)。当此条件成立时,我们定义一个新的函数 f:⋃i∈IAi→⋃i∈IBi,其图为 G=⋃i∈IGi。我们使用记号 f = glue(f i : i ∈ I)。

Given x i I A i , we can compute f ( x ) as follows. Choose any index i I such that x A i ; then f ( x ) = f i ( x ). Note that the output obtained does not depend on the choice of i ; if we choose some other index j I with x A j , we know f j ( x ) = f i ( x ) since { f i : i I } is assumed to be a gluable family. Thus f is a well-defined function.

给定 x∈⋃i∈IAi,我们可以按如下方式计算 f(x)。选择任意索引 i ∈ I,使得 x ∈ A i ;则 f(x) = f i (x)。注意,所得结果与 i 的选择无关;如果我们选择另一个索引 j ∈ I,使得 x ∈ A j ,我们知道 f j (x) = f i (x),因为 {f i : i ∈ I} 被假定为一个可粘合的族。因此,f 是一个定义良好的函数。

Here is a more formal proof that f is a function, based on checking that the graph G satisfies the three conditions in the definition of a function. Write A = i I A i and B = i I B i . First, for each i I , we know G i A i × B i A × B , so that G = i I G i is also a subset of A × B . Second, given x A , we know there exists i I with x A i . Since f i is a function, there exists y with ( x , y ) ∈ G i , and hence ( x , y ) ∈ G . Third, fix x , y , z ; assume ( x , y ) ∈ G and ( x , z ) ∈ G ; prove y = z . By definition of G , there exist i , j I with ( x , y ) ∈ G i and ( x , z ) ∈ G j . The only ordered pair in G i with first coordinate x is ( x , f i ( x )), so y = f i ( x ). Similarly, the only ordered pair in G j with first coordinate x is ( x , f j ( x )), so z = f j ( x ). We also know x A i (since ( x , y ) ∈ G i A i × B i ) and x A j (since ( x , z ) ∈ G j A j × B j ). By the gluable family condition, we deduce y = f i ( x ) = f j ( x ) = z , as needed. So f = ( A , B , G ) is indeed a function. Conversely, it can be shown that if ( A , B , G ) is a function, then { f i : i I } must be a gluable family (Exercise 25(b)).

以下是一个更正式的证明,证明 f 是一个函数,该证明基于验证图 G 满足函数定义中的三个条件。令 A=⋃i∈IAi 和 B=⋃i∈IBi。首先,对于每个 i ∈ I,我们知道 G i ⊆ A i × B i ⊆ A × B,因此 G=⋃i∈IGi 也是 A × B 的子集。其次,给定 x ∈ A,我们知道存在 i ∈ I 使得 x ∈ A i 。由于 f i 是一个函数,因此存在 y 使得 (x, y) ∈ G i ,从而 (x, y) ∈ G。第三,固定 x、y、z;假设 (x, y) ∈ G 且 (x, z) ∈ G;证明 y = z。根据 G 的定义,存在 i, j ∈ I,使得 (x, y) ∈ G i 且 (x, z) ∈ G j 。G i 中唯一以 x 为首坐标的有序对是 (x, f i (x)),因此 y = f i (x)。类似地,G j 中唯一以 x 为首坐标的有序对是 (x, f j (x)),因此 z = f j (x)。我们还知道 x ∈ A i (因为 (x, y) ∈ G i ⊆ A i × B i )且 x ∈ A j (因为 (x, z) ∈ G j ⊆ A j × B j )。根据可粘合族条件,我们推导出 y = f i (x) = f j (x) = z,符合要求。因此,f = (A, B, G) 确实是一个函数。反之,可以证明,如果 (A, B, G) 是一个函数,那么 {f i : i ∈ I} 必定是一个可粘合族(练习 25(b))。

We remark that if the sets A i are pairwise disjoint (i.e., for all i j in I , A i A j = 0 ), then { f i : i I } is automatically a gluable family.

我们注意到,如果集合 A i 两两不相交(即,对于 I 中的所有 i ≠ j,Ai∩Aj=0),则 {f i : i ∈ I} 自动是一个可粘合族。

5.69 Example: Piecewise Functions. Define a function f : R R by setting

f ( x ) = { sin x for x 0 ; x 2 for 0 x 1 ; 1 for 1 x 3 ; e x 3 for x 3.
Formally, we can build f by gluing together the following four functions:
f 1 : ( , 0 ] R given by f 1 ( x ) = sin x ; f 2 : [ 0 , 1 ] R given by f 2 ( x ) = x 2 ; f 3 : [ 1 , 3 ] R given by f 3 ( x ) = 1 ; f 4 : [ 3 , ) R given by f 4 ( x ) = e x 3 .

For all i j , f i and f j agree on the overlapping part of their domains. Thus, we can construct the overall function f = glue( f 1 , f 2 , f 3 , f 4 ) by taking the union of the graphs of the f i .

5.69 示例:分段函数。定义函数 f:R→R,令 f(x)={sin⁡x,x≤0;x²,0≤x≤1;1,1≤x≤3;ex⁻³,x≥3。形式上,我们可以通过将以下四个函数拼接在一起来构建 f:f₁:[−∞,0]→R,由 f₁(x)=sin⁡x 给出;f₂:[0,1]→R,由 f₂(x)=x² 给出;f₃:[1,3]→R,由 f₃(x)=1 给出;f₄:[3,∞)→R,由 f₄(x)=ex⁻³ 给出。对于所有 i ≠ j,f i 和 f j 的定义域重叠部分相同。因此,我们可以通过对 f i 的图取并集来构造整体函数 f = glue(f 1 , f 2 , f 3 , f 4 )。

Section Summary

章节概要

  1. Composition of Functions. Given f : A B and g : B C , the composition g f : A C is defined by ( g f )( x ) = g ( f ( x )) for all x A . To form g f , the codomain of f must equal the domain of g . Function composition is associative (meaning h ∘ ( g f ) = ( h g ) ∘ f , when both sides are defined) and satisfies the identity property f ∘Id A = f = Id B f . But function composition is not commutative: f g g f for most choices of f and g .
    函数复合。给定函数 f : A → B 和 g : B → C,复合函数 g ∘ f : A → C 定义为 (g ∘ f)(x) = g(f(x)),其中 x ∈ A。为了构成 g ∘ f,f 的值域必须等于 g 的定义域。函数复合满足结合律(即当等式两边都有定义时,h ∘ (g ∘ f) = (h ∘ g) ∘ f),并且满足恒等性质 f ∘Id A = f = Id B ∘ f。但是函数复合不满足交换律:对于大多数 f 和 g 的组合,f ∘ g ≠ g ∘ f。
  2. Restriction of a Function. Given f : A B and any subset S A , the restriction f | S : S B is defined by f | S ( x ) = f ( x ) for all x S . The graph of f | S is the intersection of the graph of f with S × B .
    函数的限制。给定函数 f : A → B 和任意子集 S ⊆ A,限制 f| S : S → B 定义为对于所有 x ∈ S,f| S (x) = f(x)。f| S 的图像是 f 的图像与 S × B 的交集。
  3. Gluing Functions. Suppose for each i in an index set I , we are given a function f i : A i B i with graph G i . These functions form a gluable family iff for all i , j I and all x A i A j , f i ( x ) = f j ( x ). In this case, we get a new function f = glue( f i : i I ) with domain i I A i , codomain i I B i , and graph i I G i . For x in the domain of f , we have f ( x ) = f i ( x ) for any choice of i I such that x A i . When the sets A i are pairwise disjoint, { f i : i I } is always a gluable family.
    粘合函数。假设对于索引集 I 中的每个 i,我们给定一个函数 f i : A i → B i ,其图为 G i 。这些函数构成一个可粘合的函数族,当且仅当对于所有 i, j ∈ I 和所有 x∈Ai∩Aj,都有 f i (x) = f j (x)。在这种情况下,我们得到一个新的函数 f = glue(f i : i ∈ I),其定义域为 ⋃i∈IAi,值域为 ⋃i∈IBi,图为 ⋃i∈IGi。对于 f 定义域中的 x,对于任意满足 x ∈ A i 的 i ∈ I,我们有 f(x) = f i (x)。当集合 A i 两两不相交时,{f i : i ∈ I} 始终是一个可粘合族。

Exercises

练习

  1. In Example 5.61 , compute f h and h f . Are these functions equal?
    在例 5.61 中,计算 f ∘ h 和 h ∘ f。这两个函数相等吗?
  2. Let A = {1, 2, 3, 4} and B = { v , w , x , y , z }. Define f : A B by f (1) = w , f (2) = z , f (3) = y , f (4) = w . Define g : B A by g ( v ) = 4, g ( w ) = 3, g ( y ) = 1, g ( x ) = 2, g ( z ) = 4. Let h : A A have graph {(1, 4), (2, 4), (3, 1), (4, 3)}. Let k : B B have graph {( v , z ), ( z , y ), ( y , v ), ( w , x ), ( x , w )}. Draw arrow diagrams for the following functions (when defined): (a) f , g , h , k (b) f f (c) f g (d) f h (e) f k (f) g f (g) g g (h) g h (i) g k (j) h f (k) h g (l) h h (m) h k (n) k f (o) k g (p) k h (q) k k .
    设 A = {1, 2, 3, 4},B = {v, w, x, y, z}。定义函数 f : A → B,其中 f(1) = w,f(2) = z,f(3) = y,f(4) = w。定义函数 g : B → A,其中 g(v) = 4,g(w) = 3,g(y) = 1,g(x) = 2,g(z) = 4。设函数 h : A → A 的图像为 {(1, 4), (2, 4), (3, 1), (4, 3)}。设函数 k : B → B 的图像为 {(v, z), (z, y), (y, v), (w, x), (x, w)}。为下列函数(如果已定义)绘制箭头图:(a)f、g、h、k(b)f ∘ f(c)f ∘ g(d)f ∘ h(e)f ∘ k(f)g ∘ f(g)g ∘ g(h)g ∘ h(i)g ∘ k(j)h ∘ f(k)h ∘ g(l)h ∘ h(m)h ∘ k(n)k ∘ f(o)k ∘ g(p)k ∘ h(q)k ∘ k。
  3. Define functions f , g , h , k : R R by setting f ( x ) = x 2 , g ( x ) = x − 3, h ( x ) = 2 x , and k ( x ) = 5 x for all x R . Find simplified formulas for the following functions. Take care not to confuse compositions like f g with pointwise products like fg . (a) f g (b) g f (c) fg (d) gf (e) f h (f) h f (g) g k (h) k g (i) k ∘ ( f + g ) (j) ( k f ) + ( k g ) (k) h ∘ ( f + g ) (l) ( h f ) + ( h g ) (m) ( h f )( h g ) (n) ( f + g ) ∘ h (o) ( f h ) + ( g h ) (p)( f h )( g h ).
    定义函数 f、g、h、k:R→R,令 f(x) = x 2 ,g(x) = x − 3,h(x) = 2 x ,k(x) = 5x,其中 x∈R。求下列函数的简化表达式。注意不要将复合函数(如 f ∘ g)与逐点乘积函数(如 fg)混淆。 (a) f ∘ g (b) g ∘ f (c) fg (d) gf (e) f ∘ h (f) h ∘ f (g) g ∘ k (h) k ∘ g (i) k ∘ (f + g) (j) (k ∘ f) + (k ∘ g) (k) h ∘ (f + g) (l) (h ∘ f) + (h ∘ g) (m) (h ∘ f)(h ∘ g) (n) (f + g) ∘ h (o) (f ∘ h) + (g ∘ h) (p)(f ∘ h)(g ∘ h).
  4. Define functions f : R > 0 R > 0 by f ( x ) = x , g : R > 0 R by g ( x ) = ln x , h : R R > 0 by h ( x ) = e x , k : R R by k ( x ) = x 2 , and j : R > 0 R by j ( x ) = x . (a) For which p , q ∈ { f , g , h , k , j } is the composition p q defined? (b) Check that h g = Id R > 0 and g h = Id R . (c) Show that k | R > 0 f = j .
    定义函数 f:R>0→R>0 为 f(x)=x,g:R>0→R 为 g(x) = lnx,h:R→R>0 为 h(x) = e,k:R→R 为 k(x) = x,j:R>0→R 为 j(x) = x。(a) 对于哪些 p, q ∈ {f, g, h, k, j},复合函数 p ∘ q 有定义?(b) 验证 h∘g=IdR>0 且 g∘h=IdR。(c) 证明 k|R>0∘f=j。
  5. (a) Prove: for all sets A , B , C , if A B C , then j B , C j A , B = j A , C . (b) Prove: for all sets A and B , if A B , then j A , B = Id B | A .
    (a) 证明:对于所有集合 A、B、C,如果 A ⊆ B ⊆ C,则 j B , C ∘ j A , B = j A , C . (b) 证明:对于所有集合 A 和 B,如果 A ⊆ B,则 j A , B = Id B | A .
  6. Let A be a set and g : A A a function. For n ≥ 0, we recursively define the powers g n : A A as follows: g 0 = Id A , and g n = g n −1 g for n ≥ 1. Let A = {1, 2, 3, 4, 5, 6, 7, 8}, and let g : A A have graph {(1, 4), (4, 7), (7, 2), (2, 5), (5, 1), (3, 8), (8, 6), (6, 3)}. (a) Compute the functions g n for 0 ≤ n ≤ 5. (b) Does there exist n > 0 such that g n = Id A ? If so, what is the smallest such n ? If not, why not?
    设 A 为集合,g : A → A 为函数。对于 n ≥ 0,我们递归地定义幂 g n : A → A 如下:g 0 = Id A ,且 g n = g n −1 ∘ g,其中 n ≥ 1。设 A = {1, 2, 3, 4, 5, 6, 7, 8},且 g : A → A 的图为 {(1, 4), (4, 7), (7, 2), (2, 5), (5, 1), (3, 8), (8, 6), (6, 3)}。 (a) 计算函数 g n ,其中 0 ≤ n ≤ 5。(b) 是否存在 n > 0 使得g n = Id A ? 如果是,那么最小的 n 是多少?如果不是,为什么?
  7. Repeat Exercise 6 for g : A A with graph
    { ( 2 , 6 ) , ( 6 , 4 ) , ( 4 , 8 ) , ( 8 , 2 ) , ( 1 , 3 ) , ( 3 , 2 ) , ( 5 , 8 ) , ( 7 , 7 ) } .

    对 g : A → A 重复练习 6,其中 g : A → A 的图像为 {(2,6),(6,4),(4,8),(8,2),(1,3),(3,2),(5,8),(7,7)}。
  8. Repeat Exercise 6 for g : A A defined by g ( x ) = x + 1 for 1 ≤ x ≤ 7, and g (8) = 1.
    重复练习 6,对于由 g(x) = x + 1(1 ≤ x ≤ 7)和 g(8) = 1 定义的 g : A → A。
  9. Let A = {1, 2, …, 10}. Find functions g : A A such that: (a) g ≠ Id A but g 2 = Id A ; (b) g 10 = Id A but g j ≠ Id A for 0 < j < 10; (c) g 21 = Id A but g j ≠ Id A for 0 < j < 21; (d) g j ≠ Id A for all j > 0.
    设 A = {1, 2, …, 10}。求函数 g : A → A,使得:(a) g ≠ Id A 但 g 2 = Id A ;(b) g 10 = Id A 但 g j ≠ Id A ,其中 0 < j < 10;(c) g 21 = Id A 但 g j ≠ Id A ,其中 0 < j < 21;(d) g j ≠ Id A ,其中 j > 0。
  10. Supply a reason for each equality in ( 5.1 ) and ( 5.2 ).
    请给出(5.1)和(5.2)中每个等式成立的理由。
  11. (a) Prove: for all functions f : A B , Id B f = f . (b) Prove: for all f : A R and all g , h : R R , ( g · h ) ∘ f = ( g f ) · ( h f ).
    (a) 证明:对于所有函数 f : A → B,Id B ∘ f = f. (b) 证明:对于所有 f:A→R 和所有 g,h:R→R,(g · h) ∘ f = (g ∘ f) · (h ∘ f).
  12. (a) Give an example of f , g , h : R R where h ∘ ( f + g ) ≠ ( h f ) + ( h g ). (b) Give an example of f , g , h : R R where h ∘ ( f · g ) ≠ ( h f ) · ( h g ).
    (a) 举例说明 f,g,h:R→R,其中 h ∘ (f + g) ≠ (h ∘ f) + (h ∘ g)。 (b) 举例说明 f,g,h:R→R,其中 h ∘ (f · g) ≠ (h ∘ f) · (h ∘ g)。
  13. Let f , g , h : R R be functions. Suppose h has the following linearity property: h ( x + y ) = h ( x ) + h ( y ) for all x , y R . Prove that h ∘ ( f + g ) = ( h f ) + ( h g ).
    设 f,g,h:R→R 为函数。假设 h 具有以下线性性质:对于所有 x,y∈R,h(x + y) = h(x) + h(y)。证明 h ∘ (f + g) = (h ∘ f) + (h ∘ g)。
  14. Let f , g , h : R R be functions. Find a condition on h that ensures that h ∘ ( f · g ) = ( h f ) · ( h g ) for all f , g : R R . Prove your answer.
    设 f, g, h: R→R 为函数。求 h 的一个条件,使得对于所有 f, g: R→R,都有 h ∘ (f · g) = (h ∘ f) · (h ∘ g)。证明你的答案。
  15. Let f : R R be defined by f ( x ) = sin x . Draw graphs of the following restrictions of f : (a) f | [− π /2, π /2] (b) f | [0,2 π ] (c) f | (−∞, π /4) (d) f | {0, π /6, π /4, π /3, π /2,3 π /4, π } (e) f | 0 (f) f | { k π / 2 : k Z } .
    设 f:R→R 由 f(x) = sin x 定义。绘制 f 的下列限制的图像:(a)f| [− π /2, π /2] (b)f| [0,2 π ] (c)f| (−∞, π /4) (d)f| {0, π /6, π /4, π /3, π /2,3 π /4, π } (e)f|0(f)f|{kπ/2:k∈Z}。
  16. Let f : R R be defined by f ( x ) = | x |. Draw graphs of the following restrictions of f : (a) f | [0,3] (b) f | ( 3 , 1 ) [ 1 , 2 ) (c) f | {0} (d) f | Z (e) f | R Z (f) f | Q .
    设 f:R→R 由 f(x) = |x| 定义。绘制 f 的下列限制的图像:(a)f| [0,3] (b)f|(−3,−1)∪[1,2)(c)f| {0} (d)f|Z(e)f|R−Z(f)f|Q。
  17. Consider the function g : {1, 2, 3, …, 8} → { v , w , x , y , z } with graph
    { ( 1 , w ) , ( 2 , z ) , ( 3 , y ) , ( 4 , w ) , ( 5 , x ) , ( 6 , x ) , ( 7 , y ) , ( 8 , x ) } .
    Draw arrow diagrams for: (a) g (b) g | {1,3,5,7} (c) g | {3,4,5} (d) g | {8} (e) g | 0 (f) g | { a A : g ( a )= x } (g) g | { a A : g ( a )= v } .
    考虑函数 g : {1, 2, 3, …, 8} → {v, w, x, y, z},其图像为 {(1,w),(2,z),(3,y),(4,w),(5,x),(6,x),(7,y),(8,x)}。绘制以下函数的箭头图:(a) g (b) g| {1,3,5,7} (c) g| {3,4,5} (d) g| {8} (e) g|0 (f) g| { a A : g ( a )= x } (g) g| { a A : g ( a )= v }
  18. Let A = [0, 2] and B = [2, ∞). Which of the following pairs of functions f : A R and g : B R are gluable? (a) f ( x ) = 2 x , g ( x ) = 4. (b) f ( x ) = x 2 , g ( x ) = 2 x . (c) f ( x ) = x 3 , g ( x ) = x 2 . (d) f ( x ) = e x −2 , g ( x ) = x /2. (e) f ( x ) = x , g ( x ) = x / 2 . (f) f ( x ) = | x |, g ( x ) = 3 − x .
    设 A = [0, 2] 且 B = [2, ∞)。下列哪些函数对 f:A→R 和 g:B→R 是可粘的? (a) f(x) = 2x, g(x) = 4. (b) f(x) = x 2 , g(x) = 2x. (c) f(x) = x 3 , g(x) = x 2 . (d) f(x) = e x −2 , g(x) = x/2. (e) f(x)=x, g(x)=x/2. (f) f(x) = |x|, g(x) = 3 − x.
  19. Consider the following four functions: f : { 1 , 2 , 3 } Z given by f (1) = 4, f (2) = 0, f (3) = 4; g : { 0 , 2 , 4 } Z given by g (0) = 2, g (2) = 2, g (4) = 2; h : { 0 , 1 , 2 } Z given by h (0) = 2, h (1) = 4, h (2) = 2; k : { 0 , 1 , 3 } Z given by k (0) = 2, k (1) = 4, k (3) = 3.

    Which of the following pairs of functions are gluable? For those that are, draw an arrow diagram illustrating the glued function. (a) f and g ; (b) f and h ; (c) f and k ; (d) g and h ; (e) g and k ; (f) h and k ; (g) f and f .


    考虑以下四个函数: f:{1,2,3}→Z,其中 f(1) = 4,f(2) = 0,f(3) = 4; g:{0,2,4}→Z,其中 g(0) = 2,g(2) = 2,g(4) = 2; h:{0,1,2}→Z,其中 h(0) = 2,h(1) = 4,h(2) = 2; k:{0,1,3}→Z,其中 k(0) = 2,k(1) = 4,k(3) = 3。 下列哪些函数对可以粘合?对于可以粘合的函数对,请绘制箭头图来表示粘合后的函数。(a) f 和 g;(b) f 和 h; (c)f 和 k;(d)g 和 h;(e)g 和 k;(f)h 和 k;(g)f 和 f。
  20. Assume f : X Y and g : Y Z are functions and S X . Prove that ( g f )| S = g ∘ ( f | S ).
    假设 f : X → Y 和 g : Y → Z 是函数,且 S ⊆ X。证明 (g ∘ f)| S = g ∘ (f| S )。
  21. Assume f : A Y and g : B Y are gluable functions and S A B . Prove that f 1 = f | A S and g 1 = g | B S are gluable functions, and that glue( f 1 , g 1 ) = glue( f , g )| S .
    假设 f : A → Y 和 g : B → Y 是可粘合函数,且 S⊆A∪B。证明 f1=f|A∩S 和 g1=g|B∩S 是可粘合函数,并且 glue(f 1 , g 1 ) = glue(f, g)| S
  22. Let X be a set. For fixed A X , show how the characteristic function χ A can be built from the constant functions 0 X and 1 X on X by using restriction and gluing operations.
    设 X 为一个集合。对于固定的 A ⊆ X,证明如何利用限制和粘合操作,从 X 上的常数函数 0 X 和 1 X 构造特征函数 χ A
  23. Suppose f : X Y is a function and S T X . (a) Prove that ( f | T )| S = f | S . (b) Prove that f | X = f . (c) Prove that f | 0 = ( 0 , Y , 0 ) .
    假设 f : X → Y 是一个函数,且 S ⊆ T ⊆ X。(a) 证明 (f| T )| S = f| S . 。(b) 证明 f| X = f。(c) 证明 f|0=(0,Y,0)。
  24. Suppose f , g : A R are functions and S A . Prove the following facts relating restriction to pointwise operations: (a) ( f + g )| S = ( f | S ) + ( g | S ); (b) ( fg )| S = ( f | S )( g | S ); (c) the restriction of a constant function on A with value c 0 is the constant function on S with value c 0 .
    假设 f,g:A→R 是函数,且 S ⊆ A。证明下列关于限制与逐点运算的关系:(a) (f + g)| S = (f| S ) + (g| S ); (b) (fg)| S = (f| S )(g| S ); (c) 在 A 上取值为 c 0 的常数函数在 S 上取值为 c 0 的常数函数。
  25. Suppose that for all i I , f i : A i B i is a function with graph G i . (a) Show that { f i : i I } is a gluable family iff for all i , j I , f i | A i A j = f j | A i A j . (b) Show that if i I G i is the graph of a function, then { f i : i I } is a gluable family.
    假设对于所有 i ∈ I,f i : A i → B i 是一个函数,其图像为 G i 。(a) 证明 {f i : i ∈ I} 是一个可粘合族当且仅当对于所有 i, j ∈ I,fi|Ai∩Aj=fj|Ai∩Aj。(b) 证明如果 ⋃i∈IGi 是一个函数的图像,则 {f i : i ∈ I} 是一个可粘合族。
  26. Suppose { f i : i I } is a gluable family, where f i : A i B i for each i I , and let h = glue( f i : i I ). (a) Prove: for all i I , h | A i = f i . (b) Prove: for any g : i I A i i I B i , if g | A i = f i for all i I , then g = h .
    假设 {f i : i ∈ I} 是一个可粘合族,其中对于每个 i ∈ I,都有 f i : A i → B i ,并令 h = glue(f i : i ∈ I)。 (a) 证明:对于所有 i ∈ I,h|Ai=fi。 (b) 证明:对于任意 g:⋃i∈IAi→⋃i∈IBi,如果对于所有 i ∈ I,g|Ai=fi,则 g = h。
  27. Suppose that h : A B is any function, and { A i : i I } is a family of sets with A = i I A i . Let h i = h | A i for i I . Prove { h i : i I } is a gluable family such that glue( h i : i I ) = h .
    假设 h : A → B 是任意函数,且 {A i : i ∈ I} 是一个集合族,其中 A=⋃i∈IAi。令 hi=h|Ai,其中 i ∈ I。证明 {h i : i ∈ I} 是一个可粘合的集合族,使得 glue(h i : i ∈ I) = h。

5.7 Direct Images and Preimages

5.7 直接像和原像

In earlier sections, we discussed the image of a set A under a relation R or the inverse relation R −1 . Intuitively, R [ A ] is the set of all outputs that are related under R to at least one input in the set A . On the other hand, R −1 [ B ] is the set of all inputs that are related under R to at least one output in the set B . Here we introduce similar constructions where the relation R is replaced by a function f .

在前面的章节中,我们讨论了集合 A 在关系 R 或逆关系 R −1 下的像。直观地说,R[A] 是所有在关系 R 下与集合 A 中至少一个输入相关的输出的集合。另一方面,R −1 [B] 是所有在关系 R 下与集合 B 中至少一个输出相关的输入的集合。这里,我们将引入类似的构造,其中关系 R 被函数 f 所取代。

Direct Images and Preimages

直接像和原像

5.70. Definition: Direct Image of a Set. Given a function f : X Y and a set A X , the direct image of A under f is f [ A ] = { f ( x ) : x A }. Equivalently, for all objects y , y f [ A ] iff x A , y = f ( x ) .

5.70. 定义:集合的直接像。给定一个函数 f : X → Y 和一个集合 A ⊆ X,A 在 f 下的直接像为 f[A] = {f(x) : x ∈ A}。等价地,对于所有对象 y,y∈f[A] 当且仅当存在 x∈A,y=f(x)。

Intuitively, f [ A ] is the set of all outputs we get by applying f to every input in the set A . If G is the graph of f , we could rephrase this definition by writing f [ A ] = G [ A ] = { y : x A , ( x , y ) G } . Thus, the direct image of a set under a function is a special case of the image of a set under a relation. In fact, this formulation is a little more general since we could apply it to any set A (not just subsets of the domain X ). However, when dealing with functions, it is sometimes easier to use the version of the definition phrased in terms of function notation.

直观地说,f[A] 是将函数 f 应用于集合 A 中的每个输入所得到的所有输出的集合。如果 G 是函数 f 的图像,我们可以将这个定义改写为 f[A]=G[A]={y:∃x∈A,(x,y)∈G}。因此,集合在函数作用下的直接像是一个特例,即集合在关系作用下的像。事实上,这个表述更具一般性,因为它可以应用于任何集合 A(而不仅仅是定义域 X 的子集)。然而,在处理函数时,有时使用函数符号来表述这个定义会更方便。

5.71. Definition: Preimage of a Set. Given a function f : X Y and a set B Y , the preimage of B under f is f −1 [ B ] = { x X : f ( x ) ∈ B }. Equivalently, for all objects x , x f 1 [ B ] iff x X and f ( x ) B . The set f −1 [ B ] is also called the inverse image of B under f .

5.71. 定义:集合的原像。给定函数 f : X → Y 和集合 B ⊆ Y,B 在函数 f 下的原像是 f −1 [B] = {x ∈ X : f(x) ∈ B}。等价地,对于任意对象 x,x∈f−1[B] 当且仅当 x∈X 且 f(x)∈B。集合 f −1 [B] 也称为 B 在函数 f 下的逆像。

Intuitively, the preimage f −1 [ B ] is the set of all inputs in the domain of f that map to an output in the set B when we apply f . As before, if f has graph G , we could have also defined f −1 [ B ] = G −1 [ B ], which is the image of B under the inverse of the relation G .

直观地说,原像 f −1 [B] 是 f 的定义域中所有在应用 f 时映射到集合 B 中的输出的输入的集合。和之前一样,如果 f 的图是 G,我们也可以定义 f −1 [B] = G −1 [B],它是 B 在关系 G 的逆变换下的像。

5.72. Warning. The notation f −1 [ B ] does not mean that the function f has an inverse function f −1 . (We discuss inverse functions in § 5.9 .) The preimage construction defined here can be applied to any function, not just those functions that have inverses. We also call your attention to our use of square brackets for images of sets under relations, direct images of sets under functions, and preimages of sets under functions. We use brackets rather than round parentheses to avoid confusion with function value notation, as shown in Example 5.74 below.

5.72. 警告。符号 f −1 [B] 并不意味着函数 f 存在反函数 f −1 。(我们将在 §5.9 中讨论反函数。)此处定义的原像构造可以应用于任何函数,而不仅仅是那些存在反函数的函数。此外,我们还要提醒您注意,我们使用方括号来表示集合在关系下的像、集合在函数下的直接像以及集合在函数下的原像。我们使用方括号而不是圆括号是为了避免与函数值表示法混淆,如下例 5.74 所示。

5.73. Example. Let f and g be the functions shown earlier in Figure 5.6 . Recall f has graph {(1, b ), (2, e ), (3, c ), (4, c ), (5, a )}, and g has graph {( a , a ), ( b , c ), ( c , d ), ( d , b ), ( e , e )}. We compute f [{3, 5}] = { c , a }, f −1 [{ c , a }] = {3, 4, 5}, f −1 [{ b , d , e }] = {1, 2}, f [{1, 2}] = { b , e }. Moreover, g [{ b , c , d }] = { b , c , d } = g −1 [{ b , c , d }], g [{ c }] = { d }, g −1 [{ a , d , e }] = { a , c , e }, and g [ 0 ] = 0 = g 1 [ 0 ] . Finally, ( g f )[{3, 4, 5}] = { d , a }, ( g f ) 1 [ { b } ] = 0 , and

5.73. 示例。设 f 和 g 为图 5.6 中所示的函数。回想一下,f 的图像为 {(1, b), (2, e), (3, c), (4, c), (5, a)},g 的图像为 {(a, a), (b, c), (c, d), (d, b), (e, e)}。我们计算 f[{3, 5}] = {c, a},f −1 [{c, a}] = {3, 4, 5},f −1 [{b, d, e}] = {1, 2},f[{1, 2}] = {b, e}。此外,g[{b, c, d}] = {b, c, d} = g −1 [{b, c, d}],g[{c}] = {d},g −1 [{a, d, e}] = {a, c, e},且 g[0]=0=g−1[0]。最后,(g ∘ f)[{3, 4, 5}] = {d, a},(g∘f)−1[{b}]=0,且

( g f f ) [ f f 1 [ { a 一个 , d d } ] ] = ( g f f ) [ { 5 } ] = { a 一个 } .

5.74. Example. Let X = { 4 , 5 , 6 , { 4 } , { 4 , 5 , 6 } , 0 } , and define a function f : X X by setting f (4) = {4}, f (5) = 4, f ( 6 ) = 0 , f ({4}) = 5, f ( { 4 , 5 , 6 } ) = 0 , and f ( 0 ) = { 4 } . An arrow diagram for this function appears below.

5.74. 示例。设 X={4,5,6,{4},{4,5,6},0},定义函数 f : X → X,令 f(4) = {4},f(5) = 4,f(6)=0,f({4}) = 5,f({4,5,6})=0,f(0)={4}。该函数的箭头图如下所示。

ufig5_19.jpg

The set X is interesting because some of the members of X (namely {4} and {4, 5, 6} and 0 ) are also subsets of X . Let us compute some direct images, function values, and preimages for f . First, f [ { 5 , 6 } ] = { 4 , 0 } since the arrows leading away from the inputs 5 and 6 lead to 4 and 0 . Similarly, f [ { 4 , { 4 } , 0 } ] = { { 4 } , 5 } . Computing some function values now, note that f (5) = 4, f (4) = {4}, f ({4}) = 5, f ( { 4 , 5 , 6 } ) = 0 , and f ( 0 ) = { 4 } . In contrast, if we switch to square brackets, we compute the direct images f [{4}] = {{4}}, f [ { 4 , 5 , 6 } ] = { { 4 } , 4 , 0 } , and f [ 0 ] = 0 . Turning to preimages, we find that f −1 [{4, 6}] = {5}, f −1 [{4, 5, 6}] = {5, {4}}, f 1 [ { { 4 , 5 , 6 } } ] = 0 , f 1 [ 0 ] = 0 , and f 1 [ { 0 } ] = { 6 , { 4 , 5 , 6 } } . Finally, f −1 [5] is not defined since 5 is not a subset of X ; on the other hand, f −1 [{5}] = {{4}}.

集合 X 很有意思,因为 X 中的一些元素(即 {4}、{4, 5, 6} 和 0)也是 X 的子集。让我们计算函数 f 的一些直接像、函数值和原像。首先,f[{5,6}]={4,0},因为从输入 5 和 6 出发的箭头分别指向 4 和 0。类似地,f[{4,{4},0}]={{4},5}。现在计算一些函数值,注意到 f(5) = 4,f(4) = {4},f({4}) = 5,f({4,5,6})=0,以及 f(0)={4}。相反,如果我们换用方括号,则直接像为 f[{4}] = {{4}},f[{4,5,6}]={{4},4,0},以及 f[0]=0。转向原像,我们发现 f −1 [{4, 6}] = {5},f −1 [{4, 5, 6}] = {5, {4}},f−1[{{4,5,6}}]=0,f−1[0]=0,且 f−1[{0}]={6,{4,5,6}}。最后,由于 5 不是 X 的子集,因此 f −1 [5] 未定义;另一方面,f −1 [{5}] = {{4}}。

5.75. Example. Define f : [ 2 π , 4 π ] R by letting f ( x ) = cos x for all x ∈ [ − 2 π , 4 π ]. The graph of f is shown below.

5.75. 示例。定义函数 f:[−2π,4π]→R,令 f(x) = cos x,其中 x ∈ [ − 2π, 4π]。函数 f 的图像如下所示。

ufig5_20.jpg

We have, for example, f [[0, π /2]] = [0, 1], f [ R > 0 ] = [ 1 , 1 ] , f −1 [{1}] = { − 2 π , 0, 2 π , 4 π }, f 1 [ R < 0 ] = ( 3 π / 2 , π / 2 ) ( π / 2 , 3 π / 2 ) ( 5 π / 2 , 7 π / 2 ) , f −1 [ f [{ π }]] = f −1 [{ − 1}] = { − π , π , 3 π }, and f 1 [ ( 1 , 2 ] ] = 0 . Looking at some restrictions of f , we have ( f | [ 0 , π ] ) 1 [ R < 0 ] = ( π / 2 , π ] , ( f | R > 0 ) 1 [ [ 1 , ) ] = { 2 π , 4 π } , and ( f | Z ) 1 [ Z ] = { 0 } .

例如,我们有 f[[0, π/2]] = [0, 1],f[R>0]=[−1,1],f −1 [{1}] = { − 2π, 0, 2π, 4π}, f−1[R<0]=(−3π/2,−π/2)∪(π/2,3π/2)∪(5π/2,7π/2),f −1 [f[{π}]] = f −1 [{ − 1}] = { − π, π, 3π},并且 f−1[(1,2]]=0。在 f 的某些限制下,我们有 (f|[0,π])−1[R<0]=(π/2,π], (f|R>0)−1[[1,∞)]={2π,4π},且 (f|Z)−1[Z]={0}。

Properties of Direct Images and Preimages

直接像和原像的性质

The Theorem on Images (for relations) specializes to give theorems on direct images and preimages of sets under functions. As we see below, the preimage construction has some especially nice additional properties.

关于像的定理(针对关系)专门用于给出函数作用下集合的直接像和原像的定理。正如我们将在下文看到的,原像构造具有一些特别优美的附加性质。

5.76. Theorem on Direct Images. Let f : X Y be a fixed function, let I be a fixed nonempty index set, and let A , C , A i (for each i I ) be subsets of X . Then:

5.76. 直接像定理。设 f : X → Y 为一个固定的函数,I 为一个固定的非空索引集,A, C, A i (对于每个 i ∈ I) 为 X 的子集。则:

(a) Empty Image: f [ 0 ] = 0 .

(a)空像:f[0]=0。

(b) Codomain Property: f [ A ] ⊆ Y .

(b)值域性质:f[A] ⊆ Y。

(c) Monotonicity: If A C , then f [ A ] ⊆ f [ C ].

(c)单调性:如果 A ⊆ C,则 f[A] ⊆ f[C]。

(d) Image of Union: f [ A C ] = f [ A ] f [ C ] and f [ i I A i ] = i I f [ A i ] .

(d)并集的图像:f[A∪C]=f[A]∪f[C] 且 f[⋃i∈IAi]=⋃i∈If[Ai]。

(e) Image of Intersection: f [ A C ] f [ A ] f [ C ] and f [ i I A i ] i I f [ A i ] . Equality is not always true.

(e) 交集的图像:f[A∩C]⊆f[A]∩f[C] 且 f[⋂i∈IAi]⊆⋂i∈If[Ai]。等式并非总是成立。

(f) Image of Set Difference: f [ A ] − f [ C ] ⊆ f [ A C ]. Equality is not always true.

(f)集合差的像:f[A] − f[C] ⊆ f[A − C]。等式并非总是成立。

(g) Image Under a Composition: For all g : Y Z , ( g f )[ A ] = g [ f [ A ]].

(g)合成下的图像:对于所有 g : Y → Z,(g ∘ f)[A] = g[f[A]]。

Proof. This theorem follows immediately from the corresponding theorem for relations, and the fact that f [ A ] = G [ A ] where G is the graph of f . To illustrate the definition of direct image using function notation, we prove part (f). Fix f : X Y and subsets A , C X . Fix an object y f [ A ] − f [ C ]; prove y f [ A C ]. We know y f [ A ] and y f [ C ] . On one hand, y f [ A ] means there is an x A with y = f ( x ). This x cannot belong to C , since otherwise y f [ C ], which is not true. So x A C . Since y = f ( x ) with x A C , we have y f [ A C ].

证明。该定理可直接由关系的相应定理以及 f[A] = G[A] 得出,其中 G 是函数 f 的图像。为了用函数符号说明正像的定义,我们证明 (f) 部分。固定 f : X → Y 和子集 A, C ⊆ X。固定一个对象 y ∈ f[A] − f[C];证明 y ∈ f[A − C]。我们已知 y ∈ f[A] 且 y∉f[C]。一方面,y ∈ f[A] 意味着存在 x ∈ A 使得 y = f(x)。这个 x 不可能属于 C,否则 y ∈ f[C],这显然不成立。所以 x ∈ A − C。由于对于 x ∈ A − C,y = f(x),因此 y ∈ f[A − C]。

To see that equality is not always true in (f), we give an example. Define f : {1, 2} → {3, 4} by f (1) = 3 = f (2). Take A = {1} and C = {2}. Then f [ A ] = f [ C ] = {3}, so f [ A ] f [ C ] = 0 . On the other hand, A C = {1}, so f [ A C ] = {3}. We see that 0 { 3 } , but equality does not hold.

为了说明等式在 (f) 中并非总是成立,我们举个例子。定义函数 f : {1, 2} → {3, 4},且 f(1) = 3 = f(2)。取 A = {1},C = {2}。则 f[A] = f[C] = {3},所以 f[A] − f[C] = 0。另一方面,A − C = {1},所以 f[A − C] = {3}。我们看到 0⊆{3},但等式不成立。

Now we consider preimages. Some parts of the next theorem follow by applying the Theorem on Images to the relation G −1 (the inverse of the graph of f ), but here some key additional properties turn out to be true.

现在我们考虑原像。下一个定理的部分内容可以通过将像定理应用于关系 G −1 (函数 f 的图像的逆函数)来得出,但在这里,一些关键的附加性质也成立。

5.77. Theorem on Preimages. Let f : X Y be a fixed function, let I be a fixed nonempty index set, and let B , D , B i (for each i I ) be subsets of Y . Then:

5.77. 原像定理。设 f : X → Y 为一个固定的函数,I 为一个固定的非空索引集,B, D, B i (对于每个 i ∈ I) 为 Y 的子集。则:

(a) Empty Preimage: f 1 [ 0 ] = 0 .

(a)空原像:f−1[0]=0。

(b) Domain Property: f −1 [ B ] ⊆ X .

(b)域属性:f −1 [B] ⊆ X。

(c) Preimage of Codomain: f −1 [ Y ] = X .

(c)余域的原像:f −1 [Y] = X。

(d) Monotonicity: If B D , then f −1 [ B ] ⊆ f −1 [ D ].

(d)单调性:如果 B ⊆ D,则 f −1 [B] ⊆ f −1 [D]。

(e) Preimage of Union: f 1 [ B D ] = f 1 [ B ] f 1 [ D ] and f 1 [ i I B i ] = i I f 1 [ B i ] .

(e)并集原像:f−1[B∪D]=f−1[B]∪f−1[D] 且 f−1[⋃i∈IBi]=⋃i∈If−1[Bi]。

(f) Preimage of Intersection: f 1 [ B D ] = f 1 [ B ] f 1 [ D ] and f 1 [ i I B i ] = i I f 1 [ B i ] .

(f)交集原像:f−1[B∩D]=f−1[B]∩f−1[D] 且 f−1[⋂i∈IBi]=⋂i∈If−1[Bi]。

(g) Preimage of Set Difference: f −1 [ B ] − f −1 [ D ] = f −1 [ B D ].

(g)集合差异的原像:f −1 [B] − f −1 [D] = f −1 [B − D].

(h) Image of Preimage: f [ f −1 [ B ]] ⊆ B . Equality is not always true.

(h)原像的像:f[f −1 [B]] ⊆ B。等式并不总是成立。

(i) Preimage of Image: For all A X , A f −1 [ f [ A ]]. Equality is not always true.

(i)像的原像:对于所有 A ⊆ X,A ⊆ f −1 [f[A]]。等式并不总是成立。

(j) Preimage Under a Composition: For all h : W X , ( f h ) −1 [ B ] = h −1 [ f −1 [ B ]].

(j)复合下的原像:对于所有 h : W → X,(f ∘ h) −1 [B] = h −1 [f −1 [B])。

We prove some parts of this theorem to illustrate the definition of preimages in terms of function notation.

我们证明该定理的某些部分,以用函数符号说明原像的定义。

5.78. Proof of Preimage Property for Intersections. With the notation of the theorem, let us prove that f 1 [ i I B i ] = i I f 1 [ B i ] via a chain proof. The sets appearing on the left and right sides of this identity are both subsets of the domain X . So, to prove they are equal, we can fix an arbitrary object x in X and show x is in the left set iff x is in the right set. For fixed x X , note x f 1 [ i I B i ] iff f ( x ) i I B i iff for all i I , f ( x ) ∈ B i iff for all i I , x f −1 [ B i ] iff x i I f 1 [ B i ] . The property f 1 [ B D ] = f 1 [ B ] f 1 [ D ] is a special case of this result (let the index set I have two elements).

5.78. 交集原像性质的证明。沿用定理的符号,我们用链式证明法证明 f−1[⋂i∈IBi]=⋂i∈If−1[Bi]。该等式左右两边的集合都是定义域 X 的子集。因此,为了证明它们相等,我们可以固定 X 中的任意对象 x,并证明 x 属于左侧集合当且仅当 x 属于右侧集合。对于固定的 x ∈ X,注意到 x∈f−1[⋂i∈IBi] 当且仅当 f(x)∈⋂i∈IBi 当且仅当对于所有 i ∈ I,f(x) ∈ B i 当且仅当对于所有 i ∈ I,x ∈ f −1 [B i ] 当且仅当 x∈⋂i∈If−1[Bi]。性质 f−1[B∩D]=f−1[B]∩f−1[D] 是该结果的一个特例(设索引集 I 有两个元素)。

5.79. Proof of Image of Preimage Property. Fix B Y ; we must prove f [ f −1 [ B ]] ⊆ B . Fix y ; assume y f [ f −1 [ B ]]; prove y B . By definition of direct image, there exists x f −1 [ B ] with y = f ( x ). By definition of preimage, x X and f ( x ) ∈ B . Since y = f ( x ), we conclude that y B .

5.79. 原像性质的证明。固定 B ⊆ Y;我们必须证明 f[f −1 [B]] ⊆ B。固定 y;假设 y ∈ f[f −1 [B]];证明 y ∈ B。根据直接像的定义,存在 x ∈ f −1 [B] 使得 y = f(x)。根据原像的定义,x ∈ X 且 f(x) ∈ B。由于 y = f(x),我们得出 y ∈ B。

Here is an example to show that f [ f −1 [ B ]] = B is not always true. Define f : {1, 2, 3} → {4, 5, 6} by letting f (1) = f (2) = f (3) = 5. Choose B = {4, 5}. Then f −1 [ B ] = {1, 2, 3}, so f [ f −1 [ B ]] = {5}. Note that {5} is a subset of B , as promised by the theorem, but this subset is unequal to B .

以下示例表明 f[f −1 [B]] = B 并非总是成立。定义函数 f : {1, 2, 3} → {4, 5, 6},令 f(1) = f(2) = f(3) = 5。选择 B = {4, 5}。则 f −1 [B] = {1, 2, 3},因此 f[f −1 [B]] = {5}。注意,{5} 是 B 的一个子集,正如定理所述,但这个子集并不等于 B。

5.80. Proof of Preimage of Image Property. Fix A X ; we must prove A f −1 [ f [ A ]]. Fix x ; assume x A ; prove x f −1 [ f [ A ]]. We must prove x X and f ( x ) ∈ f [ A ]. Since x A and A X , we see that x X is true. To prove f ( x ) ∈ f [ A ], we must prove a A , f ( x ) = f ( a ) . Choose a = x , which is in A by assumption. Then f ( x ) = f ( a ) certainly holds.

5.80. 原像性质的证明。固定 A ⊆ X;我们必须证明 A ⊆ f −1 [f[A]]。固定 x;假设 x ∈ A;证明 x ∈ f −1 [f[A]]。我们必须证明 x ∈ X 且 f(x) ∈ f[A]。由于 x ∈ A 且 A ⊆ X,我们可知 x ∈ X 成立。为了证明 f(x) ∈ f[A],我们必须证明存在 a∈A,f(x)=f(a)。选择 a = x,根据假设,a ∈ A。那么 f(x) = f(a) 显然成立。

Here is an example to show that A = f −1 [ f [ A ]] is not always true. Define f : {1, 2, 3} → {4, 5, 6} by letting f (1) = f (2) = f (3) = 5. Choose A = {3}. Then f [ A ] = {5}, so f −1 [ f [ A ]] = {1, 2, 3}, which contains A but is unequal to A .

以下示例表明 A = f −1 [f[A]] 并非总是成立。定义函数 f : {1, 2, 3} → {4, 5, 6},令 f(1) = f(2) = f(3) = 5。选择 A = {3}。则 f[A] = {5},因此 f −1 [f[A]] = {1, 2, 3},它包含 A 但不等于 A。

5.81. Example. Define a function f : R R by f ( x ) = x 2 for all x R . The graph of f is shown here.

5.81. 示例。定义函数 f:R→R,使得对于所有 x∈R,f(x) = x 2 。f 的图像如下所示。

ufig5_21.jpg

Let A = [1, 2], considered as a subset of the domain of f . We compute f [ A ] = [1, 4], and f 1 [ f [ A ] ] = f 1 [ [ 1 , 4 ] ] = [ 2 , 1 ] [ 1 , 2 ] . Thus A is a proper subset of f −1 [ f [ A ]]. Next let B = [ − 3, 4], considered as a subset of the codomain of f . We compute f −1 [ B ] = [ − 2, 2], and f [ f −1 [ B ]] = f [[ − 2, 2]] = [0, 4]. Thus f [ f −1 [ B ]] is a proper subset of B .

设 A = [1, 2],视为 f 的定义域的子集。我们计算 f[A] = [1, 4],且 f−1[f[A]] = f−1[[1,4]] = [−2,−1]∪[1,2]。因此,A 是 f −1 [f[A]] 的真子集。接下来,设 B = [ − 3, 4],视为 f 的值域的子集。我们计算 f −1 [B] = [ − 2, 2],且 f[f −1 [B]] = f[[ − 2, 2]] = [0, 4]。因此,f[f −1 [B]] 是 B 的真子集。

Section Summary

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章节概要

  1. Direct Images and Preimages. Let f : X Y be a function. For all subsets A of X and all objects y , y f [ A ] iff x A , y = f ( x ) . For all subsets B of Y and all objects x , x f −1 [ B ] iff x X and f ( x ) ∈ B . If G is the graph of f , we have f [ A ] = G [ A ] and f −1 [ B ] = G −1 [ B ]. The preimage notation f −1 [ B ] can be used for all functions, not just functions that have inverses.
    原像和原像。设 f : X → Y 为一个函数。对于 X 的所有子集 A 和所有对象 y,y ∈ f[A] 当且仅当存在 x∈A,y=f(x)。对于 Y 的所有子集 B 和所有对象 x,x ∈ f −1 [B] 当且仅当 x ∈ X 且 f(x) ∈ B。如果 G 是 f 的图像,则有 f[A] = G[A] 且 f −1 [B] = G −1 [B]。原像符号 f −1 [B] 可以用于所有函数,而不仅仅是具有逆函数的函数。
  2. Properties of Preimages. The preimage construction preserves set concepts such as union, intersection, set difference, set inclusion, and the empty set. For instance, the preimage of an arbitrary union of sets is the union of the preimages of these sets.
    原像的性质。原像构造保留了集合的概念,例如并集、交集、差集、包含关系和空集。例如,任意集合并集的原像是这些集合原像的并集。
  3. Properties of Direct Images. The direct image construction preserves set inclusions, arbitrary unions, and the empty set. But the set inclusions f [ A C ] f [ A ] f [ C ] , f [ A ] − f [ C ] ⊆ f [ A C ], f [ f −1 [ B ]] ⊆ B , and A f −1 [ f [ A ]] are not equalities in general.
    直接像的性质。直接像构造保持集合包含关系、任意并集和空集不变。但是,集合包含关系 f[A∩C]⊆f[A]∩f[C]、f[A] − f[C] ⊆ f[A − C]、f[f −1 [B]] ⊆ B 和 A ⊆ f −1 [f[A]] 一般而言并非等式。

Exercises

练习

  1. Let X = {1, 2, 3, 4, 5, 6}, and let f : X X have graph {(1, 4), (2, 3), (3, 4), (4, 1), (5, 6), (6, 4)}. Draw an arrow diagram for f . Use this to compute f [{2, 3, 4}], f −1 [{2, 3, 4}], f [{2, 5}], f −1 [{2, 5}], f [ f −1 [{4, 5, 6}]], f −1 [ f [{4, 5, 6}]], ( f f )[{1, 3, 5}] and ( f f ) −1 [{1, 3, 5}].
    设 X = {1, 2, 3, 4, 5, 6},函数 f : X → X 的图像为 {(1, 4), (2, 3), (3, 4), (4, 1), (5, 6), (6, 4)}。画出函数 f 的箭头图。使用此方法计算 f[{2, 3, 4}]、f −1 [{2, 3, 4}]、f[{2, 5}]、f −1 [{2, 5}]、f[f −1 [{4, 5, 6}]]、f −1 [f[{4, 5, 6}]]、 (f ∘ f)[{1, 3, 5}] 和 (f ∘ f) −1 [{1, 3, 5}]。
  2. For each f : R R , sketch the graph of f . Use this to compute the direct image f [( − 1, 1)], the preimage f −1 [( − 1, 1)], and f −1 [{ − 4}]. (a) f ( x ) = 3 x + 5. (b) f ( x ) = 4 x 2 . (c) f ( x ) = | x | for x ≠ 0, f (0) = −4. (d) f = χ R Z (a characteristic function). (e) f ( x ) = sin x . (f) f ( x ) = ln| x | for x ≠ 0, f (0) = 0.
    对于每个函数 f:R→R,绘制 f 的图像。利用该图像计算正像 f[( − 1, 1)]、原像 f[( − 1, 1)] 和 f[− 4]。 (a) f(x) = 3x + 5 (b) f(x) = 4x (c) f(x) = |x|,其中 x ≠ 0,f(0) = −4 (d) f=χR−Z(特征函数) (e) f(x) = sin x (f) f(x) = ln|x|当 x ≠ 0 时,f(0) = 0。
  3. For each real-valued function below, describe the preimage sets f −1 [{ c }] for each c R (these sets are called level sets of the function). Sketch the level sets for c = −1, 0, 1, 4 in each case. (a) f : R 2 R given by f ( x , y ) = y . (b) f : R 2 R given by f ( x , y ) = x + y . (c) f : R 2 R given by f ( x , y ) = 4. (d) f : R 2 R given by f ( x , y ) = x 2 + y 2 . (e) f : R 2 R given by f ( x , y ) = x 2 + y 2 . (f) f : R 2 R given by f ( x , y ) = 1 + x 2 + y 2 .
    对于下面的每个实值函数,描述每个 c∈R 的原像集 f −1 [{c}](这些集合被称为函数的水平集)。分别绘制 c = −1, 0, 1, 4 时各函数的水平集。 (a) f:R2→R 由 f(x, y) = y 给出。 (b) f:R2→R 由 f(x, y) = x + y 给出。 (c) f:R2→R 由 f(x, y) = 4 给出。 (d) f:R2→R 由 f(x, y) = x 2 + y 2 (e) f:R2→R 由 f(x,y)=x2+y2 给出。 (f) f:R2→R 由 f(x,y)=1+x2+y2 给出。
  4. Let X = { 1 , 2 , 3 } P ( { 1 , 2 } ) . Define f : X X by letting f (1) = {1}, f ( 2 ) = 0 , f (3) = 2, and f ( A ) = {1, 2} − A for all A ⊆{1, 2}. (a) Draw an arrow diagram for f . (b) Compute (if defined) f (1), f ({1}), f ({{1}}), f [1], f [{1}], and f [{{1}}]. (c) Compute (if defined) f ({1, 2}), f ({{1, 2}}), f [{1, 2}], f [{{1, 2}}], f [{1, {2}}], f [{{1}, 2}], and f [{{1}, {2}}]. (d) Compute f −1 [{2}], f −1 [{{2}}], f 1 [ 0 ] , f 1 [ { 0 } ] , f −1 [{1, {2}}], and f −1 [{{1}, 2}].
    令 X={1,2,3}∪P({1,2})。定义函数 f : X → X,令 f(1) = {1},f(2) = 0,f(3) = 2,且对于所有 A ⊆ {1, 2},f(A) = {1, 2} − A。 (a) 画出 f 的箭头图。 (b) 计算(如果已定义)f(1)、f({1})、f({{1}})、f[1]、f[{1}] 和 f[{{1}}]。 (c) 计算(如果已定义)f({1, 2})、f({{1, 2}})、f[{1, 2}]、f[{{1, 2}}]、f[{{1}, {2}}]、f[{{1}, 2}] 和 f[{{1}, {2}}]。 (d) 计算f −1 [{2}]、f −1 [{{2}}]、f−1[0]、f−1[{0}]、f −1 [{1, {2}}] 和 f −1 [{{1}, 2}]。
  5. Use the definition of direct image based on function notation to prove parts (c) and (e) of the Theorem on Direct Images.
    利用基于函数符号的直接像的定义来证明直接像定理的(c)和(e)部分。
  6. (a) Prove part (g) of the Theorem on Direct Images. (b) Prove part (j) of the Theorem on Preimages.
    (a)证明关于直接像的定理的第(g)部分。(b)证明关于原像的定理的第(j)部分。
  7. Given sets A X , let χ A : X R be the characteristic function for A . (a) Find all possible direct images χ A [ B ], as B ranges over subsets of X . Which choices of B produce each possible answer? (b) Find all possible preimages χ A 1 [ C ] , as C ranges over subsets of R . Which choices of C produce each possible answer?
    给定集合 A ⊆ X,令 χA:X→R 为 A 的特征函数。 (a) 求所有可能的直接像 χ A [B],其中 B 取遍 X 的子集。B 的哪些取值对应每个可能的答案? (b) 求所有可能的原像 χA−1[C],其中 C 取遍 R 的子集。C 的哪些取值对应每个可能的答案?
  8. Suppose f : X Y and S X . Prove: for all B Y , f | S 1 [ B ] = f 1 [ B ] S .
    假设 f : X → Y 且 S ⊆ X。证明:对于所有 B ⊆ Y,f|S−1[B]=f−1[B]∩S。
  9. Suppose f : X Y , A X , and B Y . (a) Prove: B f [ A ] = f [ f 1 [ B ] A ] . (b) Prove: f [ f 1 [ B ] A ] B f [ A ] . (c) Prove: f 1 [ B ] A f 1 [ B f [ A ] ] . (d) Prove: f 1 [ B ] A f 1 [ B f [ A ] ] .
    假设 f : X → Y,A ⊆ X,且 B ⊆ Y。 (a) 证明:B∩f[A]=f[f−1[B]∩A]. (b) 证明:f[f−1[B]∪A]⊆B∪f[A]. (c) 证明:f−1[B]∩A⊆f−1[B∩f[A]]. (d) 证明:f−1[B]∪A⊆f−1[B∪f[A]].
  10. Give examples to show that the set inclusions in parts (b), (c), and (d) of the previous exercise are not always equalities.
    举例说明上一练习中的 (b)、(c) 和 (d) 部分中的集合包含关系并不总是等式。
  11. (a) Disprove: for all relations R and all sets A , A R −1 [ R [ A ]]. (b) Prove or disprove: for all relations R and all sets B , R [ R −1 [ B ]] ⊆ B .
    (a)反证:对于所有关系 R 和所有集合 A,A ⊆ R −1 [R[A]]。(b)证明或反证:对于所有关系 R 和所有集合 B,R[R −1 [B]] ⊆ B。
  12. Prove: for all functions f : X Y and all A X , f [ f −1 [ f [ A ]]] = f [ A ].
    证明:对于所有函数 f : X → Y 和所有 A ⊆ X,f[f −1 [f[A]]] = f[A]。
  13. Prove: for all functions f : X Y and all B Y , f −1 [ f [ f −1 [ B ]]] = f −1 [ B ].
    证明:对于所有函数 f : X → Y 和所有 B ⊆ Y,f −1 [f[f −1 [B]]] = f −1 [B].
  14. Do the statements in the previous two exercises remain true if we only assume that f is a relation from X to Y ?
    如果我们只假设 f 是从 X 到 Y 的关系,那么前两个练习中的陈述是否仍然成立?
  15. (a) Find (with proof) a property of a function f : X Y that guarantees f [ A C ] = f [ A ] f [ C ] for all A , C X . (b) Find (with proof) a property of a function f : X Y that guarantees f [ A ] − f [ C ] = f [ A C ] for all A , C X .
    (a) 求函数 f : X → Y 的一个性质(并证明),该性质保证对于所有 A, C ⊆ X,都有 f[A∩C]=f[A]∩f[C]。 (b) 求函数 f : X → Y 的一个性质(并证明),该性质保证对于所有 A, C ⊆ X,都有 f[A] − f[C] = f[A − C]。
  16. (a) Find (with proof) a property of a function f : X Y that guarantees f −1 [ f [ A ]] = A for all A X . (b) Find (with proof) a property of a function f : X Y that guarantees f [ f −1 [ B ]] = B for all B Y .
    (a)找出函数 f : X → Y 的一个性质(并证明),该性质保证对于所有 A ⊆ X,都有 f −1 [f[A]] = A。(b)找出函数 f : X → Y 的一个性质(并证明),该性质保证对于所有 B ⊆ Y,都有 f[f −1 [B]] = B。

5.8 Injective, Surjective, and Bijective Functions

5.8 单射函数、满射函数和双射函数

This section studies three special kinds of functions, called injections, surjections, and bijections. Intuitively, a function is injective (one-to-one) iff distinct inputs to the function always map to distinct outputs. A function is surjective (onto) iff every potential output in the codomain of the function is actually attained as the value of the function at some input. A function is bijective iff it is both one-to-one and onto.

本节研究三种特殊类型的函数,分别称为单射、满射和双射。直观地说,一个函数是单射(一对一)的,当且仅当该函数的每个不同输入都映射到不同的输出。一个函数是满射(满射)的,当且仅当该函数值域中的每个潜在输出都对应于某个输入对应的函数值。一个函数是双射的,当且仅当它既是单射又是满射。

Definition of Injections, Surjections, and Bijections

单射、单射和双射的定义

We introduce the concepts of injective, surjective, and bijective functions using arrow diagrams. Informally, a function f : X Y is injective (or one-to-one ) iff every point in Y has at most one arrow entering it. This means that every potential output occurs as the value of the function for either zero or one inputs. The function f is surjective (or onto ) iff every point in Y has at least one arrow entering it. This means that every potential output in the codomain really does get used as an actual output assigned to one of the inputs in the domain. Finally, the function f is bijective (or a one-to-one correspondence between X and Y ) iff every point in Y has exactly one arrow entering it. This means that for every output in Y , there exists exactly one input in X mapping to that output. Compare this to the requirement in the definition of a function that every input in X has exactly one arrow leaving it. The relation between these conditions is explained in the next section when we discuss inverse functions.

我们使用箭头图来介绍单射、满射和双射函数的概念。通俗地说,函数 f : X → Y 是单射(或一一对应)当且仅当 Y 中的每个点至多只有一个箭头指向它。这意味着每个潜在的输出值都对应于零个或一个输入值。函数 f 是满射(或满射)当且仅当 Y 中的每个点至少有一个箭头指向它。这意味着值域中的每个潜在输出值都确实被用作实际的输出值,并分配给定义域中的一个输入值。最后,函数 f 是双射(或 X 和 Y 之间存在一一对应关系)当且仅当 Y 中的每个点恰好有一个箭头指向它。这意味着对于 Y 中的每个输出值,X 中恰好存在一个输入值映射到该输出值。这与函数定义中要求 X 中的每个输入值恰好有一个箭头指向它的条件相对应。这些条件之间的关系将在下一节讨论反函数时进行解释。

5.82 Example. Consider the following four arrow diagrams.

5.82 例。考虑以下四个箭头图。

ufig5_22.jpg

First of all, each of the arrow diagrams does represent a function, since each input x in the domain X has exactly one arrow leaving it, and all arrows go from X to Y . Second, f and h are injective functions, because every output in Y has at most one arrow coming in. The function g is not injective because the output c has two arrows coming in: g (2) = c = g (3). Similarly, k is not one-to-one because k (2) = b = k (3). Third, f and g are surjective functions, because every output in Y has at least one arrow coming in. The function h is not surjective because c is not hit by an arrow; similarly, k is not onto. Fourth, f is a bijection because each output in Y has exactly one arrow coming in. The other three functions are not bijective.

首先,每个箭头图都代表一个函数,因为定义域 X 中的每个输入 x 都恰好有一个箭头从它出发,并且所有箭头都从 X 指向 Y。其次,f 和 h 是单射函数,因为 Y 中的每个输出至多只有一个箭头入射。函数 g 不是单射函数,因为输出 c 有两个箭头入射:g(2) = c = g(3)。类似地,k 不是单射函数,因为 k(2) = b = k(3)。第三,f 和 g 是满射函数,因为 Y 中的每个输出至少有一个箭头入射。函数 h 不是满射函数,因为 c 没有被箭头射中;类似地,k 也不是满射。第四,f 是双射函数,因为 Y 中的每个输出恰好有一个箭头入射。其他三个函数都不是双射函数。

Next we give the formal definitions of injections, surjections, and bijections. At this time, it might be helpful to review §1.6 to recall how the phrases “at most one” and “exactly one” can be encoded using logical symbols.

接下来,我们给出单射、满射和双射的正式定义。此时,回顾一下第1.6节可能会有所帮助,以便回忆起“至多一个”和“恰好一个”这两个短语如何用逻辑符号编码。

5.83. Definition: Injective (One-to-One) Function. Given a function f : X Y , f is injective iff w X , x X , f ( w ) = f ( x ) w = x . When this holds, we also say f is one-to-one or f is an injection .

5.83. 定义:单射(一对一)函数。给定函数 f : X → Y,f 是单射当且仅当对于所有 w∈X,所有 x∈X,f(w)=f(x)⇒w=x。当此条件成立时,我们也称 f 是一对一的或 f 是单射。

The definition of injective function does not explicitly mention members of Y , but these outputs appear implicitly in the expression f ( w ) = f ( x ). The implication in the definition says that if two inputs w and x ever map to the same output under f , then the two inputs must in fact be the same. The contrapositive of the definition may also have more intuitive appeal, though it may be more difficult to use in proofs: f is one-to-one iff for all w , x X , w x implies f ( w ) ≠ f ( x ).

单射函数的定义并未明确提及集合 Y 的成员,但这些输出隐含在表达式 f(w) = f(x) 中。定义中的蕴含关系表明,如果两个输入 w 和 x 在函数 f 的作用下映射到同一个输出,那么这两个输入实际上必然相同。该定义的逆否命题可能更具直观性,尽管在证明中可能更难使用:f 是单射当且仅当对于所有 w, x ∈ X,w ≠ x 蕴含 f(w) ≠ f(x)。

5.84. Definition: Surjective (Onto) Function. Given a function f : X Y , f is surjective iff y Y , x X , y = f ( x ) .

5.84. 定义:满射(到顶)函数。给定一个函数 f : X → Y,f 是满射当且仅当 ∀y∈Y,∃x∈X,y=f(x)。

When this holds, we also say f is onto or f is a surjection .

当此条件成立时,我们也说 fis 是满射或 fis 是满射。

5.85. Definition: Bijective Function. Given a function f : X Y , f is bijective iff f is injective and f is surjective .

5.85. 定义:双射函数。给定一个函数 f : X → Y,f 是双射当且仅当 f 是单射且 f 是满射。

Equivalently, f is a bijection iff y Y , ! x X , y = f ( x ) .

等价地,f 是双射当且仅当 ∀y∈Y,∃!x∈X,y=f(x)。

5.86. Example: Denials of the Definitions.

5.86. 示例:对定义的否定。

What does it mean to say that a function f : X Y is: (a) not injective; (b) not surjective; (c) not bijective? Solution. Using the denial rules, we find:

函数 f : X → Y 满足以下条件分别意味着什么:(a) 不是单射;(b) 不是满射;(c) 不是双射? 解答:利用否定规则,我们发现:

  1. f is not injective iff w X , x X , f ( w ) = f ( x ) w x .
    f 不是单射当且仅当 ∃w∈X,∃x∈X,f(w)=f(x)∧w≠x。
  2. f is not surjective iff y Y , x X , y f ( x ) .
    f 不是满射当且仅当 ∃y∈Y,∀x∈X,y≠f(x)。
  3. f is not bijective iff f is not injective or f is not surjective .
    fis 不是双射当且仅当 fis 不是单射或者 fis 不是满射。

Examples of Injections, Surjections, and Bijections

单射、单射和双射的例子

5.87. Example: Exponential Function. Define f : R R by f ( x ) = e x for all x R . The graph of f is shown here.

5.87. 示例:指数函数。定义函数 f:R→R,使得对于所有 x∈R,f(x) = e x 。函数 f 的图像如下所示。

ufig5_23.jpg

Each vertical line x = x 0 (for x 0 R ) intersects the graph of f in exactly one point; for this reason, f is a function. On the other hand, each horizontal line y = y 0 (for y 0 R ) intersects the graph of f in at most one point. This means that every output y 0 has the form f ( x 0 ) for at most one input x 0 . So, f is injective. (Testing injectivity in this way is called the Horizontal Line Test .) But, for certain choices of y 0 R , the horizontal line y = y 0 does not intersect the graph of f anywhere. For instance, this holds for y 0 = 0, y 0 = −1, or any negative value of y 0 . This means that f is not surjective, and hence not bijective. On the other hand, consider the function f 1 : R R > 0 , given by f 1 ( x ) = e x for all x R . The function f 1 is not the same as f , because the two functions have different codomains. We see that f 1 is injective, surjective, and bijective, since every y 0 in the new codomain R > 0 has the form y 0 = f ( x 0 ) for exactly one input x 0 R , namely x 0 = ln y 0 .

每条垂直线 x = x 0 (对于 x0∈R)都与函数 f 的图像恰好相交于一点;因此,f 是一个函数。另一方面,每条水平线 y = y 0 (对于 y0∈R)都与函数 f 的图像至多相交于一点。这意味着对于至多一个输入 x 0 ,每个输出 y 0 都具有 f(x 0 ) 的形式。所以,f 是单射的。(用这种方法检验单射性称为水平线检验。)但是,对于某些 y0∈R 的取值,水平线 y = y 0 与函数 f 的图像没有任何交点。例如,当 y 0 = 0、y 0 = −1 或 y 0 的任何负值时,上述结论成立。这意味着 f 不是满射,因此也不是双射。另一方面,考虑函数 f1:R→R>0,对于所有 x∈R,有 f 1 (x) = e x 。函数 f 1 与 f 不同,因为这两个函数的值域不同。我们看到 f 1 是单射、满射和双射,因为新值域 R>0 中的每个 y 0 都有 y 0 = f(x 0 ) 的形式,其中恰好有一个输入 x0∈R,即 x 0 = ln y 0

5.88. Example: Functions from Calculus. Define relations F , G , H , K , by setting F = { ( x , x 3 ) : x R } , G = { ( x , x 2 ) : x R } , H = { ( x , sin x ) : x R } , and

5.88. 示例:微积分中的函数。定义关系 F、G、H、K,令 F={(x,x³):x∈R},G={(x,x²):x∈R},H={(x,sin⁡x):x∈R},

K K = { ( x x , y ) : x x R , tan 棕褐色 x x is defined, and 已定义,并且 y = tan 棕褐色 x x } .

The graphs of these relations are shown here:

这些关系的图表如下所示:

ufig5_24.jpg

We cannot apply the adjectives injective, surjective, or bijective to any of the graphs F , G , H , or K , because these graphs are relations, not functions. Each graph can be used to produce many different functions, depending on which domain and codomain are chosen. Consider these examples.

我们不能将单射、满射或双射这些形容词应用于图 F、G、H 或 K,因为这些图是关系,而不是函数。每个图都可以用来生成许多不同的函数,具体取决于所选择的定义域和值域。请看以下示例。

(a) f 1 : R R with graph F is injective and surjective and bijective.

(a)函数 f1:R→R,其图 F 是单射、满射和双射。

(b) f 2 : R R with graph G is not injective and not surjective and not bijective.

(b)映射 f2:R→R,图为 G,既不是单射,也不是满射,也不是双射。

(c) f 3 : R R 0 with graph G is surjective but not injective.

(c)对于图 G,f3:R→R≥0 是满射但不是单射。

(d) f 4 : R 0 R 0 with graph G is not a function, because the graph G is not a subset of R 0 × R 0 .

(d)f4:R≥0→R≥0,图为G,不是函数,因为图G不是R≥0×R≥0的子集。

(e) f 5 : R > 0 R > 0 with graph G ( R > 0 × R > 0 ) is injective and surjective and bijective. In this case, the domain is the positive x -axis, the codomain is the positive y -axis, and the graph of f 5 is the part of the parabola G in the first quadrant.

(e) 函数 f5:R>0→R>0,其图像为 G∩(R>0×R>0),且该函数是单射、满射和双射。在这种情况下,定义域为 x 轴正方向,值域为 y 轴正方向,函数 f 5 的图像是抛物线 G 在第一象限内的部分。

(f) f 6 : R R with graph H is not injective (e.g., because f 6 ( π ) = 0 = f 6 ( − π ). f 6 is not surjective (e.g., because there does not exist x 0 R with f 6 ( x 0 ) = 2). f 6 is not bijective (e.g., because it is not injective).

(f) 图 H 上的 f6:R→R 不是单射(例如,因为 f 6 (π) = 0 = f 6 ( − π)。f 6 不是满射(例如,因为不存在 x0∈R 使得 f 6 (x 0 ) = 2)。f 6 不是双射(例如,因为它不是单射)。

(g) f 7 : R [ 1 , 1 ] with graph H is surjective but not injective. Note that we obtain f 7 from f 6 by restricting the codomain.

(g) 映射 f7:R→[−1,1] 及其图 H 是满射但不是单射。注意,我们通过限制值域,从 f 6 得到 f 7

(h) f 8 : [ π / 2 , π / 2 ] R with graph H ( [ π / 2 , π / 2 ] × R ) is injective but not surjective. Note f 8 = f 6 | [− π /2, π /2] is a restriction of f 6 .

(h)函数 f8:[−π/2,π/2]→R,其图为 H∩([−π/2,π/2]×R),是单射但不是满射。注意 f 8 = f 6 | [− π /2, π /2] 是 f 6 的一个限制。

(i) f 9 : [ − π /2, π /2] → [ − 1, 1] with graph H ( [ π / 2 , π / 2 ] × R ) is injective and surjective and bijective. Note that f 9 is obtained from f 6 by restricting both the domain and the codomain; these modifications have converted a non-bijection into a bijection.

(i) 映射 f 9 : [ − π/2, π/2] → [ − 1, 1],其图为 H∩([−π/2,π/2]×R),且为单射、满射和双射。注意,f 9 是通过限制 f 6 的定义域和值域得到的;这些修改将非双射转化为双射。

(j) f 10 : [0, π /2] → [0, 1] with graph H ( [ 0 , π / 2 ] × R ) is also injective and surjective and bijective. This is a different restriction of f 6 that produces a bijection.

(j) 映射 f 10 : [0, π/2] → [0, 1],其图为 H∩([0,π/2]×R),也是单射、满射和双射。这是 f 6 的一个不同限制,它产生了一个双射。

(k) f 11 : R R with graph K is not a function, because there exist inputs x 0 in the domain R with no associated outputs. For example, for all y , ( π / 2 , y ) K , so the input π /2 has no corresponding output.

(k) 图为 K 的函数 f11:R→R 不是函数,因为在定义域 R 中存在没有对应输出的输入 x 0 。例如,对于所有 y,(π/2,y)∉K,因此输入 π/2 没有对应的输出。

(l) To fix the previous example, let A = { π / 2 + k π : k Z } be the set of real numbers x where tan x is undefined. Then f 12 : R A R with graph K is a function, and f 12 is surjective but not injective.

(l) 为了修正前面的例子,令 A={π/2+kπ:k∈Z} 为实数集 x,其中 tan x 无定义。那么,图为 K 的函数 f12:R−A→R 是一个函数,并且 f 12 是满射但不是单射。

(m) f 13 : ( π / 2 , π / 2 ) R with graph K ( ( π / 2 , π / 2 ) × R ) is a function that is injective, surjective, and bijective. The same is true of f 14 : ( π / 2 , 3 π / 2 ) R with graph K ( ( π / 2 , 3 π / 2 ) × R ) .

(m)函数 f13:(−π/2,π/2)→R,其图像为 K∩((−π/2,π/2)×R),是单射、满射和双射函数。函数 f14:(π/2,3π/2)→R,其图像为 K∩((π/2,3π/2)×R),也具有同样的性质。

These examples show that a non-surjective function f : X Y can be changed into a new, surjective function by shrinking the codomain from Y to the image f [ X ]. Similarly, we can change a non-injective function g : X Y into a new function that is injective if we are willing to shrink the domain X sufficiently. We might need to shrink the domain by a lot, however. For instance, let g : R R be the constant function given by g ( x ) = 3 for all x R . In order to manufacture an injective function from g , we need to shrink the domain all the way down to a single point! The new function is not surjective either; to obtain surjectivity, we must also shrink the codomain to the set {3}. The resulting function (mapping a one-point set to a one-point set) is not very interesting, but it is a bijection.

这些例子表明,非满射函数 f : X → Y 可以通过将值域从 Y 缩小到 f[X] 的像集来转化为新的满射函数。类似地,如果我们愿意充分缩小定义域 X,也可以将非单射函数 g : X → Y 转化为新的单射函数。然而,我们可能需要大幅缩小定义域。例如,设 g: R → R 为常数函数,对于所有 x∈ R,g(x) = 3。为了从 g 构造一个单射函数,我们需要将定义域缩小到仅剩一个点!新函数也不是满射;为了获得满射性,我们还必须将值域缩小到集合 {3}。得到的函数(将一个单点集映射到另一个单点集)虽然没什么特别之处,但它是一个双射。

Proofs of Injectivity, Surjectivity, and Bijectivity

单射性、满射性和双射性的证明

Here we consider some examples to see how to prove that a function is or is not injective, surjective, or bijective.

这里我们考虑一些例子,看看如何证明一个函数是或不是单射、满射或双射。

5.89. Example. For any set X , the identity function Id X : X X is a bijection. To prove injectivity, fix w , x X ; assume Id X ( w ) = Id X ( x ); prove w = x . Using the definition of Id X , the assumption becomes w = x , as needed. To prove surjectivity, fix y X ; we must prove x X , Id X ( x ) = y . Choose x = y ; then x X and Id X ( x ) = x = y , as needed.

5.89. 示例。对于任意集合 X,恒等函数 Id X : X → X 是一个双射。为了证明单射性,固定 w, x ∈ X;假设 Id X (w) = Id X (x);证明 w = x。利用 Id X 的定义,假设变为 w = x,满足要求。为了证明满射性,固定 y ∈ X;我们必须证明存在 x∈X,IdX(x)=y。选择 x = y;则 x ∈ X 且 Id X (x) = x = y,满足要求。

5.90. Example. Define f : ( R Z ) Z by f ( x ) = x , which is x rounded down to the nearest integer. Let us show that f is surjective but not injective. For surjectivity, fix y 0 Z . Choose x 0 = y 0 + 1/2, which does lie in the domain R Z of f . Rounding x 0 down to the nearest integer produces y 0 , so f ( x 0 ) = y 0 . Next, to prove f is not injective, choose x = 1/3 and w = 2/3, which lie in R Z . Note x w , but f ( x ) = 0 = f ( w ).

5.90. 示例。定义函数 f:(R−Z)→Z 为 f(x)=⌊x⌋,其中 x 向下取整到最接近的整数。我们来证明 f 是满射但不是单射。为了证明满射性,固定 y0∈Z。选择 x 0 = y 0 + 1/2,它位于 f 的定义域 R−Z 内。将 x 0 向下取整到最接近的整数得到 y 0 ,因此 f(x 0 ) = y 0 。接下来,为了证明 f 不是单射,选择 x = 1/3 和 w = 2/3,它们都位于 R−Z 内。注意 x ≠ w,但 f(x) = 0 = f(w)。

5.91. Example. Define g : Z 0 Z 0 by g ( n ) = 6 n . Let us show that g is injective but not surjective. For injectivity, fix n , m Z 0 and assume g ( n ) = g ( m ). We must prove n = m . The assumption means 6 n = 6 m , which is equivalent to 2 n 3 n = 2 m 3 m . By uniqueness of prime factorizations, we conclude that n = m . So g is injective. To show that g is not surjective, choose y 0 = 5. We must prove g ( x ) ≠ 5 for all x Z 0 . Consider two cases. If x = 0, then g ( x ) = 1 ≠ 5. If x > 0, then g ( x ) = 6 x ≥ 6 1 = 6 > 5, so again g ( x ) ≠ 5. So g is not surjective. (Alternatively, we can see that g ( x ) ≠ 5 by noting that g ( x ) = 2 x 3 x = 5 = 5 1 would violate the uniqueness of the prime factorization of 5.)

5.91. 示例。定义函数 g:Z≥0→Z≥0,g(n) = 6 n 。我们来证明 g 是单射但不是满射。为了证明单射性,固定 n,m∈Z≥0 并假设 g(n) = g(m)。我们需要证明 n = m。这个假设意味着 6 n = 6 m ,这等价于 2 n 3 n = 2 m 3 m 。由素因子分解的唯一性可知 n = m。因此 g 是单射。为了证明 g 不是满射,取 y 0 = 5。我们需要证明对于所有 x∈Z≥0,g(x) ≠ 5。考虑以下两种情况。如果 x = 0,则 g(x) = 1 ≠ 5。如果 x > 0,则 g(x) = 6 ≥ 6 = 6 > 5,所以 g(x) ≠ 5。因此 g 不是满射。(或者,我们可以通过注意到 g(x) = 2 ≥ 3 = 5 = 5 会破坏 5 的质因数分解的唯一性来看出 g(x) ≠ 5。)

5.92. Example. Define h : R { 3 } R { 2 } by h ( x ) = 2 x + 1 x 3 for all x ≠ 3. Let us prove that h is a bijection. First we show h is injective. Fix x , z R { 3 } ; assume h ( x ) = h ( z ); prove x = z . Our assumption means 2 x + 1 x 3 = 2 z + 1 z 3 . Using algebra to rewrite these fractions, we deduce

5.92. 例。定义 h:R−{3}→R−{2},使得对于所有 x ≠ 3,h(x)=2x+1x−3。我们来证明 h 是双射。首先,我们证明 h 是单射。固定 x,z∈R−{3};假设 h(x) = h(z);证明 x = z。我们的假设意味着 2x+1x−3=2z+1z−3。利用代数方法重写这些分数,我们推导出

2 + 7 x x 3 = 2 + 7 z z 3 .

Subtracting 2 from both sides gives 7 x 3 = 7 z 3 . Dividing by 7 gives 1 x 3 = 1 z 3 . Multiplying by the nonzero numbers z − 3 and x − 3 yields z − 3 = x − 3. Finally, adding 3 to both sides gives z = x , so x = z is indeed true.

两边同时减去 2,得到 7x−3=7z−3。两边同时除以 7,得到 1x−3=1z−3。两边同时乘以非零数 z − 3 和 x − 3,得到 z − 3 = x − 3。最后,两边同时加上 3,得到 z = x,所以 x = z 成立。

Next, we show that h is surjective. Fix y 0 R { 2 } ; we must find x R { 3 } such that h ( x ) = y 0 , or equivalently 2 + 7 x 3 = y 0 . To see which x to choose, we do a scratch calculation in which we solve for x in terms of y 0 . We get 7 x 3 = y 0 2 , then x 3 7 = 1 y 0 2 , then x 3 = 7 y 0 2 , then x = 3 + 7 y 0 2 . Returning to the official proof, we now choose x = 3 + 7 y 0 2 . Observe that this number is defined, since y 0 ≠ 2. Furthermore, 7 y 0 2 is nonzero, and so x is unequal to 3. This proves that x does belong to the domain R { 3 } of h . Finally, we conclude the proof by computing

接下来,我们证明 h 是满射。固定 y0∈R−{2};我们必须找到 x∈R−{3} 使得 h(x) = y 0 ,或者等价地,2+7x−3=y0。为了确定选择哪个 x,我们进行一个临时计算,用 y 0 表示 x。我们得到 7x−3=y0−2,然后 x−37=1y0−2,然后 x−3=7y0−2,然后 x=3+7y0−2。回到正式证明,我们现在选择 x=3+7y0−2。注意到这个数是有定义的,因为 y 0 ≠ 2。此外,7y0−2 非零,因此 x 不等于 3。这证明 x 属于 h 的定义域 R−{3}。最后,我们通过计算完成证明。

h h ( x x ) = 2 ( 3 + 7 / ( y 0 2 ) ) + 1 ( 3 + 7 / ( y 0 2 ) ) 3 = 7 + 14 / ( y 0 2 ) 7 / ( y 0 2 ) = ( 7 ( y 0 2 ) + 14 y 0 2 ) ( y 0 2 7 ) = y 0 .

Properties of Injections, Surjections, and Bijections

单射、单射和双射的性质

We conclude this section with a fundamental theorem showing how composition of functions is related to the properties of being injective, surjective, or bijective.

本节最后给出一个基本定理,该定理表明函数复合与单射、满射或双射性质之间的关系。

5.93. Theorem on Injections, Surjections, and Bijections.

5.93.关于单射、满射和双射的定理。

For all functions f : X Y and g : Y Z :

对于所有函数 f : X → Y 和 g : Y → Z:

(a) If f is injective and g is injective, then g f : X Z is injective.

(a)如果 f 是单射函数,g 也是单射函数,那么 g ∘ f : X → Z 也是单射函数。

(b) If f is surjective and g is surjective, then g f : X Z is surjective.

(b)如果 f 是满射且 g 是满射,则 g ∘ f : X → Z 是满射。

(c) If f is bijective and g is bijective, then g f : X Z is bijective.

(c)如果 f 是双射且 g 是双射,则 g ∘ f : X → Z 是双射。

(d) If g f is injective, then f is injective (but g may or may not be injective).

(d)如果 g ∘ f 是单射,则 f 也是单射(但 g 可能是单射,也可能不是单射)。

(e) If g f is surjective, then g is surjective (but f may or may not be surjective).

(e)如果 g ∘ f 是满射,那么 g 是满射(但 f 可能是满射,也可能不是满射)。

We prove parts (a), (b), and (e) here, asking you to prove (c) and (d) in the exercises. For all proofs, fix arbitrary functions f : X Y and g : Y Z ; we know g f : X Z is also a function, and ( g f )( x ) = g ( f ( x )) for all x X .

我们在此证明 (a)、(b) 和 (e) 部分,要求你在练习中证明 (c) 和 (d)。在所有证明中,固定任意函数 f : X → Y 和 g : Y → Z;我们知道 g ∘ f : X → Z 也是一个函数,并且对于所有 x ∈ X,(g ∘ f)(x) = g(f(x))。

5.94. Proof of (a). Assume f is one-to-one and g is one-to-one; prove g f is one-to-one. We have assumed

5.94. (a) 的证明。假设 f 是单射函数,g 也是单射函数;证明 g ∘ f 也是单射函数。我们已经假设

x x 1 X X , x x 2 X X , f f ( x x 1 ) = f f ( x x 2 ) x x 1 = x x 2

(5.3)

and

y 1 Y , y 2 Y , g ( y 1 ) = g ( y 2 ) y 1 = y 2 .

(5.4)

Keeping in mind that g f has domain X , we must prove

考虑到 g ∘ f 的定义域为 X,我们必须证明

w 西 1 X X , w 西 2 X X , ( g f f ) ( w 西 1 ) = ( g f f ) ( w 西 2 ) w 西 1 = w 西 2 .

(5.5)

Fix arbitrary w 1 X and w 2 X . Assume ( g f )( w 1 ) = ( g f )( w 2 ). Prove w 1 = w 2 . We have assumed g ( f ( w 1 )) = g ( f ( w 2 )). We intend to use ( 5.4 ) and the Inference Rule for ALL, taking y 1 = f ( w 1 ) and y 2 = f ( w 2 ). These choices are legitimate, because f ( w 1 ) ∈ Y and f ( w 2 ) ∈ Y (recall f maps X into Y ). Now our assumption g ( f ( w 1 )) = g ( f ( w 2 )) becomes g ( y 1 ) = g ( y 2 ). By the Inference Rule for IF and ( 5.4 ), y 1 = y 2 follows. This means f ( w 1 ) = f ( w 2 ). Now we use the rules for ALL and IF again, taking x 1 = w 1 X and x 2 = w 2 X in ( 5.3 ). We deduce w 1 = w 2 , which is the goal.

固定任意 w 1 ∈ X 和 w 2 ∈ X。假设 (g ∘ f)(w 1 ) = (g ∘ f)(w 2 )。证明 w 1 = w 2 。我们假设 g(f(w 1 )) = g(f(w 2 ))。我们打算使用 (5.4) 和 ALL 的推理规则,取 y 1 = f(w 1 ) 和 y 2 = f(w 2 )。这些选择是合理的,因为 f(w 1 ) ∈ Y 且 f(w 2 ) ∈ Y(回想一下,f 将 X 映射到 Y)。现在,我们的假设 g(f(w 1 )) = g(f(w 2 )) 变为 g(y 1 ) = g(y 2 )。根据 IF 的推理规则和 (5.4),y 1 = y 2 成立。这意味着 f(w 1 ) = f(w 2 )。现在我们再次使用 ALL 和 IF 的规则,在 (5.3) 中取 x 1 = w 1 ∈ X 和 x 2 = w 2 ∈ X。我们推导出 w 1 = w 2 ,这就是目标。

#

Here is a less formal, but potentially more intuitive, contrapositive proof of the same result. Start with two arbitrary unequal inputs w 1 w 2 in X . Since f is injective, y 1 = f ( w 1 ) and y 2 = f ( w 2 ) are unequal intermediate outputs in Y . Since g is injective, g ( f ( w 1 )) and g ( f ( w 2 )) are unequal final outputs in Z . So, w 1 w 2 implies ( g f )( w 1 ) ≠ ( g f )( w 2 ), which means that g f is one-to-one.

以下是对同一结论的一种不太正式但可能更直观的逆否命题证明。从集合 X 中任意两个不相等的输入 w 1 ≠ w 2 开始。由于 f 是单射函数,因此 y 1 = f(w 1 ) 和 y 2 = f(w 2 ) 是集合 Y 中不相等的中间输出。由于 g 是单射函数,因此 g(f(w 1 )) 和 g(f(w 2 )) 是集合 Z 中不相等的最终输出。所以,w 1 ≠ w 2 蕴含 (g ∘ f)(w 1 ) ≠ (g ∘ f)(w 2 ),这意味着 g ∘ f 是一一对应的。

5.95. Proof of (b). Assume f : X Y is onto and g : Y Z is onto; prove g f : X Z is onto. We have assumed y Y , x X , y = f ( x ) and z Z , y Y , z = g ( y ) . We must prove z Z , x X , z = ( g f ) ( x ) . To do so, fix an arbitrary z 0 Z . By our assumption on g , there is a fixed object y 0 Y with z 0 = g ( y 0 ). Now by our assumption on f , there is a fixed object x 0 X with y 0 = f ( x 0 ). We see that z 0 = g ( y 0 ) = g ( f ( x 0 )) = ( g f )( x 0 ). Thus to prove x X , z 0 = ( g f ) ( x ) , we may choose x = x 0 .

5.95. (b) 的证明。假设 f : X → Y 是满射,g : Y → Z 是满射;证明 g ∘ f : X → Z 是满射。我们假设 ∀y∈Y,∃x∈X,y=f(x) 且 ∀z∈Z,∃y∈Y,z=g(y)。我们必须证明 ∀z∈Z,∃x∈X,z=(g∘f)(x)。为此,固定一个任意的 z 0 ∈ Z。根据我们对 g 的假设,存在一个固定的对象 y 0 ∈ Y,使得 z 0 = g(y 0 )。根据我们对 f 的假设,存在一个固定对象 x 0 ∈ X,使得 y 0 = f(x 0 )。我们看到 z 0 = g(y 0 ) = g(f(x 0 )) = (g ∘ f)(x 0 )。因此,为了证明存在 x∈X,z0=(g∘f)(x),我们可以选择 x = x 0

5.96. Proof of (e). Assume g f is surjective. Prove g is surjective. We have assumed z Z , x X , z = ( g f ) ( x ) . We must prove z Z , y Y , z = g ( y ) . Fix arbitrary z 0 Z . Prove y Y , z 0 = g ( y ) . Using our assumption, we know there is some object x 0 X with z 0 = ( g f )( x 0 ). This means z 0 = g ( f ( x 0 )). Comparing to the goal, we may choose y 0 = f ( x 0 ) to make z 0 = g ( y 0 ) be true. Observe that y 0 does lie in the required set Y , since f is a function with codomain Y .

5.96. (e) 的证明。假设 g ∘ f 是满射。证明 g 是满射。我们假设对于任意 z∈Z,存在 x∈X,z=(g∘f)(x)。我们必须证明对于任意 z∈Z,存在 y∈Y,z=g(y)。固定任意 z 0 ∈ Z。证明对于任意 y∈Y,z0=g(y)。利用我们的假设,我们知道存在某个对象 x 0 ∈ X,使得 z 0 = (g ∘ f)(x 0 )。这意味着 z 0 = g(f(x 0 ))。与目标函数相比,我们可以选择 y 0 = f(x 0 ) 使得 z 0 = g(y 0 ) 为真。观察到 y 0 确实位于所需的集合 Y 中,因为 f 是一个值域为 Y 的函数。

To justify the parenthetical remark in (e), we give an example where g f is surjective but f is not surjective. Let X = {1, 2}, Y = {1, 2, 3}, and Z = {4, 5}. Define f : X Y by f (1) = 1 and f (2) = 2. Define g : Y Z by g (1) = 4 and g (2) = 5 and g (3) = 5. Now g f : X Z satisfies ( g f )(1) = 4 and ( g f )(2) = 5. It is immediate that g f is surjective (in fact, bijective), yet f is not surjective since 3 ∈ Y does not have the form f ( x ) for any x X .

为了证明(e)中的括号注释,我们给出一个例子,其中g ∘ f是满射但f不是满射。设X = {1, 2},Y = {1, 2, 3},Z = {4, 5}。定义f : X → Y,使得f(1) = 1且f(2) = 2。定义g : Y → Z,使得g(1) = 4且g(2) = 5且g(3) = 5。现在,g ∘ f : X → Z满足(g ∘ f)(1) = 4且(g ∘ f)(2) = 5。显然,g ∘ f是满射(实际上是双射),但f不是满射,因为对于任意x ∈ X,3 ∈ Y不具有f(x)的形式。

Section Summary

章节概要

  1. Injective Functions. A function f : X Y is injective (one-to-one) iff for all w , x X , if f ( w ) = f ( x ), then w = x . In an arrow diagram for an injective function, every output gets hit by at most one arrow. A real-valued function is injective iff every horizontal line y = y 0 intersects the graph in at most one point.
    单射函数。函数 f : X → Y 是单射(一对一)的,当且仅当对于所有 w, x ∈ X,如果 f(w) = f(x),则 w = x。在单射函数的箭头图中,每个输出点至多被一个箭头所击中。实值函数是单射的,当且仅当每条水平线 y = y 0 与函数图像至多相交于一个点。
  2. Surjective Functions. A function f : X Y is surjective (onto) iff y Y , x X , y = f ( x ) . In an arrow diagram for a surjective function, every output gets hit by at least one arrow. A real-valued function is surjective iff every horizontal line y = y 0 intersects the graph in at least one point.
    满射函数。函数 f : X → Y 是满射(或称全等射)当且仅当对于任意 y∈Y 和 x∈X,都有 y=f(x)。在满射函数的箭头图中,每个输出点都至少被一个箭头指向。实值函数是满射当且仅当每条水平线 y = y 0 都与函数图像至少相交于一点。
  3. Bijective Functions. A function f : X Y is bijective iff f is injective and surjective iff y Y , ! x X , y = f ( x ) . In an arrow diagram for a bijective function, every output gets hit by exactly one arrow. A real-valued function is bijective iff every horizontal line y = y 0 intersects the graph in exactly one point.
    双射函数。函数 f : X → Y 是双射当且仅当 f 是单射且满射当且仅当对于任意 y∈Y,存在 !x∈X,y=f(x)。在双射函数的箭头图中,每个输出点都恰好被一个箭头击中。实值函数是双射当且仅当每条水平线 y = y 0 与函数图像恰好相交于一点。
  4. Properties of Injections, Surjections, and Bijections. The composition of two injective functions is injective. The composition of two surjective functions is surjective. The composition of two bijective functions is bijective. If g f is injective, then f is injective. If g f is surjective, then g is surjective.
    单射、满射和双射的性质。两个单射函数的复合仍然是单射。两个满射函数的复合仍然是满射。两个双射函数的复合仍然是双射。如果 g ∘ f 是单射,那么 f 也是单射。如果 g ∘ f 是满射,那么 g 也是满射。

Exercises

练习

  1. Let A = {1, 2, 3, 4}, B = { v , w , x , y , z }, and consider the following four functions: f : A B with graph {(1, w ), (2, z ), (3, y ), (4, w )}; g : B A with graph {( v , 4), ( w , 3), ( y , 1), ( x , 2), ( z , 4)}; h : A A with graph {(1, 4), (2, 4), (3, 1), (4, 3)}; k : B B with graph {( v , z ), ( z , y ), ( y , v ), ( w , x ), ( x , w )}. Decide whether each function is injective, surjective, or bijective.
    设 A = {1, 2, 3, 4},B = {v, w, x, y, z},并考虑以下四个函数: f : A → B,其图像为 {(1, w), (2, z), (3, y), (4, w)}; g : B → A,其图像为 {(v, 4), (w, 3), (y, 1), (x, 2), (z, 4)}; h : A → A,其图像为 {(1, 4), (2, 4), (3, 1), (4, 3)}; k : B → B,其图像为 {(v, z), (z, y), (y, v), (w, x), (x, w)}。 判断每个函数是否为单射。满射,或双射。
  2. For each of the following functions f : R R , is f injective? surjective? bijective? (a) f ( x ) = cos x . (b) f ( x ) = e x . (c) f ( x ) = x 4 . (d) f ( x ) = x 5 . (e) f ( x ) = 1/ x for x ≠ 0, and f (0) = 0. (f) f ( x ) = x ( x rounded up to the nearest integer).
    对于下列每个函数 f:R→R,f 是单射、满射还是双射? (a) f(x) = cos x。 (b) f(x) = e x 。 (c) f(x) = x 4 。 (d) f(x)=x5。 (e) f(x) = 1/x,其中 x ≠ 0,且 f(0) = 0。 (f) f(x)=⌈x⌉(x 向上取整到最接近的整数)。
  3. Let a ≠ 0 and b be real constants. Prove that f : R R , given by f ( x ) = ax + b for x R , is a bijection.
    设 a ≠ 0 且 b 为实常数。证明函数 f:R→R,其中 f(x) = ax + b,x∈R,是一个双射。
  4. Define g : R 0 R 0 by g ( x ) = 1/ x for nonzero real x . Prove that g is a bijection.
    定义 g:R≠0→R≠0,其中 g(x) = 1/x,x 为非零实数。证明 g 是一个双射。
  5. Define f : R R by f ( x ) = cos x and define g : R [ 1 , 1 ] by g ( x ) = cos x . Is each of the following functions injective? surjective? bijective? (a) f (b) g (c) f | [0, π ] (d) g | [0, π ] (e) g | [0,2 π ] (f) g | [3 π /2,5 π /2] (g) f f (h) g ∘ ( f | [0,1] ).
    定义函数 f:R→R 为 f(x) = cos x,定义函数 g:R→[−1,1] 为 g(x) = cos x。下列函数是单射、满射还是双射?(a) f (b) g (c) f| [0, π ] (d) g| [0, π ] (e) g| [0,2 π ] (f) g| [3 π /2,5 π /2] (g) f ∘ f (h) g ∘ (f| [0,1] )。
  6. Give an example of a function f : Z 0 Z 0 with the stated properties. Prove that your examples work. (a) f is injective but not surjective. (b) f is surjective but not injective. (c) f is not injective, not surjective, and not constant. (d) f is injective and surjective, but f is not the identity function.
    给出一个满足下列性质的函数 f:Z≥0→Z≥0 的例子。证明你的例子成立。(a) f 是单射但不是满射。(b) f 是满射但不是单射。(c) f 既不是单射,也不是满射,也不是常数函数。(d) f 是单射且是满射,但 f 不是恒等函数。
  7. Determine whether each of the following functions is: (i) injective; (ii) surjective; (iii) bijective. Prove your answers. (a) f : R R defined by f ( x ) = x 3 x . (b) g : R [ 1 , ) defined by g ( x ) = e x 2 . (c) h : R { 3 / 7 } R { 1 / 7 } defined by h ( x ) = x + 5 7 x 3 . (d) k : R 0 R defined by k ( x ) = ln| x |. (e) p : (0, ∞) → (0, ∞) defined by p ( x ) = x x .
    判断下列函数是否为:(i) 单射;(ii) 满射;(iii) 双射。证明你的答案。 (a) f:R→R,定义为 f(x) = x 3 − x。 (b) g:R→[1,∞),定义为 g(x)=ex2。 (c) h:R−{3/7}→R−{1/7},定义为 h(x)=x+57x−3。 (d) k:R≠0→R,定义为 k(x) = ln|x|。 (e) p : (0, ∞) → (0, ∞),定义为 p(x) = x x
  8. Determine whether each of the following functions is: (i) injective; (ii) surjective; (iii) bijective. Explain informally. (a) p : Z 0 Z 0 defined by p ( n ) = n 2 for n Z 0 . (b) q : Z Z defined by q ( n ) = n 2 for n Z . (c) r : R R defined by r ( x ) = x 2 for x R . (d) s : Z Z defined by s ( m ) = m 3 for m Z . (e) t : R R defined by t ( x ) = x 3 for x R . (f) u : C C defined by u ( z ) = z 3 for z C .
    判断下列函数是否为:(i)单射;(ii)满射;(iii)双射。请用通俗易懂的方式解释。 (a) p:Z≥0→Z≥0,定义为 p(n) = n 2 ,其中 n∈Z≥0。 (b) q:Z→Z,定义为 q(n) = n 2 ,其中 n∈Z。 (c) r:R→R,定义为 r(x) = x 2 ,其中 x∈R。 (d) s:Z→Z,定义为 s(m) = m 3 ,其中 m∈Z。 (e) t:R→R,定义为 t(x) = x 3 ,其中 x∈R。 (f) u:C→C定义为 u(z) = z 3 ,其中 z∈C。
  9. Let f : X Y be a function with graph F . Prove carefully that f is injective iff for all x , y , z , ( x , y ) ∈ F and ( z , y ) ∈ F implies x = z .
    设 f : X → Y 是一个函数,其图像为 F。仔细证明 f 是单射当且仅当对于所有 x, y, z,(x, y) ∈ F 且 (z, y) ∈ F 蕴含 x = z。
  10. (a) Is the function ( 0 , 0 , 0 ) injective? surjective? bijective? (b) Repeat part (a) for the function ( 0 , B , 0 ) , where B is a nonempty set.
    (a)函数(0,0,0)是单射吗?满射吗?双射吗?(b)对函数(0,B,0)重复(a)部分,其中B是一个非空集合。
  11. For each part, find functions f , g : R R satisfying the stated conditions. (a) f and g are injective, but f + g is not injective. (b) f and g are not injective, but f + g is injective. (c) f and g are surjective, but f + g is not surjective. (d) f and g are not surjective, but f + g is surjective. (e) f and g are bijective, but f · g is not bijective. (f) f and g are not bijective, but f · g is bijective.
    对于每个部分,找出满足给定条件的函数 f,g:R→R。 (a) f 和 g 是单射,但 f + g 不是单射。 (b) f 和 g 不是单射,但 f + g 是单射。 (c) f 和 g 是满射,但 f + g 不是满射。 (d) f 和 g 不是满射,但 f + g 是满射。 (e) f 和 g 是双射,但 f · g 不是双射。 (f) f 和 g 不是双射,但 f · g 是双射。
  12. Prove parts (c) and (d) of Theorem 5.93 .
    证明定理 5.93 的 (c) 和 (d) 部分。
  13. Prove: for all f : X Y , f is injective iff for all A , C X , f [ A C ] = f [ A ] f [ C ] .
    证明:对于所有 f : X → Y,f 是单射当且仅当对于所有 A, C ⊆ X,f[A∩C]=f[A]∩f[C]。
  14. Prove: for all f : X Y , f is injective iff for all A , C X , f [ A C ] = f [ A ] − f [ C ].
    证明:对于所有 f : X → Y,f 是单射当且仅当对于所有 A, C ⊆ X,f[A − C] = f[A] − f[C]。
  15. Prove: for all f : X Y , f is injective iff for all A X , f −1 [ f [ A ]] = A .
    证明:对于所有 f : X → Y,f 是单射当且仅当对于所有 A ⊆ X,f −1 [f[A]] = A。
  16. Prove: for all f : X Y , f is surjective iff for all B Y , f [ f −1 [ B ]] = B .
    证明:对于所有 f : X → Y,f 是满射当且仅当对于所有 B ⊆ Y,f[f −1 [B]] = B。
  17. Suppose h : X X is a bijection. (a) Prove: for all functions f : X Y , f is injective iff f h is injective; f is surjective iff f h is surjective; and f is bijective iff f h is bijective. (b) Prove: for all functions g : W X , g is injective iff h g is injective; g is surjective iff h g is surjective; and g is bijective iff h g is bijective. (c) Which implications in (a) and (b) remain true if h is only assumed to be an injection? (d) Which implications in (a) and (b) remain true if h is only assumed to be a surjection?
    假设 h : X → X 是一个双射。(a) 证明:对于所有函数 f : X → Y,f 是单射当且仅当 f ∘ h 是单射;f 是满射当且仅当 f ∘ h 是满射;f 是双射当且仅当 f ∘ h 是双射。(b) 证明:对于所有函数 g : W → X,g 是单射当且仅当 h ∘ g 是单射;g 是满射当且仅当 h ∘ g 是满射;g 是双射当且仅当 h ∘ g 是双射。(c) 如果只假设 h 是单射,(a) 和 (b) 中的哪些蕴含关系仍然成立?(d) 如果只假设 h 是满射,(a) 和 (b) 中的哪些蕴含关系仍然成立?
  18. (a) Use uniqueness of prime factorization to show that g : Z 0 × Z 0 Z 0 defined by g ( a , b ) = 2 a 3 b is injective. (b) Generalize this idea to define an injection h k : Z 0 k Z 0 for any fixed integer k ≥ 1. (c) Define f : Z 0 × Z 0 Z 0 by f ( a , b ) = 2 a (2 b + 1) − 1. Prove that f is a bijection. (d) Use the function f in part (c) to recursively construct bijections p k : Z 0 k Z 0 for all k ≥ 2.
    (a) 利用素因子分解的唯一性证明由 g(a, b) = 2<sup>3</sup>(id<sub>1</sub>) 定义的 g: Z≥0×Z≥0→Z≥0 是单射。(b) 将此思想推广,定义一个单射 h<sub>k</sub>: Z≥0<sup>k</sup>→Z≥0,其中 k 为任意固定的整数,k ≥ 1。(c) 定义 f: Z≥0×Z≥0→Z≥0,其中 f(a, b) = 2<sup>2</sup>(2b + 1) − 1。证明 f 是一个双射。(d) 利用 (c) 中的函数 f,递归构造双射 p<sub>k</sub>: Z≥0<sup>k</sup>→Z≥0,其中 k ≥ 2。
  19. Suppose f : X Y is a function and S X . (a) Prove: if f is injective, then f | S is injective. (b) Prove: if f | S is surjective, then f is surjective. (c) Give an example to show that the converse of (a) can be false. (d) Give an example to show that the converse of (b) can be false.
    假设 f : X → Y 是一个函数,且 S ⊆ X。 (a) 证明:如果 f 是单射,则 f| S 也是单射。 (b) 证明:如果 f| S 是满射,则 f 也是满射。 (c) 举例说明 (a) 的逆命题可能为假。 (d) 举例说明 (b) 的逆命题可能为假。
  20. Consider a function given by the formula f ( x ) = a x + b c x + d for some real constants a , b , c , d such that ad bc ≠ 0. (a) Find the largest possible domain of f contained in R , and the smallest possible codomain corresponding to this domain. (Consider three cases: (i) a = 0; (ii) c = 0; (iii) a ≠ 0 ≠ c .) (b) Prove that the function f with domain and codomain found in part (a) is a bijection. (c) Where in your proofs did you use the assumption ad bc ≠ 0? What happens if ad bc = 0?
    考虑函数 f(x) = ax + bcx + d,其中 a、b、c、d 为实常数,且 ad − bc ≠ 0。(a)求 f 在实数集 R 中的最大定义域,以及与该定义域对应的最小值域。(考虑三种情况:(i)a = 0;(ii)c = 0;(iii)a ≠ 0 ≠ c。)(b)证明在 (a) 部分中求得的定义域和值域对应的函数 f 是一个双射。(c)在你的证明过程中,你在哪里用到了假设 ad − bc ≠ 0?如果 ad − bc = 0 会发生什么?
  21. Suppose f : A C and g : B C are gluable functions. Let h = glue ( f , g ) : A B C . (a) Prove: If f or g is surjective, then h is surjective. (b) Prove: If h is injective, then f and g are injective. (c) Show that the converse of part (a) can be false. (d) Show that the converse of part (b) can be false.
    假设 f : A → C 和 g : B → C 是可粘合函数。令 h=glue(f,g):A∪B→C。(a)证明:如果 f 或 g 是满射,则 h 也是满射。(b)证明:如果 h 是单射,则 f 和 g 都是单射。(c)证明 (a) 的逆命题可能不成立。(d)证明 (b) 的逆命题可能不成立。
  22. Suppose f : A B , g : C D , and A C = B D = 0 . Let h = glue ( f , g ) : A C B D . (a) Prove: h is injective iff f and g are injective. (b) Prove: h is surjective iff f and g are surjective. (c) Prove: h is bijective iff f and g are bijective. (d) What can be said if B D 0 ? (e) What happens if A C 0 ?
    设 f : A → B,g : C → D,且 A∩C=B∩D=0。令 h=glue(f,g):A∪C→B∪D。(a)证明:h 是单射当且仅当 f 和 g 都是单射。(b)证明:h 是满射当且仅当 f 和 g 都是满射。(c)证明:h 是双射当且仅当 f 和 g 都是双射。(d)若 B∩D≠0,结论是什么?(e)若 A∩C≠0,结论是什么?
  23. Suppose f : X Y is a function. Prove: f is injective iff for all sets W and all functions g , h : W X , f g = f h implies g = h .
    假设 f : X → Y 是一个函数。证明:f 是单射当且仅当对于所有集合 W 和所有函数 g, h : W → X,f ∘ g = f ∘ h 蕴含 g = h。
  24. Suppose f : X Y is a function. Prove: f is surjective iff for all sets Z and all functions g , h : Y Z , g f = h f implies g = h .
    假设 f : X → Y 是一个函数。证明:f 是满射当且仅当对于所有集合 Z 和所有函数 g, h : Y → Z,g ∘ f = h ∘ f 蕴含 g = h。

5.9 Inverse Functions

5.9 反函数

In calculus, it is often helpful to pass from a function f to the inverse function f −1 . The idea is that the inverse function interchanges the roles of inputs and outputs, so that for all inputs x and outputs y for f , y = f ( x ) iff x = f −1 ( y ). A key subtlety is that many functions do not have inverse functions (in contrast to relations, where the inverse of a relation R is always another relation). It turns out that the inverse of a function f exists iff f is a bijection. The optional second half of this section introduces left inverses, right inverses, and their connections to injective and surjective functions.

在微积分中,从函数 f 到其反函数 f −1 的转换常常很有帮助。其思想是,反函数交换了输入和输出的角色,因此对于 f 的所有输入 x 和输出 y,y = f(x) 当且仅当 x = f −1 (y)。一个关键的微妙之处在于,许多函数没有反函数(这与关系不同,关系 R 的逆总是另一个关系)。事实上,函数 f 的反函数存在当且仅当 f 是一个双射。本节的后半部分(可选)将介绍左逆、右逆以及它们与单射和满射函数的联系。

The Inverse of a Function

函数的反函数

The idea of interchanging inputs and outputs leads to the following definition of inverse functions.

输入和输出互换的概念引出了反函数的以下定义。

5.97. Definition: Inverse Function. Given a function f : X Y , an inverse function for f is a function g : Y X satisfying x X , y Y , y = f ( x ) x = g ( y ) . When such a function g exists, we say that f is invertible and write g = f −1 .

5.97. 定义:反函数。给定一个函数 f : X → Y,f 的反函数是一个函数 g : Y → X,满足 ∀x∈X,∀y∈Y,y=f(x)⇔x=g(y)。当这样的函数 g 存在时,我们称 f 可逆,并记为 g = f −1

The next example shows that inverse functions do not always exist. But if a function does have an inverse, it is unique, as we show later. This justifies the introduction of the notation f 1 for the inverse function.

下一个例子表明,反函数并非总是存在。但如果一个函数确实存在反函数,那么它是唯一的,我们将在后面证明这一点。这就解释了为什么需要引入符号 f 1 来表示反函数。

5.98. Example. Define f : {1, 2, 3} → { a , b , c } by f (1) = b , f (2) = c , and f (3) = a . An inverse function for f is the function g :{ a , b , c } → {1, 2, 3} given by g ( a ) = 3, g ( b ) = 1, and g ( c ) = 2. On the other hand, consider h :{1, 2, 3} → { a , b } given by h (1) = b , h (2) = a , and h (3) = b . We claim h has no inverse function. To get a contradiction, assume k :{ a , b } → {1, 2, 3} is an inverse for h . Since h (1) = b , we must have k ( b ) = 1. Since h (3) = b , we must also have k ( b ) = 3. Now k is not a function, since the input b is associated with two different outputs 1 and 3. Similarly, consider p :{1, 2} → { a , b , c } given by p (1) = a and p (2) = c . If q :{ a , b , c } → {1, 2} were an inverse function for p , we must have q ( b ) = 1 or q ( b ) = 2. On one hand, if q ( b ) = 1, we deduce that p (1) = b , which is false. On the other hand, if q ( b ) = 2, we see that p (2) = b , which is false. So p has no inverse function.

5.98. 例。定义函数 f : {1, 2, 3} → {a, b, c},其中 f(1) = b,f(2) = c,f(3) = a。f 的一个反函数是函数 g:{a, b, c} → {1, 2, 3},其中 g(a) = 3,g(b) = 1,g(c) = 2。另一方面,考虑函数 h:{1, 2, 3} → {a, b},其中 h(1) = b,h(2) = a,h(3) = b。我们断言 h 没有反函数。为了得出矛盾,假设 k:{a, b} → {1, 2, 3} 是 h 的一个反函数。由于 h(1) = b,所以 k(b) = 1。由于 h(3) = b,所以 k(b) = 3。现在 k 不是一个函数,因为输入 b 对应两个不同的输出 1 和 3。类似地,考虑集合 p:{1, 2} → {a, b, c},其中 p(1) = a 且 p(2) = c。如果 q:{a, b, c} → {1, 2} 是 p 的反函数,那么 q(b) 必须等于 1 或 q(b) = 2。一方面,如果 q(b) = 1,我们推断 p(1) = b,这是错误的。另一方面,如果 q(b) = 2,我们看到 p(2) = b,这也是错误的。因此,p 没有反函数。

The next example shows that inverse functions do not always exist . But if a function does have an inverse, it is unique, as we show later. This justifies the introduction of the notation f −1 for the inverse function.

下一个例子表明,反函数并非总是存在。但如果一个函数确实存在反函数,那么它是唯一的,我们将在后面证明这一点。这就解释了为什么需要引入符号 f −1 来表示反函数。

5.99 Theorem on Inverses and Composition. For all functions f : X Y and g : Y X , g is an inverse function for f iff f g = Id Y and g f = Id X .

5.99 关于逆函数和复合函数的定理。对于所有函数 f : X → Y 和 g : Y → X,g 是 f 的逆函数当且仅当 f ∘ g = Id Y 且 g ∘ f = Id X

Proof. Fix functions f : X Y and g : Y X . Part 1 . Assume g is an inverse function for f ; prove f g = Id Y and g f = Id X . Both f g and Id Y are functions from Y to Y . To prove they are equal, we fix y Y and show ( f g )( y ) = Id Y ( y ). Write x = g ( y ) ∈ X . By definition of inverse functions, we know y = f ( x ). Now compute ( f g )( y ) = f ( g ( y )) = f ( x ) = y = Id Y ( y ). A similar calculation shows that g f = Id X .

证明。固定函数 f : X → Y 和 g : Y → X。第一部分。假设 g 是 f 的反函数;证明 f ∘ g = Id Y 且 g ∘ f = Id X 。f ∘ g 和 Id Y 都是从 Y 到 Y 的函数。为了证明它们相等,我们固定 y ∈ Y,并证明 (f ∘ g)(y) = Id Y (y)。令 x = g(y) ∈ X。根据反函数的定义,我们知道 y = f(x)。现在计算 (f ∘ g)(y) = f(g(y)) = f(x) = y = Id Y (y)。类似的计算表明 g ∘ f = Id X

Part 2 . Assume f g = Id Y and g f = Id X ; prove x X , y Y , y = f ( x ) x = g ( y ) . Fix x X and y Y . Part 2a. Assume y = f ( x ); prove x = g ( y ). We compute x = Id X ( x ) = ( g f )( x ) = g ( f ( x )) = g ( y ), as needed. Part 2b. Assume x = g ( y ); prove y = f ( x ). Here, we compute y = Id Y ( y ) = ( f g )( y ) = f ( g ( y )) = f ( x ), as needed.     □

第二部分。假设 f ∘ g = Id Y 且 g ∘ f = Id X ;证明对于所有 x∈X,所有 y∈Y,y=f(x)⇔x=g(y)。固定 x ∈ X 和 y ∈ Y。第二部分 a。假设 y = f(x);证明 x = g(y)。我们计算 x = Id X (x) = (g ∘ f)(x) = g(f(x)) = g(y),如需要。第二部分 b。假设 x = g(y);证明 y = f(x)。这里,我们计算 y = Id Y (y) = (f ∘ g)(y) = f(g(y)) = f(x),如需要。□

5.100. Corollary: Uniqueness of Inverse Functions. Every f : X Y has at most one inverse function.

5.100. 推论:反函数的唯一性。每个函数 f : X → Y 至多有一个反函数。

Proof. Fix f : X Y ; assume g : Y X and h : Y X are both inverse functions for f ; prove g = h . [The direct approach is to fix y Y and prove g ( y ) = h ( y ). But here is a faster way using Theorem 5.99 and the identity and associativity properties from Theorem 5.62 .] We compute g = g ∘Id Y = g ∘ ( f h ) = ( g f ) ∘ h = Id X h = h .

证明。固定 f : X → Y;假设 g : Y → X 和 h : Y → X 都是 f 的反函数;证明 g = h。[直接的方法是固定 y ∈ Y 并证明 g(y) = h(y)。但这里提供了一种更快捷的方法,利用定理 5.99 以及定理 5.62 中的恒等式和结合律。] 我们计算 g = g ∘Id Y = g ∘ (f ∘ h) = (g ∘ f) ∘ h = Id X ∘ h = h。

Properties of Inverse Functions

反函数的性质

We are about to show that a function f is invertible iff f is bijective. First we give an intuitive version of the proof based on arrow diagrams. Given a function f : X Y , we know each x X has exactly one arrow leaving it. We attempt to form the inverse function g : Y X by reversing all the arrows. In order for g to be a function, it must be true that each y Y has exactly one g -arrow leaving it. Since g -arrows are the reversals of f -arrows, this means that each y Y should have exactly one f -arrow entering it. This is precisely the criterion for f to be a bijection. The next proof formalizes this argument.

我们将证明函数 f 可逆当且仅当 f 是双射。首先,我们给出一个基于箭头图的直观证明。给定一个函数 f : X → Y,我们知道每个 x ∈ X 都恰好有一个箭头离开它。我们尝试通过反转所有箭头来构造反函数 g : Y → X。为了使 g 是一个函数,必须满足每个 y ∈ Y 都恰好有一个 g 箭头离开它。由于 g 箭头是 f 箭头的反转,这意味着每个 y ∈ Y 都应该恰好有一个 f 箭头进入它。这正是 f 是双射的判别准则。接下来的证明将形式化这个论证。

5.101. Theorem on Inverses and Bijections. For all f : X Y , f has an inverse function iff f is a bijection.

5.101. 关于逆函数和双射的定理。对于所有 f : X → Y,f 存在逆函数当且仅当 f 是双射。

Proof. Fix f : X Y . Part 1 . Assume f has an inverse function g : Y X . Prove f is a bijection, which means f is surjective and injective. By Theorem 5.99 , we know g f = Id X and f g = Id Y . Now Id X is injective, so g f is injective, so f is injective by Theorem 5.93 (d). Similarly, Id Y is surjective, so f g is surjective, so f is surjective by Theorem 5.93 (e).

证明。固定函数 f : X → Y。第一部分。假设 f 有一个反函数 g : Y → X。证明 f 是双射,即 f 既是满射又是单射。根据定理 5.99,我们知道 g ∘ f = Id X 且 f ∘ g = Id Y 。现在 Id X 是单射,所以 g ∘ f 也是单射,因此根据定理 5.93(d),f 也是单射。类似地,Id Y 是满射,所以 f ∘ g 也是满射,因此根据定理 5.93(e),f 也是满射。

Part 2 Assume f is a bijection (hence surjective and injective); prove f has an inverse function. Let F = {( x , f ( x )) : x X } be the graph of f . Define g = ( Y , X , F −1 ), where F −1 = {( y , x ):( x , y ) ∈ F } is the inverse of the relation F . We prove below that g really is a function (with domain Y and codomain X ). Granting this fact, we can then conclude that for all x X and y Y , x = g ( y ) iff ( y , x ) ∈ F −1 iff ( x , y ) ∈ F iff y = f ( x ), so that g is the required inverse function for f . To prove that g is a function, we check the three conditions appearing after Definition 5.35 . First, F is a relation from X to Y (i.e., a subset of X × Y ), so we know F −1 is a relation from Y to X (i.e., a subset of Y × X ).

第二部分 假设 f 是一个双射(因此也是满射和单射);证明 f 存在反函数。令 F = {(x, f(x)) : x ∈ X} 为 f 的图像。定义 g = (Y, X, F −1 ),其中 F −1 = {(y, x):(x, y) ∈ F} 是关系 F 的逆。我们将在下文证明 g 确实是一个函数(定义域为 Y,值域为 X)。在此前提下,我们可以得出结论:对于所有 x ∈ X 和 y ∈ Y,x = g(y) 当且仅当 (y, x) ∈ F −1 当且仅当 (x, y) ∈ F 当且仅当 y = f(x),因此 g 是 f 的反函数。为了证明 g 是一个函数,我们检查定义 5.35 之后的三个条件。首先,F 是从 X 到 Y 的关系(即 X × Y 的子集),因此我们知道 F −1 是从 Y 到 X 的关系(即 Y × X 的子集)。

Second, to prove Existence of Outputs for g , we must prove y Y , x X , ( y , x ) F 1 . Note ( y , x ) ∈ F −1 iff ( x , y ) ∈ F iff y = f ( x ). So we must prove y Y , x X , y = f ( x ) . This is true since our assumption guarantees f is surjective.

其次,为了证明函数 g 的输出存在性,我们必须证明对于任意 y∈Y,存在 x∈X,(y,x)∈F−1。注意,(y, x) ∈ F −1 当且仅当 (x, y) ∈ F 当且仅当 y = f(x)。因此,我们必须证明对于任意 y∈Y,存在 x∈X,y=f(x)。这是成立的,因为我们的假设保证了 f 是满射。

Third, to prove Uniqueness of Outputs for g , we must prove for all y , x , w , if ( y , x ) ∈ F −1 and ( y , w ) ∈ F −1 , then x = w . Fix y , x , w . Assume ( y , x ) ∈ F −1 and ( y , w ) ∈ F −1 ; prove x = w . Our assumption can be rewritten ( x , y ) ∈ F and ( w , y ) ∈ F , i.e., y = f ( x ) and y = f ( w ). Then f ( x ) = f ( w ), so x = w because f is injective.      □

第三,为了证明函数 g 的输出唯一性,我们必须证明对于所有 y、x、w,如果 (y, x) ∈ F −1 且 (y, w) ∈ F −1 ,则 x = w。固定 y、x、w。假设 (y, x) ∈ F −1 且 (y, w) ∈ F −1 ;证明 x = w。我们的假设可以重写为 (x, y) ∈ F 且 (w, y) ∈ F,即 y = f(x) 且 y = f(w)。那么 f(x) = f(w),所以 x = w,因为 f 是单射。□

5.102. Example. The function f : R R , given by f ( x ) = e x for x R , is injective but not surjective, so it has no inverse. By restricting the codomain, we can obtain a bijective function exp : R R > 0 that does have an inverse. The inverse is the natural logarithm function ln : R > 0 R . In this setting, Theorem 5.99 states that ln( e x ) = x for all x R , and e ln y = y for all y R > 0 .

5.102. 示例。函数 f:R→R,定义为 f(x) = e x ,其中 x∈R,是单射但不是满射,因此没有反函数。通过限制值域,我们可以得到一个双射函数 exp:R→R>0,它有反函数。该反函数是自然对数函数 ln:R>0→R。在这种情况下,定理 5.99 指出,对于所有 x∈R,ln(e x ) = x,并且对于所有 y∈R>0,e ln y = y。

The function f : R R given by f ( x ) = x 2 for x R is not a bijection, so it has no inverse. On the other hand, f 1 : R 0 R 0 given by f 1 ( x ) = x 2 for x R 0 is a bijection; the inverse function is f 1 1 ( y ) = y for y R 0 . The function f 2 : R 0 R 0 given by f 2 ( x ) = x 2 for x R 0 is also a bijection; the inverse satisfies f 2 1 ( y ) = y for y R 0 .

函数 f:R→R,其中 f(x) = x 2 ,x∈R,不是双射,因此没有反函数。另一方面,函数 f1:R≥0→R≥0,其中 f 1 (x) = x 2 ,x∈R≥0,是双射;其反函数为 f1−1(y)=y,y∈R≥0。函数 f2:R≤0→R≥0,其中 f 2 (x) = x 2 ,x∈R≤0,也是双射;其反函数满足 f2−1(y)=−y,y∈R≥0。

Next consider h : R R given by h ( x ) = sin x . The function h is neither injective nor surjective, so it has no inverse. By restricting the domain and codomain of h , we can create a new bijective function S : [ − π /2, π /2] → [ − 1, 1] given by S ( x ) = sin x for x ∈ [ − π /2, π /2]. The inverse function S −1 :[ − 1, 1] → [ − π /2, π /2] is called the arcsine function in calculus.

接下来考虑函数 h:R→R,定义为 h(x) = sin x。函数 h 既不是单射也不是满射,因此它没有反函数。通过限制 h 的定义域和值域,我们可以构造一个新的双射函数 S : [ − π/2, π/2] → [ − 1, 1],定义为 S(x) = sin x,其中 x ∈ [ − π/2, π/2]。反函数 S:[ − 1, 1] → [ − π/2, π/2] 在微积分中被称为反正弦函数。

5.103 Theorem on Invertible Functions. For all functions f : X Y and g : Y Z :

5.103 可逆函数定理。对于所有函数 f : X → Y 和 g : Y → Z:

(a) Id X : X X is invertible, and Id X 1 = Id X .

(a)Id X : X → X 是可逆的,并且 IdX−1=IdX。

(b) If f is invertible, then f −1 is invertible and ( f −1 ) −1 = f .

(b)如果 f 是可逆的,则 f −1 是可逆的,并且 (f −1 ) −1 = f。

(c) If f and g are invertible, then g f is invertible and ( g f ) −1 = f −1 g −1 .

(c)如果 f 和 g 可逆,则 g ∘ f 可逆,且 (g ∘ f) −1 = f −1 ∘ g −1

(d) If f is a bijection, then f −1 exists and is a bijection.

(d)如果 f 是一个双射,那么 f −1 存在且是一个双射。

Proof. Fix functions f : X Y and g : Y Z . We prove each part using the criterion in Theorem 5.99 . For (a), we take Y = X and f = g = Id X in that theorem. Since “Id X ∘Id X = Id X and Id X ∘Id X = Id X ” is true, the theorem tells us that Id X is the inverse function for Id X . To prove (b), assume f is invertible. We know “ f −1 f = Id X and f f −1 = Id Y ” is true. Now use Theorem 5.99 , replacing X by Y , Y by X , f by f −1 , and g by f . We deduce that f is an inverse function for f −1 , i.e., ( f −1 ) −1 = f . To prove (c), assume f and g are invertible. By Theorem 5.99 (replacing Y by Z , f by g f , and g by f −1 g −1 ), it suffices to prove that ( g f ) ∘ ( f −1 g −1 ) = Id Z and ( f −1 g −1 ) ∘ ( g f ) = Id X . To prove the first equation, use associativity of function composition to compute

证明。固定函数 f : X → Y 和 g : Y → Z。我们使用定理 5.99 中的准则来证明每一部分。对于 (a),我们取 Y = X 且 f = g = Id X 。由于“Id X ∘Id X = Id X 且 Id X ∘Id X = Id X ”成立,因此定理告诉我们 Id X 是 Id X 的反函数。为了证明 (b),假设 f 可逆。我们知道“f −1 ∘ f = Id X 且 f ∘ f −1 = Id Y ”成立。现在使用定理 5.99,将 X 替换为 Y,Y 替换为 X,f 替换为 f −1 ,g 替换为 f。我们推断 f 是 f −1 的反函数,即 (f −1 ) −1 = f。为了证明 (c),假设 f 和 g 可逆。根据定理 5.99(将 Y 替换为 Z,f 替换为 g ∘ f,g 替换为 f −1 ∘ g −1 ),只需证明 (g ∘ f) ∘ (f −1 ∘ g −1 ) = Id Z 和 (f −1 ∘ g −1 ) ∘ (g ∘ f) = Id X 。为了证明第一个等式,利用函数复合的结合律进行计算。

( g f f ) ( f f 1 g 1 ) = g ( f f ( f f 1 g 1 ) ) = g ( ( f f f f 1 ) g 1 ) = g ( Id ID Y g 1 ) = g g 1 = Id ID Z Z .

The second equation is proved similarly. Finally, part (d) follows from part (b) and the fact that f is invertible iff f is bijective.     □

第二个方程的证明类似。最后,(d) 部分由 (b) 部分以及 f 可逆当且仅当 f 是双射这一事实得出。□

Left Inverses (Optional)

左反转(可选)

Theorem 5.99 tells us that g is an inverse function for f iff f g and g f are both identity functions. We now investigate what happens when only one of these conditions holds.

定理 5.99 告诉我们,g 是 f 的反函数当且仅当 f ∘ g 和 g ∘ f 都是恒等函数。现在我们来研究当这两个条件中只有一个成立时会发生什么。

5.104. Definition: Left Inverses. For all functions f : X Y and g : Y X , g is a left inverse for f iff g f = Id X .

5.104. 定义:左逆。对于所有函数 f : X → Y 和 g : Y → X,g 是 f 的左逆当且仅当 g ∘ f = Id X

5.105. Example. Figure 5.10 gives arrow diagrams for two functions f and h . A left inverse for f is a function g : { a , b , c , d , e } → {1, 2, 3} satisfying ( g f )( x ) = Id X ( x ) = x for all x ∈ {1, 2, 3}. Taking x = 1, 2, 3, we see that g must satisfy g ( d ) = 1, g ( b ) = 2, g ( a ) = 3. However, no requirement has been imposed on g ( c ) and g ( e ), since c and e are not in the image of f . We obtain one left inverse for f by setting g ( c ) = 1 and g ( e ) = 1. We could build other left inverses for f by choosing other values for g ( c ) and g ( e ). In this example, f has nine different left inverses. So, unlike the two-sided inverses studied earlier, left inverses need not be unique . Also, no left inverse g can satisfy f g = Id Y , since f is not a bijection. For instance, the g chosen above has ( f g )( c ) = f ( g ( c )) = f (1) = d c = Id Y ( c ).

5.105. 示例。图 5.10 给出了两个函数 f 和 h 的箭头图。f 的左逆函数是一个函数 g : {a, b, c, d, e} → {1, 2, 3},满足 (g ∘ f)(x) = Id X (x) = x,其中 x ∈ {1, 2, 3}。取 x = 1, 2, 3,我们发现 g 必须满足 g(d) = 1,g(b) = 2,g(a) = 3。然而,由于 c 和 e 不在 f 的像中,因此对 g(c) 和 g(e) 没有要求。令 g(c) = 1 和 g(e) = 1,我们得到 f 的一个左逆函数。通过选择 g(c) 和 g(e) 的其他值,我们可以构造 f 的其他左逆函数。在本例中,f 有九个不同的左逆函数。因此,与之前研究的双边逆不同,左逆不一定是唯一的。此外,没有左逆 g 能满足 f ∘ g = Id Y ,因为 f 不是双射。例如,上面选择的 g 有 (f ∘ g)(c) = f(g(c)) = f(1) = d ≠ c = Id Y (c)。

fig5_10

Figure 5.10

图 5.10

Arrow diagrams of two functions.

两个函数的箭头图。

Furthermore, left inverses do not always exist . Suppose we try to find a left inverse g : { a , b , c } → {1, 2, 3, 4, 5, 6} for the function h in the figure. Writing out the requirement g h = Id X , we obtain (among other conditions) g ( h (2)) = 2 and g ( h (5)) = 5. Now h (2) = c = h (5), so we must have g ( c ) = 2 and g ( c ) = 5. But then g is not a function, since input c has two associated outputs. This shows that h does not have any left inverses. The trouble arises because h was not injective: distinct inputs 2 and 5 to h both mapped to the same output c .

此外,左逆函数并非总是存在。假设我们试图找到图中函数 h 的左逆函数 g : {a, b, c} → {1, 2, 3, 4, 5, 6}。写出要求 g ∘ h = Id X ,我们得到(除其他条件外)g(h(2)) = 2 和 g(h(5)) = 5。现在 h(2) = c = h(5),所以我们必须有 g(c) = 2 和 g(c) = 5。但这样 g 就不是一个函数,因为输入 c 有两个对应的输出。这表明 h 没有左逆函数。问题在于 h 不是单射函数:不同的输入 2 和 5 都映射到同一个输出 c。

The previous example illustrates the following characterization of which functions have left inverses.

前面的例子说明了哪些函数具有左逆函数的以下特征。

5.106. Theorem on Left Inverses. Let X be a nonempty set. For all functions f : X Y , f has a left inverse iff f is injective.

5.106. 左逆定理。设 X 为非空集合。对于所有函数 f : X → Y,f 存在左逆当且仅当 f 是单射。

Proof. Fix f : X Y , where X is nonempty. First assume f has a left inverse g : Y X . Since g f = Id X is an injective function, Theorem 5.93 (d) tells us that f is injective.

证明。固定 f : X → Y,其中 X 非空。首先假设 f 有一个左逆函数 g : Y → X。由于 g ∘ f = Id X 是一个单射函数,根据定理 5.93(d),f 也是单射函数。

Conversely, assume f is injective; we must construct a left inverse g : Y X . To do so, let Z = f [ X ] ⊆ Y by the image of f , and define f 1 : X Z by f 1 ( x ) = f ( x ) for all x X . The function f 1 (obtained by restricting the codomain of f ) is injective and surjective, so it has a two-sided inverse g 1 : Z X . In particular, g 1 f 1 = Id X . To build the function g : Y X , select one fixed element x 0 X ; this is possible since X is nonempty. Now define g ( y ) = g 1 ( y ) for y Z , and g ( y ) = x 0 for y Y Z . You can check that g really is a function from Y to X . For instance, this is a special case of the gluing construction in Definition 5.68 ; here we are gluing together g 1 and the constant function with domain Y Z and constant value x 0 .

反之,假设 f 是单射;我们需要构造一个左逆函数 g : Y → X。为此,令 Z = f[X] ⊆ Y 为 f 的像,并定义 f 1 : X → Z,使得对于所有 x ∈ X,都有 f 1 (x) = f(x)。函数 f 1 (通过限制 f 的值域得到)是单射且满射,因此它有一个双侧逆函数 g 1 : Z → X。特别地,g 1 ∘ f 1 = Id X 。为了构造函数 g : Y → X,选择一个固定的元素 x 0 ∈ X;这是可能的,因为 X 非空。现在定义 g(y) = g 1 (y),其中 y ∈ Z;定义 g(y) = x 0 ,其中 y ∈ Y − Z。你可以验证 g 确实是从 Y 到 X 的函数。例如,这是定义 5.68 中粘合构造的一个特例;这里我们将 g 1 与定义域为 Y − Z、值为常数 x 0 的常函数粘合在一起。

To finish, we must check that g f = Id X . Fix x X , and set y = f ( x ), which is in the image Z of f . We now compute ( g f )( x ) = g ( f ( x )) = g ( y ) = g 1 ( y ) = g 1 ( f 1 ( x )) = ( g 1 f 1 )( x ) = Id X ( x ).

最后,我们必须验证 g ∘ f = Id X 。固定 x ∈ X,并令 y = f(x),它在 f 的像 Z 中。现在我们计算 (g ∘ f)(x) = g(f(x)) = g(y) = g 1 (y) = g 1 (f 1 (x)) = (g 1 ∘ f 1 )(x) = Id X (x)。

Right Inverses (Optional)

右逆元(可选)

5.107. Definition: Right Inverses. For all functions f : X Y and g : Y X , g is a right inverse for f iff f g = Id Y .

5.107. 定义:右逆。对于所有函数 f : X → Y 和 g : Y → X,g 是 f 的右逆当且仅当 f ∘ g = Id Y

5.108 Example. Consider once again the functions f and h in Figure 5.10 . A function g : { a , b , c } → {1, 2, 3, 4, 5, 6} is a right inverse for h iff h g = Id { a , b , c } iff h ( g ( a )) = a and h ( g ( b )) = b and h ( g ( c )) = c . Now, h ( x ) = a is true when x is 1, 4, or 6. So we could define g ( a ) to be any one of these three values and thereby satisfy h ( g ( a )) = a . Similarly, h ( x ) = b is solved by x = 3, and h ( x ) = c is true when x is 2 or 5. Thus, g is a right inverse for h iff g ( a ) ∈ {1, 4, 6} = h −1 [{ a }] and g ( b ) ∈ {3} = h −1 [{ b }] and g ( c ) ∈ {2, 5} = h −1 [{ c }]. For example, one right inverse is given by g ( a ) = 1, g ( b ) = 3, and g ( c ) = 2. Another right inverse is given by g 1 ( a ) = 4, g 1 ( b ) = 3, and g 1 ( c ) = 5. In this example, there are 3 · 1 · 2 = 6 possible right inverses for h . So, right inverses need not be unique . Also, no right inverse g can satisfy g h = Id X , since h is not a bijection. For instance, the right inverse g chosen above has ( g h )(4) = g ( h (4)) = g ( a ) = 1 ≠ 4 = Id X (4).

5.108 例。再次考虑图 5.10 中的函数 f 和 h。函数 g : {a, b, c} → {1, 2, 3, 4, 5, 6} 是 h 的右逆函数当且仅当 h ∘ g = Id { a , b , c } 当且仅当 h(g(a)) = a 且 h(g(b)) = b 且 h(g(c)) = c。现在,当 x 为 1、4 或 6 时,h(x) = a 成立。因此,我们可以将 g(a) 定义为这三个值中的任意一个,从而满足 h(g(a)) = a。类似地,当 x = 3 时,h(x) = b 成立;当 x 为 2 或 5 时,h(x) = c 成立。因此,g 是 h 的右逆当且仅当 g(a) ∈ {1, 4, 6} = h −1 [{a}] 且 g(b) ∈ {3} = h −1 [{b}] 且 g(c) ∈ {2, 5} = h −1 [{c}]。例如,一个右逆元由 g(a) = 1、g(b) = 3 和 g(c) = 2 给出。另一个右逆元由 g(a) = 4、g(b) = 3 和 g(c) = 5 给出。在这个例子中,h 有 3 × 1 × 2 = 6 个可能的右逆元。因此,右逆元不一定是唯一的。此外,没有右逆元 g 能满足 g ∘ h = Id(id_00007),因为 h 不是双射。例如,上面选择的右逆元 g 满足 (g ∘ h)(4) = g(h(4)) = g(a) = 1 ≠ 4 = Id(id_00008)(4)。

Furthermore, right inverses do not always exist . Suppose we try to find a right inverse g : { a , b , c , d , e } → {1, 2, 3} for the function f in the figure. Writing out the requirement f g = Id Y , we obtain (among other conditions) f ( g ( c )) = c . But there is no x in the domain of f solving f ( x ) = c , since f 1 [ { c } ] = 0 , so it is impossible to define g ( c ) in a way that makes f ( g ( c )) = c true. So f does not have any right inverses. This happened because f was not surjective: there exist points (such as c and e ) in the codomain of f that do not have the form f ( x ) for any x in the domain of f .

此外,右逆函数并非总是存在。假设我们试图找到图中函数 f 的右逆函数 g : {a, b, c, d, e} → {1, 2, 3}。写出条件 f ∘ g = Id Y ,我们得到(除其他条件外)f(g(c)) = c。但是 f 的定义域中不存在满足 f(x) = c 的 x,因为 f−1[{c}]=0,所以无法定义 g(c) 使得 f(g(c)) = c 成立。因此,f 没有右逆函数。这是因为 f 不是满射:在 f 的值域中存在一些点(例如 c 和 e),它们对于 f 定义域中的任何 x 都不满足 f(x) 的形式。

The previous example illustrates the following characterization of which functions have right inverses.

前面的例子说明了哪些函数具有右逆函数的以下特征。

5.109. Theorem on Right Inverses. For all functions f : X Y , f has a right inverse iff f is surjective.

5.109. 右逆定理。对于所有函数 f : X → Y,f 存在右逆当且仅当 f 是满射。

Proof. Fix f : X Y . First assume f has a right inverse g : Y X . Since f g = Id Y is a surjective function, Theorem 5.93 (e) tells us that f is surjective.

证明。固定 f : X → Y。首先假设 f 有一个右逆函数 g : Y → X。由于 f ∘ g = Id Y 是一个满射函数,定理 5.93(e) 告诉我们 f 是满射。

Conversely, assume f is surjective; we must construct a right inverse g : Y X . Informally, we can build g as follows. For each y Y , we know that the preimage f −1 [{ y }] is a nonempty subset of X , because f is surjective. Define g ( y ) to be any particular element of this nonempty set (chosen arbitrarily). To check that f g = Id Y , fix y Y , and compute ( f g )( y ) = f ( g ( y )) = y = Id Y ( y ). The equality f ( g ( y )) = y is true because g ( y ) ∈ f −1 [{ y }]. Thus the proof appears to be complete.

反之,假设 f 是满射;我们需要构造一个右逆 g : Y → X。非正式地,我们可以按如下方式构造 g。对于每个 y ∈ Y,我们知道原像 f −1 [{y}] 是 X 的一个非空子集,因为 f 是满射。定义 g(y) 为这个非空集合中的任意元素(任意选取)。为了验证 f ∘ g = Id Y ,固定 y ∈ Y,并计算 (f ∘ g)(y) = f(g(y)) = y = Id Y (y)。等式 f(g(y)) = y 成立,因为 g(y) ∈ f −1 [{y}]。因此,证明似乎完成。

[This paragraph should be considered optional.] There is a technical subtlety, however. To give a completely rigorous definition of g , we must construct the graph of g , which is a certain set of ordered pairs G = {( y , g ( y )) : y Y }. To prove that this set exists working within axiomatic set theory, we must use the Axiom of Choice (see §3.7). Specifically, let Z be the collection of all sets f −1 [{ y }], for y Y . The set Z exists by the Power Set Axiom and the Axiom of Specification, since we can write Z = { S P ( X ) : y Y , S = f 1 [ { y } ] } . Since f is surjective, the empty set is not a member of Z . As discussed in §3.7, the Axiom of Choice provides us with a set G ′ ⊆ Z × X such that for all ( S , x ) ∈ G ′, x S , and for each S Z , there exists a unique x X with ( S , x ) ∈ G ′. (The set G ′ is the formal mechanism by which we choose one element x from each nonempty set S = f −1 [{ y }].) Now define G = { ( y , x ) Y × X : S Z , S = f 1 [ { y } ] ( S , x ) G } . It must now be checked that g = ( Y , X , G ) is a function; the key point is that for each y Y , there exists a unique S Z with S = f −1 [{ y }], and hence there exists a unique x X with ( y , x ) ∈ G . Finally, by definition of G and Z , ( y , x ) ∈ G implies x f −1 [{ y }], so that f ( g ( y )) = f ( x ) = y for each y Y .

[本段可视为可选。] 然而,这里存在一个技术上的微妙之处。为了给出 g 的完全严格的定义,我们必须构造 g 的图,即有序对集合 G = {(y, g(y)) : y ∈ Y}。为了在公理化集合论框架内证明该集合的存在,我们必须使用选择公理(参见 §3.7)。具体来说,令 Z 为所有集合 f −1 [{y}] 的集合,其中 y ∈ Y。集合 Z 由幂集公理和规范公理确定存在,因为我们可以写成 Z={S∈P(X):∃y∈Y,S=f−1[{y}]}。由于 f 是满射,空集不属于 Z。如 §3.7 所述,选择公理为我们提供了一个集合 G′ ⊆ Z × X,使得对于所有 (S, x) ∈ G′,x ∈ S,以及对于每个 S ∈ Z,都存在唯一的 x ∈ X 使得 (S, x) ∈ G′。(集合 G′ 是我们从每个非空集合 S = f −1 [{y}] 中选择一个元素 x 的形式机制。)现在定义 G={(y,x)∈Y×X:∃S∈Z,S=f−1[{y}]∧(S,x)∈G′}。现在必须验证 g = (Y, X, G) 是一个函数;关键在于,对于每个 y ∈ Y,存在唯一的 S ∈ Z,使得 S = f −1 [{y}],因此存在唯一的 x ∈ X,使得 (y, x) ∈ G。最后,根据 G 和 Z 的定义,(y, x) ∈ G 蕴含 x ∈ f −1 [{y}],因此对于每个 y ∈ Y,都有 f(g(y)) = f(x) = y。

Section Summary

章节概要

  1. Inverse Functions. Given f : X Y , a function g : Y X is an inverse function for f iff x X , y Y , y = f ( x ) x = g ( y ) . Equivalently, g is an inverse function for f iff f g = Id Y and g f = Id X . Inverse functions are unique when they exist; the inverse of f is denoted f −1 .
    反函数。给定函数 f : X → Y,函数 g : Y → X 是 f 的反函数当且仅当 ∀x∈X,∀y∈Y,y=f(x)⇔x=g(y)。等价地,g 是 f 的反函数当且仅当 f ∘ g = Id Y 且 g ∘ f = Id X 。反函数存在时是唯一的;f 的反函数记为 f −1
  2. Properties of Invertibility. A function f : X Y is invertible iff f is a bijection. Identity functions are invertible. The inverse of an invertible function is invertible. The composition of invertible functions f : X Y and h : Y Z is invertible, and ( h f ) −1 = f −1 h −1 .
    可逆性的性质。函数 f : X → Y 可逆当且仅当 f 是双射。恒等函数可逆。可逆函数的逆函数也是可逆的。可逆函数 f : X → Y 和 h : Y → Z 的复合函数也是可逆的,且 (h ∘ f) −1 = f −1 ∘ h −1
  3. Left Inverses. A left inverse of f : X Y is a function g : Y X such that g f = Id X . Assuming X is nonempty, f has a left inverse iff f is injective. Left inverses are not unique in general.
    左逆。函数 f : X → Y 的左逆是一个函数 g : Y → X,使得 g ∘ f = Id X 。假设 X 非空,则 f 存在左逆当且仅当 f 是单射。一般来说,左逆不唯一。
  4. Right Inverses. A right inverse of f : X Y is a function g : Y X such that f g = Id Y . A function f has a right inverse iff f is surjective. Right inverses are not unique in general.
    右逆函数。函数 f : X → Y 的右逆函数是函数 g : Y → X,使得 f ∘ g = Id Y 。函数 f 存在右逆函数当且仅当 f 是满射。一般来说,右逆函数不唯一。

Exercises

练习

  1. Let A = {1, 2, 3, 4}, B = { v , w , x , y , z }, and consider the following four functions:
    f : A B defined by f ( 1 ) = w , f ( 2 ) = z , f ( 3 ) = y , f ( 4 ) = w ; g : B A defined by g ( v ) = 4 , g ( w ) = 3 , g ( y ) = 1 , g ( x ) = 2 , g ( z ) = 4 ; h : A A defined by h ( 1 ) = 4 , h ( 2 ) = 4 , h ( 3 ) = 1 , h ( 4 ) = 3 ; k : B B defined by k ( v ) = z , k ( z ) = y , k ( y ) = v , k ( w ) = x , k ( x ) = w .
    Which functions have: (a) a left inverse? (b) a right inverse? (c) a two-sided inverse? Give an example of each type of inverse when it exists.
    设 A = {1, 2, 3, 4},B = {v, w, x, y, z},并考虑以下四个函数:f:A→B,定义为 f(1)=w, f(2)=z, f(3)=y, f(4)=w;g:B→A,定义为 g(v)=4, g(w)=3, g(y)=1, g(x)=2, g(z)=4;h:A→A,定义为 h(1)=4, h(2)=4, h(3)=1, h(4)=3;k:B→B,定义为 k(v)=z, k(z)=y, k(y)=v, k(w)=x, k(x)=w。哪些函数具有:(a) 左逆函数?(b) 右逆函数?(c) 双侧逆函数?如果存在,请分别举例说明。
  2. For each of the following functions from calculus, say whether the function has a left, right, or two-sided inverse. Give an example of each type of inverse when it exists. (a) f : R R 0 given by f ( x ) = x 2 . (b) f : R R given by f ( x ) = cos x . (c) f : R [ 1 , 1 ] given by f ( x ) = cos x . (d) f : ( 0 , π / 2 ) R given by f ( x ) = cos x . (e) f : [0, π ] → [ − 1, 1] given by f ( x ) = cos x . (f) f : R R given by f ( x ) = x 3 x . (g) f : R 0 R given by f ( x ) = 1/(1 + x 2 ). (h) f : R 0 ( 0 , 1 ] given by f ( x ) = 1/(1 + x 2 ).
    对于下列微积分函数,请说明该函数是否存在左逆函数、右逆函数或双侧逆函数。如果存在,请分别举例说明每种逆函数。 (a) f:R→R≥0,由 f(x) = x 2 . 给出 (b) f:R→R,由 f(x) = cos x. 给出 (c) f:R→[−1,1],由 f(x) = cos x. 给出 (d) f:(0,π/2)→R,由 f(x) = cos x. 给出 (e) f:[0,π]→[−1,1],由 f(x) = cos x. 给出 (f) f:R→R,由 f(x) = x 3 − x. 给出 (g) f:R≥0→R,由 f(x) = 1/(1 + x 2 ). (h) f:R≤0→(0,1] 由 f(x) = 1/(1 + x 2 ).
  3. Let A = {1, 2, 3, 4, 5, 6} and B = { w , x , y , z }. Consider the functions:
    f : A B with graph { ( 1 , x ) , ( 2 , z ) , ( 3 , w ) , ( 4 , x ) , ( 5 , z ) , ( 6 , y ) } ; g : B A with graph { ( w , 4 ) , ( x , 6 ) , ( y , 1 ) , ( z , 3 ) } ; h : A A with graph { ( 1 , 3 ) , ( 2 , 4 ) , ( 3 , 6 ) , ( 4 , 2 ) , ( 5 , 1 ) , ( 6 , 5 ) } ; k : B B with graph { ( w , w ) , ( x , w ) , ( y , w ) , ( z , z ) } .
    (a) For each function, find all possible left inverses for the function, or explain why none exist. (b) Repeat part (a) for right inverses. (c) Repeat part (a) for two-sided inverses.
    设 A = {1, 2, 3, 4, 5, 6},B = {w, x, y, z}。考虑以下函数:f:A→B,其图像为 {(1,x),(2,z),(3,w),(4,x),(5,z),(6,y)};g:B→A,其图像为 {(w,4),(x,6),(y,1),(z,3)};h:A→A,其图像为 {(1,3),(2,4),(3,6),(4,2),(5,1),(6,5)};k:B→B,其图像为 {(w,w),(x,w),(y,w),(z,z)}。(a) 对于每个函数,找出其所有可能的左逆函数,或者解释为什么不存在左逆函数。(b) 对右逆函数重复(a)部分。(c) 对双侧逆函数重复(a)部分。
  4. For each of the following functions, say whether the function has a left, right, or two-sided inverse. Give an example of each type of inverse when it exists. (a) f : Z Z given by f ( n ) = 2 n . (b) f : Z 0 Z 0 given by f ( n ) = n + 1. (c) f : Z Z given by f ( n ) = n + 1. (d) f : Z Z given by f ( n ) = − n . (e) f : Z 0 Z 0 given by f ( n ) = n 2 . (f) f : Z 0 Z given by f ( n ) = n . (g) f : Z 0 Z 0 given by f (0) = 0 and f ( n ) = n − 1 for n > 0.
    对于下列每个函数,说明该函数是否有左逆函数、右逆函数或双侧逆函数。给出每种逆函数的例子(如果存在)。 (a) f:Z→Z,其中 f(n) = 2n。 (b) f:Z≥0→Z≥0,其中 f(n) = n + 1。 (c) f:Z→Z,其中 f(n) = n + 1。 (d) f:Z→Z,其中 f(n) = −n。 (e) f:Z≥0→Z≥0,其中 f(n) = n。 2 (f) f:Z≥0→Z,其中 f(n) = n。 (g) f:Z≥0→Z≥0,其中 f(0) = 0 且 f(n) =对于 n > 0,n − 1。
  5. Define f : Z 0 { 0 , 1 , 2 , , 9 } by letting f ( n ) be the last digit in the base-10 expansion of n . For example, f (1407) = 7. Describe three different right inverses for f .
    定义函数 f:Z≥0→{0,1,2,…,9},其中 f(n) 为 n 的十进制展开式的最后一位数字。例如,f(1407) = 7。描述 f 的三个不同的右逆函数。
  6. (a) Finish Part 1 of the proof of Theorem 5.99 by showing that g f = Id X . (b) Finish the proof of part (c) of Theorem 5.103 .
    (a)完成定理 5.99 的证明第 1 部分,证明 g ∘ f = Id X 。(b)完成定理 5.103 的证明第 (c) 部分。
  7. Let B be a fixed set, and let f : 0 B be the empty function. Under what conditions does f have a left, right, or two-sided inverse?
    设 B 为一个固定的集合,f:0→B 为空函数。在什么条件下 f 存在左逆、右逆或双侧逆?
  8. Suppose f : A R is a function. Find necessary and sufficient conditions to ensure that there exists a function g : A R such that f · g = g · f = 1. (The dot denotes pointwise product of functions, and 1 denotes the constant function on A with value 1.)
    假设 f:A→R 是一个函数。求保证存在函数 g:A→R 使得 f · g = g · f = 1 的充要条件。(点表示函数的逐点乘积,1 表示在 A 上值为 1 的常数函数。)
  9. Suppose f : R R is a function. Find necessary and sufficient conditions to ensure that there exists a function g : R R such that f g = g f = Id R .
    假设 f:R→R 是一个函数。求保证存在函数 g:R→R 使得 f⋅g=g⋅f=IdR 的充要条件。
  10. Suppose f : R R is a function. (a) Under what conditions does there exist g : R R such that f g = 1 (the constant function 1)? Describe all functions g that satisfy this property (when such g exist). (b) Under what conditions does there exist g : R R such that g f = 1? Describe all functions g that satisfy this property (when such g exist).
    假设 f:R→R 是一个函数。(a) 在什么条件下存在 g:R→R 使得 f ∘ g = 1(常数函数 1)?描述所有满足此性质的函数 g(如果存在)。(b) 在什么条件下存在 g:R→R 使得 g ∘ f = 1?描述所有满足此性质的函数 g(如果存在)。
  11. Suppose f : X Y and g , h : Y Z are functions. Prove: if f has a right inverse and g f = h f , then g = h . Give an example to show that this right cancellation property can fail if f does not have a right inverse.
    假设 f : X → Y 和 g, h : Y → Z 是函数。证明:如果 f 有右逆函数且 g ∘ f = h ∘ f,则 g = h。举例说明当 f 没有右逆函数时,右消去性质可能不成立。
  12. Suppose f : X Y and g , h : W X are functions. Prove: if f has a left inverse and f g = f h , then g = h . Give an example to show that this left cancellation property can fail if f does not have a left inverse.
    假设 f : X → Y 和 g, h : W → X 是函数。证明:如果 f 有左逆函数且 f ∘ g = f ∘ h,则 g = h。举例说明当 f 没有左逆函数时,左约分性质可能不成立。
  13. (a) Suppose f : X Y and g : Y Z both possess left inverses. Show that g f also has a left inverse. (b) Prove a similar result for right inverses.
    (a) 假设函数 f : X → Y 和 g : Y → Z 都存在左逆函数。证明 g ∘ f 也存在左逆函数。(b) 证明右逆函数也存在类似的结论。
  14. Suppose n Z > 0 , X 0 , …, X n are sets, and f i : X i −1 X i are bijections for 1 ≤ i n . Use induction on n to prove that ( f n ∘ · · · ∘ f 1 ) −1 exists, and ( f n f 2 f 1 ) 1 = f 1 1 f 2 1 f n 1 .
    假设 n∈Z>0,X 0 , …, X n 是集合,且 f i : X i −1 → X i 是 1 ≤ i ≤ n 的双射。利用归纳法证明 (f n ∘ · · · ∘ f 1 ) −1 存在,且 (fn∘⋯∘f2∘f1)−1=f1−1∘f2−1⋯∘fn−1。
  15. Prove: for all sets X , Y and all f : X Y , g : Y X , g is a left inverse of f iff ( x X , y Y , y = f ( x ) x = g ( y )).
    证明:对于所有集合 X、Y 和所有 f : X → Y、g : Y → X,g 是 f 的左逆当且仅当 (∀x∈X,∀y∈Y,y=f(x)⇒x=g(y))。
  16. Prove: for all sets X , Y and all f : X Y , g : Y X , g is a right inverse of f iff ( x X , y Y , x = g ( y ) y = f ( x )).
    证明:对于所有集合 X、Y 和所有 f : X → Y、g : Y → X,g 是 f 的右逆当且仅当 (∀x∈X,∀y∈Y,x=g(y)⇒y=f(x))。
  17. Use right inverses to prove: for all nonempty sets X , Y , Z , all f : X Y , and all g : Y Z , if g f is one-to-one and f is onto, then g is one-to-one.
    利用右逆证明:对于所有非空集合 X、Y、Z,所有 f : X → Y,以及所有 g : Y → Z,如果 g ∘ f 是单射且 f 是满射,则 g 也是单射。
  18. Use left inverses to prove: for all nonempty sets X , Y , Z , all f : X Y , and all g : Y Z , if g f is surjective and g is injective, then f is surjective.
    利用左逆证明:对于所有非空集合 X、Y、Z,所有 f : X → Y,以及所有 g : Y → Z,如果 g ∘ f 是满射且 g 是单射,则 f 是满射。
  19. Suppose f is an injection mapping an n -element set into an m -element set, where m > n . How many left inverses does f have?
    假设 f 是一个单射,将一个 n 元素集合映射到一个 m 元素集合,其中 m > n。f 有多少个左逆?
  20. Let f : X Y be a surjection. Suppose Y = { y 1 , y 2 , …, y m }, and suppose that, for 1 ≤ i m , there are exactly n i elements x X with f ( x ) = y i . How many right inverses does f have?
    设 f : X → Y 为满射。假设 Y = {y 1 , y 2 , …, y m ,并且对于 1 ≤ i ≤ m,恰好存在 n i 个元素 x ∈ X 使得 f(x) = y i 。f 有多少个右逆?
  21. Let X be an n -element set. How many bijections are there from X onto X ?
    设 X 是一个包含 n 个元素的集合。从 X 到 X 的双射有多少个?
  22. Suppose f : X Y is a function with graph F . Prove that the inverse relation F −1 is the graph of a function (not necessarily a function from Y to X ) iff f is injective.
    假设 f : X → Y 是一个函数,其图像为 F。证明反关系 F −1 是一个函数的图像(不一定是从 Y 到 X 的函数),当且仅当 f 是单射。
  23. Suppose f : X Y is a function with graph F . Prove that the inverse relation F −1 contains the graph of some function g : Y X iff f is surjective.
    假设 f : X → Y 是一个函数,其图像为 F。证明反关系 F −1 包含某个函数 g : Y → X 的图像当且仅当 f 是满射。
  24. Prove: if f : X Y has exactly one right inverse g , then f is invertible and g = f −1 .
    证明:如果 f : X → Y 恰好有一个右逆 g,则 f 可逆且 g = f −1
  25. If f : X Y has exactly one left inverse, must f be invertible? What if X has more than one element?
    如果函数 f : X → Y 恰好有一个左逆函数,那么 f 一定是可逆的吗?如果 X 包含多个元素呢?
#

6

Equivalence Relations and Partial Orders

等价关系和偏序关系

6.1  Reflexive, Symmetric, and Transitive Relations

6.1 自反关系、对称关系和传递关系

In this chapter, we continue our study of relations. So far, we have been thinking of a relation R as a generalization of a function, where ( x , y ) ∈ R means that the input x is associated with the output y under R . Now, we change our viewpoint slightly, interpreting ( x , y ) ∈ R to mean that x is “related” to y in some way, without insisting on the notion that x is the “input” and y is the “output.” For example, ( x , y ) ∈ R might mean x is equal to y , or x is greater than y , or x is a subset of y , or x is a sibling of y , and so on. We begin this section by developing a new way of visualizing R , called a digraph , that emphasizes this “relational” viewpoint. Later, we will be especially interested in relations that share certain properties with the equality relation, called reflexivity, symmetry, and transitivity. We define these properties here and give many examples.

在本章中,我们将继续学习关系。到目前为止,我们一直将关系 R 视为函数的推广,其中 (x, y) ∈ R 表示在 R 下输入 x 与输出 y 相关联。现在,我们稍微改变一下视角,将 (x, y) ∈ R 解释为 x 与 y 之间存在某种“关联”,而不再强调 x 是“输入”,y 是“输出”。例如,(x, y) ∈ R 可能表示 x 等于 y,或者 x 大于 y,或者 x 是 y 的子集,或者 x 是 y 的兄弟元素,等等。本节首先介绍一种新的关系可视化方法,称为有向图,它强调了这种“关系”视角。稍后,我们将重点关注与相等关系共享某些性质的关系,这些性质称为自反性、对称性和传递性。我们将在此定义这些性质并给出许多示例。

Digraph of a Relation

关系的有向图

Previously, we looked at relations R from X to Y , where X was an “input set” and Y was an “output set.” Now, we focus attention on a single set of objects X , thinking of R as specifying a relationship that may or may not hold between any two given objects in X . This leads to the following definitions.

之前,我们研究了从 X 到 Y 的关系 R,其中 X 是“输入集”,Y 是“输出集”。现在,我们将注意力集中在单个对象集 X 上,并将 R 视为指定 X 中任意两个给定对象之间可能存在也可能不存在的关系。由此得出以下定义。

6.1. Definition: Relation on a Set X .. For all sets X and R , R is a relation on X iff R X × X . For all objects x and y , we write x R y iff ( x , y ) R . The notation xRy is called infix notation and can be read “ x is related to y under R .”

6.1 定义:集合 X 上的关系。对于任意集合 X 和 R,R 是 X 上的关系当且仅当 R⊆X×X。对于任意对象 x 和 y,我们记 xRy 当且仅当 (x,y)∈R。记号 xRy 称为中缀记号,可以读作“x 在 R 下与 y 相关”。

6.2 Example. Let X = {1, 2, 3}. The following sets of ordered pairs are relations on X :

6.2 示例。设 X = {1, 2, 3}。下列有序对集合是 X 上的关系:

R 1 = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) } ; R 2 = { ( 1 , 2 ) , ( 2 , 3 ) , ( 1 , 3 ) } ; R 3 = R 1 R 2 ;
R 4 = { ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 ) } ; R 5 = { ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) } ; R 6 = { ( 1 , 3 ) , ( 2 , 3 ) } .

Relation R 1 was formerly called the identity relation I X on the set X . Now, we can think of R 1 as the equality relation on this set, since for all x , y X , ( x , y ) ∈ R 1 iff x = y . Similarly, R 2 is the less-than relation on X , since for all x , y X , ( x , y ) ∈ R 2 iff x < y . Next, for all x , y X , xR 3 y iff x y . We see that the notation xRy for relations imitates the way we write arithmetical relations such as =, <, and ≤.

关系 R 1 以前被称为集合 X 上的恒等关系 I X 。现在,我们可以将 R 1 视为该集合上的相等关系,因为对于所有 x, y ∈ X,(x, y) ∈ R 1 当且仅当 x = y。类似地,R 2 是 X 上的小于关系,因为对于所有 x, y ∈ X,(x, y) ∈ R 2 当且仅当 x < y。接下来,对于所有 x, y ∈ X,xR 3 y 当且仅当 x ≤ y。我们可以看到,关系的符号 xRy 模仿了我们书写算术关系(例如 =、< 和 ≤)的方式。

Turning to R 4 , we have xR 4 y iff x + y = 4, for all x , y X . As for R 5 , note xR 5 y iff x = 1, so that this “relation” between x and y does not actually involve y . Relation R 6 does not describe any particularly meaningful relationship between x and y ; it illustrates that a relation on X can be any set of ordered pairs contained in X × X .

对于关系 R 4 ,我们有 xR 4 y 当且仅当对于所有 x, y ∈ X,x + y = 4。对于关系 R 5 ,注意到 xR 5 y 当且仅当 x = 1,因此 x 和 y 之间的这种“关系”实际上并不涉及 y。关系 R 6 并没有描述 x 和 y 之间任何特别有意义的关系;它只是说明 X 上的关系可以是包含在 X × X 中的任意有序对集合。

Earlier, we introduced arrow diagrams and (Cartesian) graphs as ways to visualize relations from X to Y . When R is a relation on a finite set X , we can instead draw a modified version of the arrow diagram called the digraph (directed graph) of the relation. We make a single copy of the set X , drawing a dot labeled x for each object x X . For each ordered pair ( x , y ) ∈ R , we draw an arrow from x to y . If y = x , this arrow becomes a little loop that starts and ends at x . Figure 6.1 displays the digraphs for the six relations in Example 6.2 .

前面我们介绍了箭头图和(笛卡尔)图,作为可视化从 X 到 Y 的关系的方法。当 R 是有限集合 X 上的关系时,我们可以绘制箭头图的改进版本,称为该关系的有向图(digraph)。我们复制集合 X,并在其中为每个对象 x ∈ X 绘制一个标记为 x 的点。对于每个有序对 (x, y) ∈ R,我们绘制一条从 x 到 y 的箭头。如果 y = x,则该箭头变成一个以 x 为起点和终点的环。图 6.1 展示了例 6.2 中六个关系的有向图。

fig6_1

Figure 6.1

图 6.1

Digraphs for six relations on X = {1, 2, 3}.

X = {1, 2, 3} 上的六个关系的有向图。

6.3 Example. Define a relation R on the set X = {1, 2, 3, 4, 5, 6} by

6.3 示例。定义集合 X = {1, 2, 3, 4, 5, 6} 上的关系 R,如下所示:

R = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( 4 , 4 ) , ( 2 , 5 ) , ( 5 , 2 ) , ( 1 , 3 ) , ( 3 , 4 ) , ( 4 , 1 ) , ( 2 , 6 ) } .

Figure 6.2 displays both the arrow diagram and the digraph for this relation. We will see that certain structural features of R can be seen more easily in the digraph picture.

图 6.2 展示了该关系的箭头图和有向图。我们将看到,在有向图中可以更清晰地观察到关系 R 的某些结构特征。

fig6_2

Figure 6.2

图 6.2

Arrow diagram and digraph for a relation R .

关系 R 的箭头图和有向图。

Reflexive Relations

反身关系

Now we introduce three properties of abstract relations that are based on corresponding properties of logical equality. The first property, called reflexivity , comes from the fact that x = x for all objects x .

现在我们介绍抽象关系的三个性质,它们都基于逻辑相等性的相应性质。第一个性质称为自反性,它源于对于所有对象 x,都有 x = x。

6.4. Definition: Reflexive Relations. For all relations R and all sets X :

6.4. 定义:自反关系。对于所有关系 R 和所有集合 X:

R is reflexive 反身 on X X iff 当……时 x x X X , ( x x , x x ) R .

Using infix notation, we can also say that R is reflexive on X iff for all x X , xRx .

使用中缀表示法,我们还可以说 R 在 X 上是自反的,当且仅当对于所有 x ∈ X,xRx。

Informally, R is reflexive on X iff every object x in X is related to itself under R . This means that the digraph for R has a loop at every vertex in the set X . Applying the negation rules, we see that R is n o t reflexive on X iff x X , ( x , x ) R .

通俗地说,R 在 X 上是自反的,当且仅当 X 中的每个对象 x 在 R 下都与自身相关。这意味着 R 的有向图在集合 X 的每个顶点处都有一个环。应用否定规则,我们看到 R 在 X 上不是自反的,当且仅当存在 x∈X,(x,x)∉R。

6.5 Example. Relations R 1 and R 3 in Figure 6.1 are reflexive on X = {1, 2, 3}, but the other four relations are not. In particular, even though (2, 2) ∈ R 4 , R 4 is not reflexive on X because (1, 1) is not in R 4 . Similarly, the relation R in Figure 6.2 is not reflexive on the set X = {1, 2, 3, 4, 5, 6} because ( 5 , 5 ) R . However, R is reflexive on the smaller set X ′ = {1, 2, 3, 4}, although R is not a relation on X ′. Thus, reflexivity is a property of both the relation R and the set X . Usually, however, we are dealing with relations on a fixed set X , and we are only interested in reflexivity with respect to the full set X . In such circumstances, we often say “ R is reflexive” as an abbreviation for “ R is reflexive on X .”

6.5 示例。图 6.1 中的关系 R 1 和 R 3 在 X = {1, 2, 3} 上具有自反性,但其他四个关系则不具有自反性。特别地,尽管 (2, 2) ∈ R 4 ,但 R 4 在 X 上不具有自反性,因为 (1, 1) 不属于 R 4 。类似地,图 6.2 中的关系 R 在集合 X = {1, 2, 3, 4, 5, 6} 上不具有自反性,因为 (5, 5) ∉ R。然而,R 在较小的集合 X′ = {1, 2, 3, 4} 上具有自反性,尽管 R 不是 X′ 上的关系。因此,自反性既是关系 R 的属性,也是集合 X 的属性。然而,通常情况下,我们处理的是定义在固定集合 X 上的关系,并且我们只关心关系 R 对整个集合 X 的自反性。在这种情况下,我们通常用“R 是自反的”来简写“R 在 X 上是自反的”。

6.6 Example. Suppose X is any collection of sets, and R is the relation on X defined by ( A , B ) ∈ R iff A B , for all A , B X . This “subset relation” R is reflexive on X , because A A for every set A .

6.6 示例。假设 X 是任意集合族,R 是 X 上的关系,定义为 (A, B) ∈ R 当且仅当 A⊆ B,对于所有 A, B ∈ X。这个“子集关系”R 在 X 上是自反的,因为对于每个集合 A,都有 A⊆ A。

6.7 Example. For any set X , the equality (or identity) relation I X = {( x , x ): x X } is reflexive on X . The product set X × X is a relation on X that is reflexive on X . The empty set is a relation on X that is not reflexive on X , assuming X itself is nonempty.

6.7 示例。对于任意集合 X,等式(或恒等)关系 I X = {(x, x):x ∈ X} 在 X 上是自反的。乘积集 X × X 是 X 上的一个关系,它在 X 上是自反的。空集 ∅ 是 X 上的一个关系,它不是自反的,假设 X 本身非空。

6.8 Example. Define relations R = { ( x , y ) Z 2 : x + y is even } , S = {( x , y ) ∈ [ − 1, 1] 2 : x 2 + y 2 = 1}, and T = { ( A , B ) P ( Z ) 2 : 8 A B } . Determine (with proof) which of these relations are reflexive. Solution. First, we show R is reflexive on Z . Fix x Z ; we must prove ( x , x ) ∈ R , which means x + x is even. Since x + x = 2 x where x is an integer, x + x is even. Second, we show S is not reflexive on [ − 1, 1]. Pick x = 1 ∈ [ − 1, 1], and note ( x , x ) S because x 2 + x 2 = 2 ≠ 1. Third, we show T is reflexive on P ( Z ) . Fix A P ( Z ) ; we must prove ( A , A ) ∈ T , which means 8 A A . Since A A = by the Theorem on Set Equality, 8 is not in this set.

6.8 例。定义关系 R={(x,y)∈Z²:x+y为偶数},S = {(x,y)∈[-1,1] 2 :x 2 + y 2 = 1},以及 T={(A,B)∈P(Z)²:8∉A−B}。确定(并证明)这些关系中哪些是自反的。解:首先,我们证明 R 在 Z 上是自反的。固定 x∈Z;我们必须证明 (x,x)∈R,这意味着 x + x 是偶数。因为 x + x = 2x,其中 x 是整数,所以 x + x 是偶数。其次,我们证明 S 在 [-1,1] 上不是自反的。选取 x = 1 ∈ [ − 1, 1],并注意到 (x,x)∉S,因为 x 2 + x 2 = 2 ≠ 1。第三,我们证明 T 在 P(Z) 上是自反的。固定 A∈P(Z);我们必须证明 (A, A) ∈ T,这意味着 8∉A−A。由于根据集合等式定理,A−A=∅,因此 8 不属于该集合。

Symmetric Relations

对称关系

We next generalize the symmetry property of equality, which says that for all objects x , y , if x = y then y = x .

接下来,我们将等式的对称性质推广,即对于所有对象 x、y,如果 x = y,则 y = x。

6.9. Definition: Symmetric Relations.. For all relations R :

6.9 定义:对称关系。对于所有关系 R:

R is symmetric 对称 iff 当……时 x x , y , ( x x , y ) R ( y , x x ) R .

Using infix notation, we can also say R is symmetric iff for all x and y , xRy implies yRx .

使用中缀表示法,我们还可以说 R 是对称的,当且仅当对于所有 x 和 y,xRy 蕴含 yRx。

Thus, symmetry of R means that whenever x is related to y , y is also related to x . Pictorially, this means that for any arrow from x to y in the digraph for R , the reversed arrow from y back to x must also appear in the digraph. Applying the negation rules, we see that R is n o t symmetric iff x , y , ( x , y ) R ( y , x ) R .

因此,R 的对称性意味着,只要 x 与 y 相关,y 也与 x 相关。形象地说,这意味着对于 R 的有向图中从 x 到 y 的任何箭头,从 y 到 x 的反向箭头也必须出现在该有向图中。应用否定规则,我们发现 R 非对称当且仅当 ∃x,∃y,(x,y)∈R∧(y,x)∉R。

6.10 Example. In Figure 6.1 , relations R 1 and R 4 are symmetric, but the other relations are not. For instance, R 2 is not symmetric because (1, 2) ∈ R 2 but ( 2 , 1 ) R 2 . The relation R in Figure 6.2 is not symmetric because, for instance, (2, 6) ∈ R but ( 6 , 2 ) R .

6.10 示例。在图 6.1 中,关系 R 1 和 R 4 是对称的,但其他关系不是。例如,R 2 不是对称的,因为 (1, 2) ∈ R 2 但 (2,1)∉R2。图 6.2 中的关系 R 也不是对称的,例如,(2, 6) ∈ R 但 (6,2)∉R。

6.11 Example. Let X be the set of all subsets of {1, 2, 3}, and let R be the subset relation given by aRb iff a b (for a , b X ). R is not symmetric, since {1}⊆{1, 2} but { 1 , 2 } { 1 } .

6.11 例。设 X 为 {1, 2, 3} 的所有子集的集合,R 为子集关系,定义为 aRb 当且仅当 a⊆ b(对于 a, b ∈ X)。R 不是对称的,因为 {1}⊆{1, 2} 但 {1,2}⊈{1}。

6.12 Example. For any set X , the equality relation I X is symmetric. X × X is also symmetric. The empty relation is symmetric, since the implication ( x , y ) ( y , x ) has a false hypothesis and is therefore true. The relation < on R is not symmetric since 3 < 5 is true but 5 < 3 is false.

6.12 示例。对于任意集合 X,等式关系 I X 是对称的。X × X 也是对称的。空关系 ∅ 是对称的,因为蕴含式 (x,y)∈∅⇒(y,x)∈∅ 的假设为假,因此为真。R 上的关系 < 不是对称的,因为 3 < 5 为真,但 5 < 3 为假。

6.13 Example. Decide (with proof) whether the relations R , S , and T from Example 6.8 are symmetric. Solution. First, we prove R is symmetric. Fix arbitrary ( x , y ); assume ( x , y ) ∈ R ; prove ( y , x ) ∈ R . We have assumed x , y Z and x + y is even; we must prove x , y Z and y + x is even. Since x + y = y + x , the conclusion holds. Second, we prove S is symmetric. Fix arbitrary ( x , y ) R 2 ; assume ( x , y ) ∈ S ; prove ( y , x ) ∈ S . We have assumed x 2 + y 2 = 1; we must prove y 2 + x 2 = 1. Since x 2 + y 2 = y 2 + x 2 , this conclusion holds. Third, we prove T is not symmetric. Choose A = {1, 2, 3} and B = {6, 7, 8}, which are elements of P ( Z ) . Note ( A , B ) ∈ T , since A B = A , and 8 A . But ( B , A ) T , since B A = B , and 8 ∈ B .

6.13 例。判断(并证明)例 6.8 中的关系 R、S 和 T 是否对称。解。首先,我们证明 R 是对称的。固定任意 (x, y);假设 (x, y) ∈ R;证明 (y, x) ∈ R。我们假设 x, y ∈ Z 且 x + y 为偶数;我们必须证明 x, y ∈ Z 且 y + x 为偶数。由于 x + y = y + x,结论成立。其次,我们证明 S 是对称的。固定任意 (x, y) ∈ R²;假设 (x, y) ∈ S;证明 (y, x) ∈ S。我们假设 x 2 + y 2 = 1;我们必须证明 y 2 + x 2 = 1。由于 x 2 + y 2 = y 2 + x 2 ,因此该结论成立。第三,我们证明 T 不是对称的。选取 A = {1, 2, 3} 和 B = {6, 7, 8},它们都是 P(Z) 的元素。注意 (A, B) ∈ T,因为 A − B = A,且 8∉ A。但 (B,A)∉ T,因为 B − A = B,且 8 ∈ B。

Transitive Relations

传递关系

The transitive property of equality says that for all x , y , z , if x = y and y = z , then x = z . This leads to the following definition.

等式的传递性表明,对于任意 x、y、z,如果 x = y 且 y = z,则 x = z。由此可得出以下定义。

6.14. Definition: Transitive Relations.. For all relations R :

6.14. 定义:传递关系。对于所有关系 R:

R is transitive 及物动词 iff 当……时 x x , y , z z , if 如果 ( x x , y ) R and ( y , z z ) R , then 然后 ( x x , z z ) R .

Using infix notation, we can say R is transitive iff for all x , y , z , if xRy and yRz , then xRz .

使用中缀表示法,我们可以说 R 是传递的,当且仅当对于所有 x、y、z,如果 xRy 且 yRz,则 xRz。

Transitivity means that whenever x is related to y and y is related to z , then x must be related to z . To detect transitivity from a digraph, we must check that whenever there are arrows from x to y to z , there is also an arrow directly from x to z . We must be sure to check all cases, including those where x = z . By the negation rules, R is n o t transitive iff x , y , z , ( x , y ) R ( y , z ) R ( x , z ) R .

传递性是指,当 x 与 y 相关且 y 与 z 相关时,x 必然与 z 相关。为了从有向图中检测传递性,我们必须检查是否存在从 x 到 y 再到 z 的箭头,并且是否存在一条从 x 直接指向 z 的箭头。我们必须确保检查所有情况,包括 x = z 的情况。根据否定规则,R 不具有传递性当且仅当存在 x, y, z,且 (x, y)∈R ∧ (y, z)∈R ∧ (x, z)∉R。

6.15 Example. All of the relations in Figure 6.1 are transitive except for R 4 . We see that R 4 is not transitive because (1, 3) ∈ R 4 and (3, 1) ∈ R 4 , but ( 1 , 1 ) R 4 . Relation R 6 is transitive because there are no choices for x , y , z that make the hypothesis ( x , y ) ∈ R 6 and ( y , z ) ∈ R 6 true; so the IF-statement defining transitivity is true. The relation R in Figure 6.2 is not transitive because, for example, (5, 2) ∈ R and (2, 6) ∈ R but ( 5 , 6 ) R .

6.15 示例。图 6.1 中的所有关系除了 R 4 之外都是传递的。我们看到 R 4 不是传递的,因为 (1, 3) ∈ R 4 且 (3, 1) ∈ R 4 ,但 (1,1)∉R4。关系 R 6 是传递的,因为不存在任何 x、y、z 的组合使得假设 (x, y) ∈ R 6 且 (y, z) ∈ R 6 为真;因此定义传递性的 IF 语句为真。图 6.2 中的关系 R 不是传递的,例如,(5, 2) ∈ R 且 (2, 6) ∈ R,但 (5,6)∉R。

6.16 Example. If X is any collection of sets, the subset relation ⊆ on X is transitive since A B and B C imply A C . For any set X , the equality relation I X is transitive, as is the relation X × X . The empty set is a transitive relation, by the definition of IF. The relation < on R is transitive as well.

6.16 示例。如果 X 是任意集合族,则 X 上的子集关系 ⊆ 是传递的,因为 A⊆ B 且 B⊆ C 蕴含 A⊆ C。对于任意集合 X,等式关系 I X 是传递的,关系 X × X 也是如此。根据 IF 的定义,空集是传递关系。R 上的关系 < 也是传递的。

6.17 Example. Decide (with proof) whether the relations R , S , and T from Example 6.8 are transitive. Solution. First, we show R is transitive. Fix x , y , z Z , assume ( x , y ) ∈ R and ( y , z ) ∈ R , and prove ( x , z ) ∈ R . We have assumed x + y is even and y + z is even, which means x + y = 2 a and y + z = 2 b for some integers a , b . We must prove x + z is even, which means c Z , x + z = 2 c . Compute x + z = (2 a y ) + (2 b y ) = 2( a + b y ), where a + b y Z . So we may choose c = a + b y to see that x + z is even. Second, we show S is not transitive. Choose x = 0, y = 1, and z = 0. Then ( x , y ) ∈ S because 0 2 + 1 2 = 1; ( y , z ) ∈ S because 1 2 + 0 2 = 1; but ( x , z ) S because 0 2 + 0 2 ≠ 1.

6.17 例。判断(并证明)例 6.8 中的关系 R、S 和 T 是否具有传递性。解:首先,我们证明 R 具有传递性。固定 x, y, z∈Z,假设 (x, y) ∈ R 且 (y, z) ∈ R,并证明 (x, z) ∈ R。我们假设 x + y 为偶数且 y + z 为偶数,这意味着对于某些整数 a 和 b,有 x + y = 2a 且 y + z = 2b。我们必须证明 x + z 为偶数,这意味着存在 c∈Z,x + z = 2c。计算 x + z = (2a − y) + (2b − y) = 2(a + b − y),其中 a + b − y∈Z。因此,我们可以选择 c = a + b − y 来证明 x + z 为偶数。其次,我们证明 S 不具有传递性。选择 x = 0,y = 1,z = 0。则 (x, y) ∈ S,因为 0 2 + 1 2 = 1;(y, z) ∈ S,因为 1 2 + 0 2 = 1;但 (x,z)∉S,因为 0 2 + 0 2 ≠ 1。

Third, we show T is transitive using a proof by contradiction. Assume, to get a contradiction, that there exist A , B , C P ( Z ) with ( A , B ) ∈ T and ( B , C ) ∈ T and ( A , C ) T . By definition, this means 8 A B and 8 B C and 8 ∈ A C . The last condition tells us that 8 ∈ A and 8 C . Since 8 is in A but not in A B , we must have 8 ∈ B . But now 8 is in B and not in C , so 8 ∈ B C . This contradicts the assumption 8 B C .

第三,我们用反证法证明 T 是传递的。假设存在 A, B, C∈P(Z),使得 (A, B) ∈ T 且 (B, C) ∈ T 且 (A, C) ∉ T。根据定义,这意味着 8 ∉ A−B 且 8 ∉ B−C 且 8 ∈ A − C。最后一个条件告诉我们 8 ∈ A 且 8 ∉ C。由于 8 属于 A 但不属于 A − B,因此 8 必须属于 B。但现在 8 属于 B 但不属于 C,所以 8 ∈ B − C。这与假设 8 ∉ B−C 矛盾。

6.18 Remark. A relation R on a fixed set X may be reflexive or not, symmetric or not, and transitive or not. In general, all eight possible combinations can occur (see the exercises).

6.18 备注。定义在固定集合 X 上的关系 R 可以是自反的,也可以不是;可以是对称的,也可以不是;可以是传递的,也可以不是。一般来说,所有八种可能的组合都有可能出现(参见练习)。

Properties of Reflexivity, Symmetry, and Transitivity

自反性、对称性和传递性的性质

The next theorem lists some facts about reflexivity, symmetry, and transitivity.

下一个定理列举了一些关于自反性、对称性和传递性的事实。

6.19 Theorem on Reflexivity, Symmetry, and Transitivity. For all relations R and all sets X :

6.19 关于自反性、对称性和传递性的定理。对于所有关系 R 和所有集合 X:

(a) R is reflexive on X iff I X R .

(a)R 对 X 是自反的,当且仅当 I X ⊆ R。

(b) R is reflexive on X iff R −1 is reflexive on X .

(b)R 对 X 具有自反性当且仅当 R −1 对 X 具有自反性。

(c) Any union or intersection of reflexive relations on X is reflexive on X .

(c)X 上自反关系的任何并集或交集都是 X 上的自反关系。

(d) R is symmetric iff R −1 is symmetric.

(d)R 是对称的当且仅当 R −1 是对称的。

(e) R is symmetric iff R R −1 iff R = R −1 .

(e)R 是对称的当且仅当 R⊆ R −1 当且仅当 R = R −1

(f) Any union or intersection of symmetric relations is symmetric.

(f)对称关系的任何并集或交集都是对称的。

(g) R is transitive iff R −1 is transitive.

(g)R 是传递的当且仅当 R −1 是传递的。

(h) R is transitive iff R R R .

(h)R 是传递的当且仅当 R ○ R⊆ R。

(i) Any intersection of transitive relations is transitive.

(i)传递关系的任何交集都是传递的。

Proof . We prove a few parts here and leave the rest as exercises. Fix a relation R and a set X . To prove (b), we first assume R is reflexive on X and prove R −1 is reflexive on X . Fix x X ; prove ( x , x ) ∈ R −1 . Because R is reflexive on X , we know ( x , x ) ∈ R . Switching the order of the components, we conclude ( x , x ) ∈ R −1 as needed. The converse is proved similarly, or it can be deduced from the IF-statement we just proved by replacing the arbitrary relation R by R −1 and recalling ( R −1 ) −1 = R .

证明。我们在此证明部分内容,其余部分留作练习。固定一个关系 R 和一个集合 X。为了证明 (b),我们首先假设 R 对 X 自反,并证明 R −1 对 X 自反。固定 x ∈ X;证明 (x, x) ∈ R −1 。因为 R 对 X 自反,我们知道 (x, x) ∈ R。交换各部分的顺序,我们即可得出 (x, x) ∈ R −1 。逆命题的证明类似,或者也可以通过将任意关系 R 替换为 R −1 并记住 (R −1 ) −1 = R,从我们刚刚证明的 IF 语句中推导出来。

Next we prove that the union of any collection of symmetric relations is symmetric (part of (f)). Let { R i : i I } be a collection of relations, and define S = i I R i . Assume R i is symmetric for all i I ; prove S is symmetric. Fix objects x , y ; assume ( x , y ) ∈ S ; prove ( y , x ) ∈ S . We know there exists i I with ( x , y ) ∈ R i . Since R i is symmetric, it follows that ( y , x ) ∈ R i . By definition of union, ( y , x ) ∈ S , as needed.

接下来,我们证明任意对称关系集合的并集仍然是对称的((f) 的一部分)。设 {R i :i ∈ I} 为一个关系集合,并定义 S=⋃i∈IRi。假设对于所有 i ∈ I,R i 都是对称的;证明 S 也是对称的。固定对象 x, y;假设 (x, y) ∈ S;证明 (y, x) ∈ S。我们知道存在 i ∈ I 使得 (x, y) ∈ R i 。由于 R i 是对称的,因此 (y, x) ∈ R i 。根据并集的定义,(y, x) ∈ S,满足要求。

Finally, we prove that if R is transitive, then R R R (part of (h)). We have assumed x , y , z , ( x , y ) R ( y , z ) R ( x , z ) R . To prove R R R , fix an ordered pair ( u , v ) ∈ R R and prove ( u , v ) ∈ R . By definition of composition of relations, there exists w with ( u , w ) ∈ R and ( w , v ) ∈ R . By transitivity of R , ( u , v ) ∈ R follows. □

最后,我们证明如果 R 是传递的,则 R ○ R ⊆ R((h) 的一部分)。我们假设对于所有 x, y, z,(x, y) ∈ R ∧ (y, z) ∈ R ⇒ (x, z) ∈ R。为了证明 R ○ R ⊆ R,固定一个有序对 (u, v) ∈ R 并证明 (u, v) ∈ R。根据关系复合的定义,存在 w 使得 (u, w) ∈ R 且 (w, v) ∈ R。由 R 的传递性,(u, v) ∈ R 成立。□

6.20 Example. It is false that the union of transitive relations must always be transitive. For example, let R = {(1, 2)} and S = {(2, 3)}, so R S = { ( 1 , 2 ) , ( 2 , 3 ) } . Both R and S are transitive relations, but their union is not.

6.20 例。传递关系的并集不一定是传递的。例如,设 R = {(1, 2)} 且 S = {(2, 3)},则 R∪S={(1,2),(2,3)}。R 和 S 都是传递关系,但它们的并集不是传递关系。

Section Summary

章节概要

  1. Relations and Digraphs. R is a relation on a set X iff R X × X . In this case, the digraph of R consists of a point for each x X , and an arrow from x to y for each ordered pair ( x , y ) ∈ R .
    关系与有向图。R 是集合 X 上的关系当且仅当 R⊆ X × X。在这种情况下,R 的有向图由每个 x ∈ X 的一个点和每个有序对 (x, y) ∈ R 的从 x 到 y 的箭头组成。
  2. Reflexive Relations. R is a reflexive relation on X iff for all x X , ( x , x ) ∈ R .
    自反关系。R 是 X 上的自反关系当且仅当对于所有 x ∈ X,(x, x) ∈ R。
  3. Symmetric Relations. A relation R is symmetric iff for all a , b , if ( a , b ) ∈ R , then ( b , a ) ∈ R .
    对称关系。关系 R 是对称的,当且仅当对于所有 a、b,如果 (a, b) ∈ R,则 (b, a) ∈ R。
  4. Transitive Relations. A relation R is transitive iff for all a , b , c , if ( a , b ) ∈ R and ( b , c ) ∈ R , then ( a , c ) ∈ R .
    传递关系。关系 R 是传递的,当且仅当对于所有 a、b、c,如果 (a, b) ∈ R 且 (b, c) ∈ R,则 (a, c) ∈ R。
  5. Infix Notation. Given a relation R , aRb means ( a , b ) ∈ R . In this notation, R is reflexive on X iff x X , x R x ; R is symmetric iff a , b , a R b b R a ; and R is transitive iff a , b , c , ( a R b b R c ) a R c .
    中缀表示法。给定一个关系 R,aRb 表示 (a, b) ∈ R。在这种表示法中,R 对 X 具有自反性当且仅当 ∀x∈X,xRx;R 具有对称性当且仅当 ∀a,b,aRb⇒bRa;R 具有传递性当且仅当 ∀a,b,c,(aRb∧bRc)⇒aRc。
  6. Properties of Reflexivity, Symmetry, and Transitivity. If R is reflexive on X , or symmetric, or transitive, then R −1 has the same property. Reflexivity, symmetry, and transitivity are inherited by intersections of relations. Reflexivity and symmetry are inherited by unions of relations, but transitivity is not. R is reflexive on X iff I X R ; R is symmetric on X iff R R −1 iff R = R −1 ; R is transitive on X iff R R R .
    自反性、对称性和传递性的性质。如果关系 R 在 X 上具有自反性、对称性或传递性,则 R −1 也具有相同的性质。关系的交集继承自反性、对称性和传递性。关系的并集继承自自反性和对称性,但传递性不继承。关系 R 在 X 上具有自反性当且仅当 I X ⊆ R;关系 R 在 X 上具有对称性当且仅当 R⊆ R −1 当且仅当 R = R −1 ;关系 R 在 X 上具有传递性当且仅当 R ○ R⊆ R。

Exercises

练习

  1. The digraphs of various relations are shown in Figure 6.3 . Is each relation reflexive on X ? symmetric? transitive?

    fig6_3

    Figure 6.3

    Digraphs for some relations.


    图 6.3 显示了各种关系的有向图。每个关系关于 X 是自反的吗?对称的吗?传递的吗? fig6_3 图 6.3 一些关系的有向图。
  2. Draw an arrow diagram and a digraph for each relation. Is each relation reflexive on X = {1, 2, 3, 4}? symmetric? transitive? (a) R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 4), (1, 4), (2, 4)}. (b) R = {(1, 2), (2, 1), (3, 4), (4, 3)}. (c) R = ( X × X ) − I X . (d) R = {(1, 1), (3, 3), (1, 3), (3, 1), (2, 2), (4, 4), (2, 4), (4, 2)}.
    为每个关系绘制箭头图和有向图。每个关系在 X = {1, 2, 3, 4} 上是否具有自反性?是否对称?传递性? (a) R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 4), (1, 4), (2, 4)}. (b) R = {(1, 2), (2, 1), (3, 4), (4, 3)}. (c) R = (X × X) − I X . (d) R = {(1, 1), (3, 3), (1, 3), (3, 1), (2, 2), (4, 4), (2, 4), (4, 2)}.
  3. Is each relation reflexive on Z ? symmetric? transitive? Explain. (a) R = { ( a , b ) Z 2 : a < b } . (b) R = { ( a , b ) Z 2 : a 2 b 2 } . (c) R = { ( a , b ) Z 2 : a b is even } . (d) R = { ( a , b ) Z 2 : | a b | 1 } . (e) R = { ( a , b ) Z 2 : a divides b } . (f) R = { ( a , b ) Z 2 : 5 divides a b } . (g) R = { ( a , b ) Z 2 : 7 divides a b } .
    每个关系在 Z 上是自反的吗?对称的吗?传递的吗?请解释。 (a) R={(a,b)∈Z2:a<b}. (b) R={(a,b)∈Z2:a2≤b2}. (c) R={(a,b)∈Z2:abiseven}. (d) R={(a,b)∈Z2:|a−b|≤1}. (e) R={(a,b)∈Z2:adividesb}. (f) R={(a,b)∈Z2:5dividesab}. (g) R={(a,b)∈Z2:7dividesa−b}.
  4. Let S be a relation on R . (a) Describe how to determine if S is reflexive on R by visual inspection of the graph of S in the xy -plane. (b) Similarly, describe how to determine if S is symmetric by inspection of its graph.
    设 S 是 R 上的一个关系。(a) 描述如何通过观察 S 在 xy 平面上的图像来判断 S 是否在 R 上是自反的。(b) 类似地,描述如何通过观察 S 的图像来判断 S 是否是对称的。
  5. Is each relation reflexive on R ? symmetric? transitive? Explain. (a) R = { ( x , y ) R 2 : cos x = cos y } . (b) R = { ( x , y ) R 2 : x y = 1 } . (c) R = { ( x , y ) R 2 : y = x y = x x 2 + y 2 = 1 } . (d) R = { ( x , y ) R 2 : x y < 0 } . (e) R = { ( x , y ) R 2 : x y 0 } . (f) R = { ( x , y ) R 2 : x y > 0 } { ( 0 , 0 ) } . (g) R = { ( x , y ) R 2 : y x Z } .
    每个关系在 R 上都是自反的吗?对称的吗?传递的吗?解释。 (a) R={(x,y)∈R2:cos⁡x=cos⁡y}. (b) R={(x,y)∈R2:xy=1}. (c) R={(x,y)∈R2:y=x∨y=−x∨x²+y²=1}. (d) R={(x,y)∈R2:xy<0}. (e) R={(x,y)∈R2:xy≥0}. (f) R={(x,y)∈R2:xy>0}∪{(0,0)}. (g) R={(x,y)∈R2:y−x∈Z}。
  6. Is each relation reflexive on R 2 ? symmetric? transitive? Explain. (a) ( a , b ) R ( c , d ) means a 2 + b 2 = c 2 + d 2 . (b) ( a , b ) R ( c , d ) means ad = bc . (c) ( a , b ) R ( c , d ) means a + d = b + c . (d) ( a , b ) R ( c , d ) means a < c or ( a = c and b d ).
    每个关系在 R2 上是自反的吗?对称的吗?传递的吗?请解释。 (a) (a, b)R(c, d) 表示 a 2 + b 2 = c 2 + d 2 (b) (a, b)R(c, d) 表示 ad = bc。 (c) (a, b)R(c, d) 表示 a + d = b + c。 (d) (a, b)R(c, d) 表示 a < c 或 (a = c 且 b ≤ d)。
  7. How can you decide by looking at the digraph of a relation G on X whether G is the graph of a function f : X X ?
    如何通过观察关系 G 在 X 上的有向图来判断 G 是否是函数 f:X → X 的图像?
  8. Let X = { a , b , c }. For each part, give two different examples of relations on X with the indicated properties (if possible). (a) R is reflexive on X and symmetric and transitive. (b) R is reflexive on X and symmetric and not transitive. (c) R is reflexive on X and not symmetric and transitive. (d) R is reflexive on X and not symmetric and not transitive. (e) R is not reflexive on X and symmetric and transitive. (f) R is not reflexive on X and symmetric and not transitive. (g) R is not reflexive on X and not symmetric and transitive. (h) R is not reflexive on X and not symmetric and not transitive.
    令 X = {a, b, c}。对于每个部分,请给出两个关于集合 X 的关系示例,并尽可能给出指定的属性。 (a) R 在 X 上具有自反性、对称性和传递性。 (b) R 在 X 上具有自反性、对称性,但不具有传递性。 (c) R 在 X 上具有自反性、不具有对称性和传递性。 (d) R 在 X 上具有自反性、不具有对称性和传递性。 (e) R 在 X 上不具有自反性、对称性和传递性。 (f) R 在 X 上不具有自反性、对称性和传递性。 (g) R 在 X 上不具有自反性、不具有对称性和传递性。 (h) R 在 X 上不具有自反性、不具有对称性和传递性。及物动词。
  9. Which parts of the previous exercise can be solved (with either one or two examples) using relations on the two-element set X = {1, 2}?
    上一题的哪些部分可以用二元集合 X = {1, 2} 上的关系式来解决(用一个或两个例子即可)?
  10. Prove (a) and (c) of Theorem 6.19 .
    证明定理 6.19 的 (a) 和 (c)。
  11. Prove (d), (e), and the second part of (f) in Theorem 6.19 .
    证明定理 6.19 中的 (d)、(e) 和 (f) 的第二部分。
  12. Prove (g), (i), and the second part of (h) in Theorem 6.19 .
    证明定理 6.19 中的 (g)、(i) 和 (h) 的第二部分。
  13. (a) Suppose R is a reflexive relation on a nonempty set X . What can you say about whether ( X × X ) − R is reflexive on X ? (b) Repeat (a) replacing “reflexive” by “symmetric.” (c) Repeat (a) replacing “reflexive” by “transitive.”
    (a) 假设 R 是非空集合 X 上的自反关系。关于 (X × X) − R 在 X 上是否自反,你能得出什么结论? (b) 重复 (a),将“自反”替换为“对称”。 (c) 重复 (a),将“自反”替换为“传递”。
  14. (a) If R and S are reflexive relations on X , must R S be reflexive on X ? (b) If R and S are symmetric relations, must R S be symmetric? (c) If R and S are transitive relations, must R S be transitive?
    (a) 如果 R 和 S 是 X 上的自反关系,那么 R ○ S 一定是 X 上的自反关系吗? (b) 如果 R 和 S 是对称关系,那么 R ○ S 一定是对称关系吗? (c) 如果 R 和 S 是传递关系,那么 R ○ S 一定是传递关系吗?
  15. Prove: for any relation R and any set X , R R 1 I X is reflexive on X and symmetric.
    证明:对于任意关系 R 和任意集合 X,R∪R−1∪IX 在 X 上是自反的且对称的。
  16. (a) For a relation R on X , prove that the intersection T of all transitive relations on X containing R is a transitive relation on X containing R . (b) Recursively define R 1 = R and R n +1 = R n R for n Z 1 . Prove: for all n , m Z 1 , R n + m = R n R m . (c) Prove that T = n = 1 R n .
    (a) 对于集合 X 上的关系 R,证明 X 上所有包含 R 的传递关系的交集 T 也是 X 上包含 R 的传递关系。(b) 递归定义 R 1 = R 和 R n +1 = R n ○ R,其中 n∈Z≥1。证明:对于所有 n,m∈Z≥1,R n + m = R n ○ R m 。(c) 证明 T=⋃n=1∞Rn。

6.2 Equivalence Relations

6.2 等价关系

In mathematics, it often happens that two objects are equivalent in a certain respect, even though the objects in question are not equal to each other. For instance, two triangles (thought of as subsets of R 2 ) can be geometrically congruent even if they are not equal as sets of points. As further examples, two unequal shapes may be geometrically similar; two unequal vectors may have the same length; two unequal lines may have the same slope; two unequal times (differing by a multiple of 24 hours) may represent the same hour of the day; two unequal real numbers (differing by a multiple of 2 π ) may represent the same geometric angle; and so on. The abstract definition of an equivalence relation gives a precise mathematical formulation of various intuitive notions of equivalence. We define equivalence relations in this section and consider many examples, including congruence of integers modulo n and the equivalence relation induced by a function.

在数学中,两个对象在某些方面等价的情况并不少见,即便它们本身并不相等。例如,两个三角形(可以看作是 R² 的子集)即使作为点集并不相等,它们在几何上也可能全等。再举几个例子:两个不等的形状可能几何相似;两个不等的向量可能长度相等;两条不等的直线可能斜率相等;两个不等的时间(相差 24 小时的倍数)可能表示一天中的同一小时;两个不等的实数(相差 2π 的倍数)可能表示同一个几何角度;等等。等价关系的抽象定义为各种直观的等价概念提供了精确的数学表述。本节将定义等价关系,并考虑许多例子,包括整数模 n 的同余关系以及由函数诱导的等价关系。

Definition and Examples of Equivalence Relations

等价关系的定义和示例

Any intuitive concept of “equivalence” ought to share some of the basic properties of logical equality. It turns out that reflexivity, symmetry, and transitivity are the key properties we need.

任何关于“等价”的直观概念都应该具备逻辑等价的一些基本属性。事实证明,自反性、对称性和传递性是我们需要的关键属性。

6.21. Definition: Equivalence Relations.. Given a relation R on a set X ,

6.21. 定义:等价关系。给定集合 X 上的关系 R,

R is an 一个 equivalence 等价 relation 关系 on X X iff 当……时 R is reflexive 反身 on X X , symmetric 对称 , and transitive 及物动词

Expanding this definition, we see that R is an equivalence relation on X iff the following three conditions hold for all a , b , c X : (i) aRa ; (ii) if aRb then bRa ; (iii) if aRb and bRc , then aRc .

扩展此定义,我们看到 R 是 X 上的等价关系当且仅当对于所有 a、b、c ∈ X,以下三个条件成立:(i)aRa;(ii)如果 aRb 则 bRa;(iii)如果 aRb 且 bRc 则 aRc。

6.22 Example: Equality Relation. For any set X , the identity relation I X = {( x , x ): x X } is an equivalence relation on X . If we write x = y instead of xI X y , the three conditions above become: (i) a = a ; (ii) if a = b , then b = a ; (iii) if a = b and b = c , then a = c .

6.22 示例:等价关系。对于任意集合 X,恒等关系 I X = {(x, x):x ∈ X} 是 X 上的等价关系。如果我们用 x = y 代替 xI X y,则上述三个条件变为:(i) a = a;(ii) 如果 a = b,则 b = a;(iii) 如果 a = b 且 b = c,则 a = c。

6.23 Example. For any set X , R = X × X is an equivalence relation on X . In this case, xRy is true for all pairs of objects x , y X . So it is immediate that conditions (i), (ii), and (iii) hold for this R .

6.23 例。对于任意集合 X,R = X × X 是 X 上的一个等价关系。在这种情况下,对于所有对象对 x, y ∈ X,xRy 都成立。因此,显然条件 (i)、(ii) 和 (iii) 对该 R 成立。

6.24 Example. Let X = Z and R = { ( x , y ) Z 2 : x + y is even } . We saw in the last section that R is reflexive on X , symmetric, and transitive. Thus, R is an equivalence relation on X . On the other hand, let S = { ( x , y ) Z 2 : x + y is odd } . S is not reflexive on Z , since ( 1 , 1 ) S , so S is not an equivalence relation on Z .

6.24 例。设 X=Z,R={(x,y)∈Z²:x+y 为偶数}。我们在上一节中看到,R 在 X 上是自反的、对称的和传递的。因此,R 是 X 上的一个等价关系。另一方面,设 S={(x,y)∈Z²:x+y 为奇数}。S 在 Z 上不是自反的,因为 (1,1)∉S,所以 S 不是 Z 上的一个等价关系。

6.25 Example. Define T = { ( x , y ) R 2 : x y 0 } . You can check that T is reflexive on R and symmetric. But T is not transitive, because (1, 0) ∈ T and (0, − 1) ∈ T but ( 1 , 1 ) T . So T is not an equivalence relation on R .

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6.25 例。定义 T={(x,y)∈R²:xy≥0}。可以验证 T 在 R 上是自反的且对称的。但 T 不是传递的,因为 (1, 0) ∈ T 且 (0, -1) ∈ T,但 (1, -1)∉ T。所以 T 不是 R 上的等价关系。

6.26 Example. Figure 6.4 shows the digraphs for six relations R 1 , …, R 6 on the set X = {1, 2, 3, 4, 5, 6}. We see from inspection of the figure that R 1 , R 2 , and R 3 are equivalence relations on X , but R 4 , R 5 , and R 6 are not. R 4 is not symmetric because (2, 3) ∈ R 4 but ( 3 , 2 ) R 4 . R 5 is not reflexive on X because ( 5 , 5 ) R 5 . R 6 is not transitive because (1, 3) ∈ R 6 and (3, 6) ∈ R 6 , but ( 1 , 6 ) R 6 .

6.26 示例。图 6.4 显示了集合 X = {1, 2, 3, 4, 5, 6} 上六个关系 R 1 , …, R 6 的有向图。从图中可以看出,R 1 、R 2 和 R 3 是 X 上的等价关系,但 R 4 、R 5 和 R 6 不是。R 4 不是对称的,因为 (2, 3) ∈ R 4 但 (3, 2) ∉ R4。R 5 不是 X 上的自反关系,因为 (5, 5) ∉ R5。 R 6 不是传递的,因为 (1, 3) ∈ R 6 且 (3, 6) ∈ R 6 ,但 (1,6)∉R6。

fig6_4

Figure 6.4

图 6.4

Digraphs for some equivalence relations and non-equivalence relations.

表示一些等价关系和非等价关系的有向图。

In Figure 6.1 from the previous section, only R 1 (the equality relation) is an equivalence relation on X = {1, 2, 3}. Digraphs of equivalence relations have special structural features that we will explore later, when we discuss equivalence classes and set partitions.

在上一节的图 6.1 中,只有 R 1 (相等关系)是 X = {1, 2, 3} 上的等价关系。等价关系的有向图具有特殊的结构特征,我们将在后面讨论等价类和集合划分时进行探讨。

Equivalence Relation Induced by a Function

由函数诱导的等价关系

Given a function f : X Y , we can create an equivalence relation on the domain of f by the following construction.

给定一个函数 f:X → Y,我们可以通过以下构造在 f 的定义域上创建一个等价关系。

6.27. Definition: Equivalence Relation Induced by a Function.. Given f : X Y , the equivalence relation induced by f is R f = {( a , b ) ∈ X × X : f ( a ) = f ( b )}. We often denote R f by the symbol ∼ f ; thus, for all a , b X , a f b means f ( a ) = f ( b ) .

6.27. 定义:由函数诱导的等价关系。给定函数 f:X → Y,由 f 诱导的等价关系为 R f = {(a, b) ∈ X × X:f(a) = f(b)}。我们通常用符号 ∼ f 表示 R f ;因此,对于所有 a, b ∈ X,a∼fb 表示 f(a)=f(b)。

Let us check that ∼ f always is an equivalence relation. Fix an arbitrary function f : X Y , and fix a , b , c X . We have a f a because f ( a ) = f ( a ), so ∼ f is reflexive on X . To prove symmetry, assume a f b and prove b f a . We have assumed f ( a ) = f ( b ); we must prove f ( b ) = f ( a ); this follows since equality is symmetric. To prove transitivity, assume a f b and b f c ; prove a f c . We have assumed f ( a ) = f ( b ) and f ( b ) = f ( c ); we must prove f ( a ) = f ( c ). This follows since equality is transitive. We have now proved ∼ f is an equivalence relation on X .

让我们验证 ∼ f 始终是一个等价关系。固定任意函数 f:X → Y,并固定 a, b, c ∈ X。我们有 a ∼ f a,因为 f(a) = f(a),所以 ∼ f 在 X 上是自反的。为了证明对称性,假设 a ∼ f b,并证明 b ∼ f a。我们已经假设 f(a) = f(b);我们必须证明 f(b) = f(a);这成立,因为等式是对称的。为了证明传递性,假设 a ∼ f b 且 b ∼ f c;证明 a ∼ f c。我们已经假设 f(a) = f(b) 且 f(b) = f(c);我们必须证明 f(a) = f(c)。由于等式具有传递性,因此结论成立。我们现在已经证明了∼ f 是X上的一个等价关系。

By choosing different functions f , we can use this construction to build a wide variety of equivalence relations. (In fact, we will see later that all equivalence relations arise in this way for an appropriate choice of f .) Here are some typical examples.

通过选择不同的函数 f,我们可以利用这种构造方法构建各种各样的等价关系。(事实上,我们稍后会看到,对于合适的 f 值,所有等价关系都可以通过这种方式产生。)以下是一些典型的例子。

6.28 Example. For f : R R given by f ( x ) = | x |, we get an equivalence relation ∼ f on R such that x f y iff | x | = | y |. So, for instance, −3 ∼ f 3, 4 ∼ f 4, 0 ∼ f 0, but 3 f 5 . Intuitively, this relation regards two real numbers as equivalent iff they have the same magnitude (disregarding sign). The function g : R R given by g ( x ) = x 2 induces the same equivalence relation as f . For, given any x , y R , x g y iff x 2 = y 2 iff x 2 = y 2 iff | x | = | y | iff x f y .

6.28 例。对于函数 f:R→R,定义为 f(x) = |x|,我们在 R 上得到一个等价关系 ∼ f ,使得 x ∼ f y 当且仅当 |x| = |y|。例如,−3 ∼ f 3,4 ∼ f 4,0 ∼ f 0,但 −3≁f5。直观地说,该关系将两个实数视为等价当且仅当它们具有相同的绝对值(忽略符号)。函数 g:R→R,定义为 g(x) = x 2 ,也导出与 f 相同的等价关系。对于任意 x,y∈R,x ∼ g y 当且仅当 x 2 = y 2 当且仅当 x2=y2 当且仅当 |x| = |y| 当且仅当 x ∼ f y。

6.29 Example. Define sgn : R { 1 , 0 , 1 } by setting sgn( x ) = −1 if x < 0, sgn( x ) = 0 if x = 0, and sgn( x ) = 1 if x > 0. The equivalence relation ∼ sgn regards two real numbers as equivalent iff they have the same sign. For example, −3 ∼ sgn − 7, 0 ∼ sgn 0, 2 ∼ sgn 7, but 2 sgn 2 .

6.29 例。定义函数 sgn:R→{−1,0,1},令当 x < 0 时 sgn(x) = −1,当 x = 0 时 sgn(x) = 0,当 x > 0 时 sgn(x) = 1。等价关系 ∼ sgn 认为两个实数等价当且仅当它们符号相同。例如,−3 ∼ sgn − 7,0 ∼ sgn 0,2 ∼ sgn 7,但 2≁sgn−2。

6.30 Example. Define f : R R 2 by f ( t ) = (cos t , sin t ) for all t R . From trigonometry, we know that for all s , t R , f ( s ) = f ( t ) iff s = t + 2 πk for some integer k . Intuitively, ∼ f regards two real numbers as equivalent iff those numbers represent the same angle going around the unit circle. The individual sine and cosine functions induce more complicated equivalence relations on R . For instance, π /3 ∼ sin 2 π /3 and π /3 ∼ cos π /3, although no two of the numbers − π /3, π /3, 2 π /3 are related under ∼ f .

6.30 例。定义函数 f:R→R²,其中 f(t) = (cos t, sin t),t∈R。根据三角学,我们知道对于所有 s,t∈R,f(s) = f(t) 当且仅当 s = t + 2πk,其中 k 为某个整数。直观地说,∼ f 认为两个实数等价当且仅当它们表示单位圆上相同的角度。正弦和余弦函数在 R 上诱导出更复杂的等价关系。例如,π/3 ∼ sin 2π/3 和 π/3 ∼ cos − π/3,尽管 −π/3、π/3 和 2π/3 这三个数中没有两个在 ∼ f 下相关。

6.31 Example. Define f : Z > 0 { 0 , 1 , , 9 } by letting f ( n ) be the rightmost digit in the decimal representation of n . For instance, f (5247) = 7, f (10 3 − 1) = 9, and f (10 k ) = 0 for any k Z > 0 . The relation ∼ f regards two positive integers as equivalent iff they have the same last digit.

6.31 示例。定义函数 f:Z>0→{0,1,…,9},其中 f(n) 表示 n 的十进制表示的最右边数字。例如,f(5247) = 7,f(10 3 − 1) = 9,且对于任意 k∈Z>0,f(10k) = 0。关系 ∼ f 表示两个正整数等价当且仅当它们的最后一位数字相同。

Congruence Modulo n

模 n 全等

We now introduce a family of equivalence relations that play a fundamental role in number theory.

现在我们引入一类在数论中起着根本作用的等价关系。

6.32. Definition: Congruence Modulo n .. For each positive integer n ≥ 1, define a relation ≡ n on Z as follows: for all a , b Z , a n b iff n divides a b . When a n b , we also use the notation a b ( mod n ) , which can be read “ a is congruent to b modulo n .”

6.32. 定义:模 n 同余。对于每个正整数 n ≥ 1,定义 Z 上的关系 ≡ n 如下:对于所有 a,b∈Z,a≡nb 当且仅当 n 整除 a−b。当 a ≡ n b 时,我们也使用符号 a≡b(modn),它可以读作“a 与 b 模 n 同余”。

The next theorem says that congruence modulo n is an equivalence relation that is compatible with the operations of integer addition and multiplication.

下一个定理指出,模 n 同余是一种等价关系,它与整数加法和乘法运算相容。

6.33 Theorem on Congruence Modulo n . For all n Z 1 and all a , b , c , d Z :

6.33 模 n 同余定理。对于所有 n∈Z≥1 和所有 a,b,c,d∈Z:

(a) ≡ n is an equivalence relation on Z .

(a)≡ n 是 Z 上的一个等价关系。

(b) If a n b and c n d , then a + c n b + d .

(b)如果 a ≡ n b 且 c ≡ n d,则 a + c ≡ n b + d。

(c) If a n b and c n d , then ac n bd .

(c)如果 a ≡ n b 且 c ≡ n d,则 ac ≡ n bd。

Proof . For (a), we must show ≡ n is reflexive on Z and symmetric and transitive. To prove ≡ n is reflexive on Z , fix a Z . We prove a n a , which means n divides a a . Since a a = 0 = 0 n where 0 is an integer, n does divide a a . To prove symmetry, fix a , b Z ; assume a n b ; prove b n a . We have assumed n divides a b , so a b = kn for some integer k . We must prove n divides b a . Since b a = −( a b ) = −( kn ) = ( − k ) n where − k is an integer, n does divide b a . To prove transitivity, fix a , b , c Z ; assume a n b and b n c ; prove a n c . We have assumed n divides a b , so a b = dn for some integer d ; and n divides b c , so b c = en for some integer e . We must prove n divides a c . Note a c = ( a + b ) + ( b c ) = dn + en = ( d + e ) n where d + e is an integer; so n does divide a c .

证明。对于 (a),我们必须证明 ≡ n 在 Z 上是自反的,并且是对称的和传递的。为了证明 ≡ n 在 Z 上是自反的,固定 a∈Z。我们证明 a ≡ n a,这意味着 n 整除 a − a。由于 a − a = 0 = 0n,其中 0 是整数,所以 n 确实整除 a − a。为了证明对称性,固定 a,b∈Z;假设 a ≡ n b;证明 b ≡ n a。我们已经假设 n 整除 a − b,所以对于某个整数 k,a − b = kn。我们必须证明 n 整除 b − a。由于 b − a = −(a − b) = −(kn) = ( − k)n,其中 −k 是整数,所以 n 确实整除 b − a。为了证明传递性,固定 a,b,c∈Z;假设 a ≡ b 且 b ≡ c;证明 a ≡ c。我们假设 n 整除 a − b,所以对于某个整数 d,有 a − b = dn;并且 n 整除 b − c,所以对于某个整数 e,有 b − c = en。我们必须证明 n 整除 a − c。注意 a − c = (a + b) + (b − c) = dn + en = (d + e)n,其中 d + e 为整数;因此 n 确实整除 a − c。

We prove (c), leaving (b) as an exercise. Assume a n b and c n d , so a b = jn and c d = kn for some j , k Z . We must prove ac n bd , i.e., ac bd = pn for some p Z . Compute

我们证明了 (c),将 (b) 作为练习。假设 a ≡ b 且 c ≡ d,因此对于某些 j,k∈Z,有 a − b = jn 且 c − d = kn。我们必须证明 ac ≡ bd,即对于某些 p∈Z,有 ac − bd = pn。计算

a 一个 c c b b d d = a 一个 c c a 一个 d d + a 一个 d d b b d d = a 一个 ( c c d d ) + ( a 一个 b b ) d d = a 一个 ( k k n n ) + ( j j n n ) d d = ( a 一个 k k + j j d d ) n n .

Choosing p = ak + jd , which is an integer, we then have ac bd = pn as needed.      □

选择 p = ak + jd,这是一个整数,那么我们就有 ac − bd = pn,符合要求。□

6.34 Example. When n = 2, a 2 b iff 2 divides a b . By considering cases, we see that for all a , b Z , a 2 b iff a and b are both even, or a and b are both odd. Thus, ≡ 2 regards two integers as being equivalent iff these integers have the same parity (odd or even).

6.34 例。当 n = 2 时,a ≡ 2 b 当且仅当 2 能整除 a − b。通过考虑各种情况,我们发现对于所有 a, b∈Z,a ≡ 2 b 当且仅当 a 和 b 都是偶数,或者 a 和 b 都是奇数。因此,≡ 2 认为两个整数等价当且仅当它们具有相同的奇偶性(奇数或偶数)。

6.35 Example. Taking n = 10, we have 5247 ≡ 10 7, 115 ≡ 10 9015, but 4441 10 4444 . For all positive integers a , b , we have a 10 b iff a and b have the same last digit written in decimal iff a f b , where f is the function from Example 6.31 . However, we must be careful when negative integers appear. For instance, 17 ≡ 10 − 3 since 10 divides 17 − ( − 3) = 20, but 17 10 7 since 10 does not divide 17 − ( − 7) = 24.

6.35 例。取 n = 10,则 5247 ≡ 10 7,115 ≡ 10 9015,但 4441≢104444。对于所有正整数 a 和 b,a ≡ 10 b 当且仅当 a 和 b 的十进制个位数相同,a ∼ f b,其中 f 是例 6.31 中的函数。然而,当出现负整数时,我们必须格外小心。例如,17 ≡ 10 − 3,因为 10 整除 17 − ( − 3) = 20,但 17≢10−7,因为 10 不整除 17 − ( − 7) = 24。

Further Examples of Equivalence Relations

等价关系的更多示例

When considering an equivalence relation (or any relation) on a set X , the objects in X can be anything. In the next example, the objects are ordered pairs (points or vectors in the xy -plane), so that the relation itself is a set of ordered pairs where each component is an ordered pair. In this situation, infix notation greatly improves readability.

在考虑集合 X 上的等价关系(或任何关系)时,X 中的对象可以是任何事物。在下一个例子中,对象是有序对(xy 平面上的点或向量),因此关系本身是一个有序对的集合,其中每个分量都是一个有序对。在这种情况下,中缀表示法可以极大地提高可读性。

6.36 Example. Let X = R × R . Define a relation R on X by setting ( a , b ) R ( c , d ) iff there exists r R > 0 with c = ra and d = rb . We now prove that R is an equivalence relation on X . For reflexivity, fix ( a , b ) ∈ X . To prove ( a , b ) R ( a , b ), we must prove r R > 0 , a = ra and b = rb . Choosing r = 1, these equations do hold. For symmetry, fix ( a , b ), ( c , d ) ∈ X , assume ( a , b ) R ( c , d ), and prove ( c , d ) R ( a , b ). By assumption, there is r R > 0 with c = ra and d = rb . We must prove s R > 0 , a = sc and b = sd . Choose s = r −1 , which exists and is a positive real number since r is positive. The assumptions c = ra and d = rb become a = r −1 c = sc and b = r −1 d = sd , as needed. For transitivity, fix ( a , b ), ( c , d ), ( e , f ) ∈ X , assume ( a , b ) R ( c , d ) and ( c , d ) R ( e , f ), and prove ( a , b ) R ( e , f ). By assumption, there exist t , u R > 0 with c = ta , d = tb , e = uc , and f = ud . Combining these equations, we see that e = uc = u ( ta ) = ( ut ) a and f = ud = u ( tb ) = ( ut ) b , where u t R > 0 since ut is a product of two positive real numbers. We now see that ( a , b ) R ( e , f ), as needed.

6.36 例。设 X=R×R。定义 X 上的关系 R,令 (a, b)R(c, d) 当且仅当存在 r∈R>0 使得 c = ra 且 d = rb。现在我们证明 R 是 X 上的等价关系。为了证明自反性,固定 (a, b) ∈ X。为了证明 (a, b)R(a, b),我们必须证明存在 r∈R>0,使得 a = ra 且 b = rb。取 r = 1,这些等式成立。为了证明对称性,固定 (a, b),(c, d) ∈ X,假设 (a, b)R(c, d),并证明 (c, d)R(a, b)。根据假设,存在 r∈R>0 使得 c = ra 且 d = rb。我们必须证明存在 s∈R>0,使得 a = sc 且 b = sd。选择 s = r −1 ,由于 r 为正数,因此 s 存在且为正实数。假设 c = ra 和 d = rb 变为 a = r −1 c = sc 和 b = r −1 d = sd,满足需要。为了证明传递性,固定 (a, b), (c, d), (e, f) ∈ X,假设 (a, b)R(c, d) 且 (c, d)R(e, f),并证明 (a, b)R(e, f)。根据假设,存在 t,u∈R>0,使得 c = ta,d = tb,e = uc,f = ud。结合这些方程,我们发现 e = uc = u(ta) = (ut)a 和 f = ud = u(tb) = (ut)b,其中 ut∈R>0,因为 ut 是两个正实数的乘积。现在我们看到 (a, b)R(e, f),正如我们所需要的那样。

In the next example, the elements of X are functions.

在下一个例子中,X 的元素是函数。

6.37 Example. Let X be the set of all functions f : R R . Define a relation R on X by setting fRg iff f (3) = g (3) for all f , g X . It is routine to prove that R is an equivalence relation by verifying the definition. Alternatively, we can deduce this using a previous construction, as follows. Define an “evaluation function” E : X R by setting E ( f ) = f (3) for all f X . Note that E is a function that takes another function as its input and produces a real number as its output. For all f , g X , fRg iff f (3) = g (3) iff E ( f ) = E ( g ) iff f E g . Thus R is the equivalence relation induced by the function E .

6.37 例。设 X 为所有函数 f:R→R 的集合。定义 X 上的关系 R,使得 fRg 当且仅当对于所有 f, g ∈ X,f(3) = g(3)。通过验证定义,可以很容易地证明 R 是一个等价关系。或者,我们可以使用先前的构造来推导,如下所示。定义“求值函数”E:X→R,使得对于所有 f ∈ X,E(f) = f(3)。注意,E 是一个以另一个函数作为输入并产生一个实数作为输出的函数。对于所有 f, g ∈ X,fRg 当且仅当 f(3) = g(3) 当且仅当 E(f) = E(g) 当且仅当 f ∼ E g。因此,R 是由函数 E 诱导的等价关系。

In the next example, the elements of X are sets.

在下一个例子中,X 的元素是集合。

6.38 Example. Let X = P ( Z ) , and for A , B X , define A B iff A Z > 0 = B Z > 0 . The relation ≈ is an equivalence relation on X . In fact, ≈ is the equivalence relation ∼ f induced by the function f : X P ( Z ) such that f ( A ) = A Z > 0 for all A X .

6.38 例。设 X=P(Z),对于 A, B ∈ X,定义 A ≈ B 当且仅当 A∩Z>0=B∩Z>0。关系 ≈ 是 X 上的一个等价关系。事实上,≈ 是由函数 f:X→P(Z) 诱导的等价关系 ∼ f ,使得对于所有 A ∈ X,f(A)=A∩Z>0。

We now give a method for obtaining a new equivalence relation from existing equivalence relations.

现在我们给出一种从现有等价关系中获得新的等价关系的方法。

6.39 Lemma on Intersection of Equivalence Relations. For all sets X , the intersection of any nonempty set of equivalence relations on X is an equivalence relation on X .

6.39 等价关系交集引理。对于任意集合 X,X 上任意非空等价关系集的交集仍然是 X 上的等价关系。

Proof . Fix a set X , and let { R i : i I } be a nonempty set of equivalence relations on X . We must prove R = i I R i is also an equivalence relation on X . Since every R i is reflexive on X , the intersection R is also reflexive on X by Theorem 6.19 . Similarly, by the same theorem, R is symmetric since all R i are symmetric, and R is transitive since all R i are transitive. So R is an equivalence relation on X . □

证明。固定一个集合 X,令 {R i :i ∈ I} 为 X 上的一个非空等价关系集合。我们需要证明 R=⋂i∈IRi 也是 X 上的一个等价关系。由于每个 R i 在 X 上都是自反的,根据定理 6.19,交集 R 在 X 上也是自反的。类似地,根据同一定理,由于所有 R i 都是对称的,所以 R 也是对称的;由于所有 R i 都是传递的,所以 R 也是传递的。因此,R 是 X 上的一个等价关系。□

6.40 Example. Given X = Z , we claim that 3 5 equals ≡ 15 . Writing R n for the relation ≡ n , we are claiming that the two sets of ordered pairs R 3 R 5 and R 15 are equal. [We prove the set equality using a chain proof, observing that all members of both sets are ordered pairs of integers.] Fix an arbitrary pair ( a , b ) Z × Z . Note that ( a , b ) R 3 R 5 iff ( a , b ) ∈ R 3 and ( a , b ) ∈ R 5 iff 3 divides a b and 5 divides a b iff 15 divides a b iff ( a , b ) ∈ R 15 . The next-to-last equivalence can be proved by considering the prime factorization of a b . More generally, you can check that for all m , n Z > 0 , m n equals ≡ lcm( m , n ) .

6.40 例。给定 X=Z,我们断言 ≡3∩≡5 等于 ≡ 15 。记 R n 为关系 ≡ n ,我们断言两个有序对集合 R3∩R5 和 R 15 相等。[我们使用链式证明来证明集合相等性,注意到两个集合的所有元素都是整数有序对。] 固定任意一对 (a,b)∈Z×Z。注意,(a,b)∈R3∩R5 当且仅当 (a, b) ∈ R 3 ,(a, b) ∈ R 5 当且仅当 3 整除 a − b,5 整除 a − b 当且仅当 15 整除 a − b 当且仅当 (a, b) ∈ R 15 。倒数第二个等价性可以通过考虑 a − b 的质因数分解来证明。更一般地,你可以验证对于所有 m,n∈Z>0,≡m∩≡n 等于 ≡ lcm( m , n )

Section Summary

章节概要

  1. Equivalence Relations. R is an equivalence relation on a set X iff R is a relation on X that is reflexive on X , symmetric, and transitive. The intersection of any nonempty set of equivalence relations on X is also an equivalence relation on X .
    等价关系。关系 R 是集合 X 上的等价关系,当且仅当 R 是 X 上的一个关系,且该关系在 X 上是自反的、对称的和传递的。X 上任意非空等价关系集合的交集也是 X 上的一个等价关系。
  2. Equivalence Relation Induced by a Function. Given any function f : X Y , there is an induced equivalence relation ∼ f on X defined by a f b iff f ( a ) = f ( b ) (where a , b X ).
    由函数诱导的等价关系。给定任意函数 f:X → Y,存在一个在 X 上诱导的等价关系 ∼ f ,定义为 a ∼ f b 当且仅当 f(a) = f(b)(其中 a, b ∈ X)。
  3. Congruence mod n . For each n ≥ 1, the relation ≡ n defined by a n b iff n divides a b (for a , b Z ) is an equivalence relation on Z . This relation is compatible with addition and multiplication: for all a , a , b , b Z , if a n a ′ and b n b ′, then a + b n a ′ + b ′ and ab n a b ′. When a n b , we also write a b ( mod n ) .
    模 n 同余。对于每个 n ≥ 1,由 a ≡ b 当且仅当 n 整除 a − b(对于 a, b∈Z)定义的等价关系 ≡ n 是 Z 上的一个等价关系。该关系与加法和乘法相容:对于所有 a, a′, b, b′∈Z,如果 a ≡ a′ 且 b ≡ b′,则 a + b ≡ a′ + b′ 且 ab ≡ a′b′。当 a ≡ b 时,我们也记作 a≡b(mod n)。

Exercises

练习

  1. (a) Which relations shown in Figure 6.3 are equivalence relations on X ? (b) Repeat (a) replacing each relation R i by R i I X .
    (a)图 6.3 中所示的哪些关系是 X 上的等价关系?(b)重复(a),将每个关系 R i 替换为 Ri∪IX。
  2. Let X = {1, 2, 3, 4, 5, 6}. For each function f : X → { a , b , c , d , e }, describe the equivalence relation ∼ f as a digraph and as a set of ordered pairs. (a) f ( x ) = c for all x X . (b) f ( x ) = a for odd x X , f ( x ) = e for even x X . (c) f (1) = b , f (2) = e , f (3) = a , f (4) = b , f (5) = e , f (6) = a . (d) f (1) = a , f (2) = b , f (3) = c , f (4) = d , f (5) = e , f (6) = b .
    令 X = {1, 2, 3, 4, 5, 6}。对于每个函数 f:X → {a, b, c, d, e},将等价关系 ∼ f 描述为有向图和有序对集合。 (a) 对于所有 x ∈ X,f(x) = c。 (b) 对于奇数 x ∈ X,f(x) = a;对于偶数 x ∈ X,f(x) = e。 (c) f(1) = b,f(2) = e,f(3) = a,f(4) = b,f(5) = e,f(6) = a。 (d) f(1) = a,f(2) = b,f(3) = c,f(4) = d,f(5) = e,f(6) = b。
  3. Let X = {1, 2, 3, 4, 5, 6, 7, 8}. It is known that R is an equivalence relation on X containing these ordered pairs: (2, 4), (5, 6), (1, 3), (4, 6), (8, 3). Find the relation R (satisfying these conditions) such that R has the smallest possible number of elements. Display R as a set of ordered pairs and as a digraph. [ Hint: Since (2, 4) ∈ R and R is symmetric, R must also contain (4, 2). Make further deductions of this kind to see what other ordered pairs must be in R .]
    设 X = {1, 2, 3, 4, 5, 6, 7, 8}。已知 R 是 X 上的一个等价关系,包含以下有序对:(2, 4), (5, 6), (1, 3), (4, 6), (8, 3)。求满足上述条件的关系 R,使得 R 的元素个数最少。将 R 表示为有序对集合和有向图。[提示:由于 (2, 4) ∈ R 且 R 是对称的,因此 R 也必须包含 (4, 2)。继续进行类似的推导,找出 R 中还包含哪些其他有序对。]
  4. Draw the digraph of the smallest equivalence relation R on the set X = {1, 2, 3, 4, 5, 6, 7, 8} such that R contains the ordered pairs (1, 3), (1, 5), (1, 7), (2, 2), and (8, 6).
    在集合 X = {1, 2, 3, 4, 5, 6, 7, 8} 上绘制最小等价关系 R 的有向图,使得 R 包含有序对 (1, 3), (1, 5), (1, 7), (2, 2), 和 (8, 6)。
  5. Show that each relation R is an equivalence relation on X by finding a function f : X Y such that R is ∼ f . (a) X = Z , R = I Z . (b) X = Z , R = Z × Z . (c) X = Z , R is ≡ 2 . (d) X = R 2 , ( a , b ) R ( c , d ) iff ( a , b ) and ( c , d ) lie on the same vertical line. (e) X = R 2 , ( a , b ) R ( c , d ) iff ( a , b ) and ( c , d ) lie on the same circle centered at the origin. (f) X = R > 0 2 , ( a , b ) R ( c , d ) iff ( a , b ) and ( c , d ) lie on the same line through the origin. (g) X = R , xRy iff x , y R belong to the same interval ( n , n + 1] with n Z .
    证明每个关系 R 都是 X 上的等价关系,方法是找到一个函数 f: X → Y,使得 R ∼ f . (a) X=Z, R=IZ. (b) X=Z, R=Z×Z. (c) X=Z, R ≡ 2 . (d) X=R2, (a, b)R(c, d) 当且仅当 (a, b) 和 (c, d) 位于同一条垂直线上. (e) X=R2, (a, b)R(c, d) 当且仅当 (a, b) 和 (c, d) 位于以原点为中心的同一个圆上. (f) X=R>02, (a, b)R(c, d) 当且仅当 (a, b) 和 (c, d) 位于过原点的同一条直线上。 (g) X=R, xRy 当且仅当 x,y∈R 属于同一区间 (n, n + 1],其中 n∈Z。
  6. (a) Find all b Z such that 7 ≡ 2 b . (b) Find all b Z such that 7 ≡ 5 b . (c) Find all b Z such that 7 ≡ 10 b .
    (a)找出所有满足 7 ≡ 2 b 的 b∈Z。(b)找出所有满足 7 ≡ 5 b 的 b∈Z。(c)找出所有满足 7 ≡ 10 b 的 b∈Z。
  7. Find all n such that 111 ≡ n 471.
    找到所有 n,使得 111 ≡ n 471。
  8. Prove part (b) of the Theorem on Congruence Modulo n .
    证明模 n 同余定理的 (b) 部分。
  9. Fix n ≥ 1. (a) Prove: for all a , b Z , if a n b , then − a n b . (b) Prove: for all a , b , c , d Z , if a n b and c n d , then a c n b d .
    固定 n ≥ 1。(a)证明:对于所有 a,b∈Z,如果 a ≡ n b,则 −a ≡ n − b。(b)证明:对于所有 a,b,c,d∈Z,如果 a ≡ n b 且 c ≡ n d,则 a − c ≡ n b − d。
  10. Prove by induction on k : for all n Z 1 and all k Z 0 and all a , b Z , if a b ( mod n ) , then a k b k ( mod n ) .
    对 k 进行归纳证明:对于所有 n∈Z≥1 和所有 k∈Z≥0 和所有 a,b∈Z,如果 a≡b(modn),则 ak≡bk(modn)。
  11. Draw the digraphs of all equivalence relations on the set X = {1, 2, 3}.
    画出集合 X = {1, 2, 3} 上所有等价关系的有向图。
  12. Give a specific example to show that the union of two equivalence relations on the set {1, 2, 3, 4} need not be an equivalence relation on this set.
    举一个具体的例子来说明集合 {1, 2, 3, 4} 上的两个等价关系的并集不一定是该集合上的等价关系。
  13. Let X = Z × Z . Define a relation R on X by setting ( a , b ) R ( c , d ) iff a + d = b + c . Prove that R is an equivalence relation on X .
    设 X=Z×Z。在 X 上定义关系 R,使得 (a, b)R(c, d) 当且仅当 a + d = b + c。证明 R 是 X 上的等价关系。
  14. Let X = Z × Z 0 . Define a relation R on X by setting ( a , b ) R ( c , d ) iff ad = bc . Prove that R is an equivalence relation on X . Would this still be true if we had used X = Z × Z ?
    设 X=Z×Z≠0。定义 X 上的关系 R,使得 (a, b)R(c, d) 当且仅当 ad = bc。证明 R 是 X 上的等价关系。如果 X=Z×Z,这个结论还成立吗?
  15. Define a relation ∼ on R n by setting v w iff there exists r R 0 with w = r v (i.e., w i = rv i for i = 1, 2, …, n ). Prove that ∼ is an equivalence relation on R n . Would this still be true if we had used r R instead of r R 0 ?
    在 Rn 上定义关系 ∼,令 v ∼ w 当且仅当存在 r∈R≠0 使得 w = rv(即,对于 i = 1, 2, …, n,w i = rv i )。证明 ∼ 是 Rn 上的等价关系。如果我们用 r∈R 代替 r∈R≠0,这个结论是否仍然成立?
  16. Given sets A and X with A X , let R = ( A × A ) ( ( X A ) × ( X A ) ) . Prove R is an equivalence relation on X .
    给定集合 A 和 X,其中 A⊆ X,令 R=(A×A)∪((X−A)×(X−A))。证明 R 是 X 上的一个等价关系。
  17. Let S be any relation from X to Y . Define a relation ∼ S on X by setting (for a , b X ) a S b iff S [{ a }] = S [{ b }]. Is ∼ S an equivalence relation on X ?
    设 S 是从 X 到 Y 的任意关系。在 X 上定义关系 ∼ S ,使得(对于 a, b ∈ X)a ∼ S b 当且仅当 S[{a}] = S[{b}]。∼ S 是 X 上的等价关系吗?
  18. Fix n ≥ 1, and define f : Z { 0 , 1 , , n 1 } by letting f ( a ) be the unique remainder in {0, 1, …, n − 1} when a is divided by n . Prove that ≡ n is the equivalence relation ∼ f .
    固定 n ≥ 1,定义 f:Z→{0,1,…,n−1},使得 f(a) 为 a 除以 n 在 {0, 1, …, n − 1} 中的唯一余数。证明 ≡ n 是等价关系 ∼ f
  19. Let X be any collection of functions with a common domain D . Let S be a fixed subset of D . For f , g X , define f g iff f ( s ) = g ( s ) for all s S . Prove ∼ is an equivalence relation on X .
    设 X 为具有公共定义域 D 的任意函数族。设 S 为 D 的一个固定子集。对于 f, g ∈ X,定义 f ∼ g 当且仅当对于所有 s ∈ S,f(s) = g(s)。证明 ∼ 是 X 上的等价关系。
  20. Let X be any collection of sets. For A , B X , define A B iff there exists a bijection f : A B . Prove ∼ is an equivalence relation on X .
    设 X 为任意集合族。对于 A, B ∈ X,定义 A ∼ B 当且仅当存在双射 f:A → B。证明 ∼ 是 X 上的等价关系。
  21. (a) Given functions f : X Y and g : Y Z , prove that ∼ f is a subset of ∼ g f . (b) Prove that if g is injective, then ∼ f equals ∼ g f .
    (a)给定函数 f:X → Y 和 g:Y → Z,证明 ∼ f 是 ∼ g f 的子集。(b)证明如果 g 是单射函数,则 ∼ f 等于 ∼ g f
  22. Write R m for the relation ≡ m . Given positive integers n 1 , …, n k , prove that i = 1 k R n i = R N , where N = lcm( n 1 , …, n k ).
    用 R m 表示关系 ≡ m 。给定正整数 n 1 , …, n k ,证明 ⋂i=1kRni=RN,其中 N = lcm(n 1 , …, n k )。
  23. Given any relation S on a set X , let R be the intersection of all equivalence relations on X containing S . (a) Show that R is an equivalence relation on X containing S , such that R R ′ for all equivalence relations R ′ on X containing S . (b) Let T = n = 1 ( S S 1 I X ) n (this notation is defined in Exercise 16 of § 6.1 ). Prove R = T .
    给定集合 X 上的任意关系 S,令 R 为 X 上所有包含 S 的等价关系的交集。(a) 证明 R 是 X 上包含 S 的等价关系,使得对于 X 上所有包含 S 的等价关系 R′,R⊆ R′。(b) 令 T=⋃n=1∞(S∪S−1∪IX)n(此符号定义见 §6.1 的练习 16)。证明 R = T。

6.3 Equivalence Classes

6.3 等价类

Intuitively, an equivalence relation R on a set X lets us “clump together” all the objects in X that are the same in some way, creating a collection of subsets of X . For instance, congruence modulo 2 decomposes the set of integers into the subset of odd integers and the subset of even integers. The subset of X consisting of all objects in X related under R to a given object b is called the equivalence class of b determined by R . We can gain a lot of insight into an equivalence relation by studying its equivalence classes. We consider many examples of equivalence classes in this section. We then show that any two equivalence classes of R are either equal or do not overlap at all, so that every element of X belongs to exactly one equivalence class.

直观地说,集合 X 上的等价关系 R 允许我们将 X 中所有以某种方式相同的对象“聚集”在一起,从而创建 X 的子集集合。例如,模 2 同余将整数集合分解为奇数子集和偶数子集。由 R 下与给定对象 b 相关的 X 中的所有对象组成的子集称为由 R 确定的 b 的等价类。通过研究等价类,我们可以深入了解等价关系。本节将考虑许多等价类的例子。然后,我们将证明 R 的任意两个等价类要么相等,要么完全不重叠,因此 X 中的每个元素都恰好属于一个等价类。

Definition and Examples of Equivalence Classes

等价类的定义和示例

Consider again the digraphs of the three equivalence relations shown in the top row of Figure 6.4 , which are redrawn in Figure 6.5 below. We notice that the arrows in the digraph divide the points of X into different clumps, where all pairs of points in a given clump are connected by relation arrows, but no arrow goes between points in two different clumps. The next definition helps us understand this situation better by giving names to the clumps — they are the equivalence classes of the given equivalence relation.

再次考虑图 6.4 顶行所示的三个等价关系的有向图,它们在图 6.5 中重新绘制。我们注意到,有向图中的箭头将 X 中的点划分成不同的簇,同一簇内的所有点对都通过关系箭头连接,但不同簇之间的点之间没有箭头连接。下一个定义通过给这些簇命名来帮助我们更好地理解这种情况——它们是给定等价关系的等价类。

fig6_5

Figure 6.5

图 6.5

Digraphs for some equivalence relations.

某些等价关系的有向图。

6.41. Definition: Equivalence Classes.. Let R be an equivalence relation on a set X . For all b X , the equivalence class of b determined by R , denoted [ b ] R , is the set of all objects related to b by R . In symbols,

6.41. 定义:等价类。设 R 是集合 X 上的一个等价关系。对于所有 b ∈ X,由 R 确定的 b 的等价类,记为 [b] R ,是所有与 b 相关的对象的集合。用符号表示,

[ b b ] R = { x x X X : ( x x , b b ) R } .

Since R is symmetric, we can restate this definition in infix notation as follows:

由于 R 是对称的,我们可以用中缀表示法重新表述这个定义如下:

for all 对所有 b b , x x X X , x x [ b b ] R ( x x , b b ) R x x R b b b b R x x .

If R is understood from context, we may write [ b ] to abbreviate [ b ] R . Given any set S , S is an equivalence class of R iff b X , S = [ b ] R .

如果 R 可以从上下文中理解,我们可以用 [b] 来缩写 [b] R 。给定任意集合 S,SisanequivalenceclassofR 当且仅当 ∃b∈X,S=[b]R。

6.42 Example. Let us compute the equivalence classes for the equivalence relations shown in Figure 6.5 . For R 1 , we find that

6.42 示例。让我们计算图 6.5 所示等价关系的等价类。对于 R 1 ,我们发现

[ 1 ] R 1 = { 1 , 4 } , [ 2 ] R 1 = { 2 , 3 } , [ 3 ] R 1 = { 2 , 3 } , [ 4 ] R 1 = { 1 , 4 } , [ 5 ] R 1 = { 5 , 6 } , [ 6 ] R 1 = { 5 , 6 } .

Notice that [ 1 ] R 1 = [ 4 ] R 1 , [ 2 ] R 1 = [ 3 ] R 1 , and [ 5 ] R 1 = [ 6 ] R 1 . In general, for an equivalence relation R , it often happens that a b and yet [ a ] R = [ b ] R . This means that a given equivalence class S may have several different names [ a ] R for various choices of a X . In fact, we will prove shortly that for all a , b X , [ a ] R = [ b ] R iff aRb . In the case of R 1 , there are only three distinct equivalence classes, specifically {1, 4} and {2, 3} and {5, 6}, but each equivalence class has two different names. The fact that a given equivalence class may have multiple names is a confusing but fundamental feature of this concept.

注意,[1]R1=[4]R1,[2]R1=[3]R1,且[5]R1=[6]R1。一般来说,对于等价关系 R,经常会出现 a ≠ b 的情况,但 [a] R = [b] R 。这意味着对于给定的等价类 S,对于不同的 a ∈ X,可能存在多个不同的名称 [a] R 。事实上,我们稍后将证明,对于所有 a, b ∈ X,[a] R = [b] R 当且仅当 aRb。对于 R 1 ,只有三个不同的等价类,分别是 {1, 4}、{2, 3} 和 {5, 6},但每个等价类都有两个不同的名称。一个给定的等价类可以有多个名称,这是该概念一个令人困惑但又至关重要的特征。

Turning to R 2 , inspection of the digraph shows that

转向 R 2 ,检查有向图可知:

[ 1 ] R 2 = { 1 , 2 } = [ 2 ] R 2 , [ 3 ] R 2 = { 3 } , [ 4 ] R 2 = { 4 , 5 , 6 } = [ 5 ] R 2 = [ 6 ] R 2 .

We see that there are three distinct equivalence classes of R 2 , namely {1, 2} and {3} and {4, 5, 6}. The class {1, 2} has two members and two names; the class {3} has one member and one name; the class {4, 5, 6} has three members and three names. In general, if S is an equivalence class of an equivalence relation R , each member a S gives us a possible name [ a ] R for the equivalence class S . For this reason, each a S is called a representative of the equivalence class S .

我们看到 R 2 有三个不同的等价类,分别是 {1, 2}、{3} 和 {4, 5, 6}。类 {1, 2} 有两个成员和两个名称;类 {3} 有一个成员和一个名称;类 {4, 5, 6} 有三个成员和三个名称。一般来说,如果 S 是等价关系 R 的一个等价类,则 S 中的每个成员 a 都给出了等价类 S 的一个可能名称 [a] R 。因此,每个 a ∈ S 都被称为等价类 S 的一个代表元。

Finally, R 3 has three equivalence classes: [ 2 ] R 3 = { 2 } , [ 5 ] R 3 = { 5 } , and [ 1 ] R 3 = { 1 , 3 , 4 , 6 } = [ 3 ] R 3 = [ 4 ] R 3 = [ 6 ] R 3 .

最后,R 3 有三个等价类:[2]R3={2}、[5]R3={5}、[1]R3={1,3,4,6}=[3]R3=[4]R3=[6]R3。

6.43 Example. For any set X , consider the equality relation I X = {( x , x ): x X }. For all a , x X , xI X a holds iff x = a , so that [ a ] I X = { a } . In this case, each equivalence class consists of a single element. Intuitively, the relation of logical equality never clumps together two unequal objects. (See the first digraph in Figure 6.1 .) At the other extreme, consider the equivalence relation R = X × X . For all a , x X , xRa is true, so [ a ] R = X for all a X . In this case, there is a single equivalence class consisting of every element of X . This equivalence relation clumps together all objects in X .

6.43 示例。对于任意集合 X,考虑等式关系 I X = {(x, x):x ∈ X}。对于所有 a, x ∈ X,xI X a 成立当且仅当 x = a,因此 [a]IX={a}。在这种情况下,每个等价类都包含一个元素。直观地说,逻辑等式关系永远不会将两个不相等的对象聚集在一起。(参见图 6.1 中的第一个有向图。)在另一个极端,考虑等价关系 R = X × X。对于所有 a, x ∈ X,xRa 为真,因此对于所有 a ∈ X,[a] R = X。在这种情况下,存在一个包含 X 中所有元素的等价类。此等价关系将 X 中的所有对象聚集在一起。

Equivalence Classes of ∼ f

f 的等价类

We saw in § 6.2 that for any function f : X Y , there is an associated equivalence relation ∼ f on X such that for all a , b X , a f b iff f ( a ) = f ( b ). So for each a X , the equivalence class of a is

我们在 §6.2 中看到,对于任意函数 f:X → Y,在 X 上存在一个相关的等价关系 ∼ f ,使得对于所有 a, b ∈ X,a ∼ f b 当且仅当 f(a) = f(b)。因此,对于每个 a ∈ X,a 的等价类是

[ a 一个 ] f f = { x x X X : x x f f a 一个 } = { x x X X : f f ( x x ) = f f ( a 一个 ) } = f f 1 [ { f f ( a 一个 ) } ] .

This equivalence class is sometimes called the fiber of f over f ( a ). The figure in the next example explains this terminology.

这个等价类有时被称为 f 在 f(a) 上的纤维。下一个例子中的图解释了这个术语。

6.44 Example. Let X = {1, 2, 3, 4, 5, 6}, Y = { r , s , t , u }, and define f : X Y by f (1) = f (2) = f (3) = s , f (4) = f (6) = t , and f (5) = u . Figure 6.6 shows the arrow diagram for f and the digraph for ∼ f in a single picture. The arrows leading into a given point in the image of f resemble threads in a fiber starting at that point and leading back into the domain of f . The distinct equivalence classes of ∼ f are [ 1 ] f = { 1 , 2 , 3 } = [ 2 ] f = [ 3 ] f = f 1 [ { s } ] , [ 4 ] f = { 4 , 6 } = [ 6 ] f = f 1 [ { t } ] , and [ 5 ] f = { 5 } = f 1 [ { u } ] . As before, we can specify an equivalence class by listing its elements or by picking a representative b and using the name [ b ] f . In this situation, we can also use fiber notation for an equivalence class, e.g., denoting {1, 2, 3} as f −1 [{ s }]. The fiber f −1 [{ r }] is the empty set, which is not an equivalence class.

6.44 例。设 X = {1, 2, 3, 4, 5, 6},Y = {r, s, t, u},并定义函数 f:X → Y,其中 f(1) = f(2) = f(3) = s,f(4) = f(6) = t,f(5) = u。图 6.6 展示了函数 f 的箭头图和 ∼ f 的有向图,两者绘制在同一幅图中。指向 f 图像中某一点的箭头类似于从该点出发并返回 f 定义域的纤维。 ∼ f 的不同等价类为 [1]∼f={1,2,3}=[2]∼f=[3]∼f=f−1[{s}]、[4]∼f={4,6}=[6]∼f=f−1[{t}] 和 [5]∼f={5}=f−1[{u}]。与之前一样,我们可以通过列出等价类的元素或选择一个代表元素 b 并使用名称 [b]∼f 来指定等价类。在这种情况下,我们还可以使用纤维表示法来表示等价类,例如,将 {1, 2, 3} 表示为 f −1 [{s}]。纤维 f −1 [{r}] 是空集,它不是一个等价类。

fig6_6

Figure 6.6

图 6.6

A function f and its associated equivalence relation ∼ f .

函数 f 及其相关的等价关系 ∼ f

6.45 Example. For f : R R given by f ( x ) = | x |, the equivalence classes (fibers) are [ x ] f = { x , x } for each x R . For example, [ 3 ] f = { 3 , 3 } = [ 3 ] f , [ 7 ] f = { 7 , 7 } = [ 7 ] f , and [ 0 ] f = { 0 , 0 } = { 0 } . The function g : R R given by g ( x ) = x 2 has the same induced equivalence relation and hence the same equivalence classes. The function sgn : R R has three distinct equivalence classes: the set R < 0 of negative numbers, the set {0}, and the set R > 0 of positive numbers. For instance, [ 8 ] sgn = [ 0.001 ] sgn = [ 2 π ] sgn = R < 0 .

6.45 例。对于由 f(x) = |x| 给出的函数 f:R→R,其等价类(纤维)为 [x]∼f={x,−x},其中 x∈R。例如,[3]∼f={3,−3}=[−3]∼f,[−7]∼f={−7,7}=[7]∼f,以及 [0]∼f={0,−0}={0}。由 g(x) = x 2 给出的函数 g:R→R 具有相同的诱导等价关系,因此也具有相同的等价类。函数 sgn:R→R 有三个不同的等价类:负数集 R<0、集合 {0} 和正数集 R>0。例如,[−8]∼sgn=[−0.001]∼sgn=[−2π]∼sgn=R<0。

6.46 Example. Define f : R R 2 by f ( t ) = (cos t , sin t ) for t R . By trigonometry, f ( s ) = f ( t ) iff s = t + 2 πk for some integer k . Therefore,

6.46 例。定义函数 f:R→R2 为 f(t) = (cos t, sin t),其中 t∈R。根据三角函数,f(s) = f(t) 当且仅当 s = t + 2πk,其中 k 为某个整数。因此,

[ t t ] f f = { s s R : f f ( s s ) = f f ( t t ) } = { s s R : k k Z Z , s s = t t + 2 π p k k } .

For instance, [ 0 ] f = { 0 , ± 2 π , ± 4 π , } = [ 2 π ] f , and [ π / 3 ] f = { π / 3 , 7 π / 3 , 13 π / 3 , , 5 π / 3 , 11 π / 3 , } . In this example, each equivalence class has infinitely many members (representatives) and hence infinitely many names.

例如,[0]∼f={0,±2π,±4π,…}=[2π]∼f,且[π/3]∼f={π/3,7π/3,13π/3,…,−5π/3,−11π/3,…}。在这个例子中,每个等价类都有无穷多个成员(代表元),因此也有无穷多个名称。

More Examples of Equivalence Classes

等价类的更多示例

6.47 Example. Fix n ≥ 1, and consider congruence modulo n on the set Z . For brevity, let us write [ a ] n instead of [ a ] n . For each a Z , we have

6.47 例。固定 n ≥ 1,考虑集合 Z 上模 n 的同余。为简便起见,我们用 [a] n 代替 [a]≡n。对于每个 a∈Z,我们有

[ a 一个 ] n n = { x x Z Z : x x n n a 一个 } = { x x Z Z : n n divides 分割 x x a 一个 } = { x x Z Z : k k Z Z , x x a 一个 = k k n n } = { x x Z Z : k k Z Z , x x = a 一个 + k k n n } = { a 一个 + k k n n : k k Z Z } .

So, for example, [0] 5 = {0, ± 5, ± 10, ± 15, …} = [5] 5 = [10] 5 = [ − 15] 5 = [5 k ] 5 for any k in Z ; [1] 4 = {1, 5, 9, 13, …, − 3, − 7, − 11, …}; [777] 10 = {7, 17, 27, 37, …, − 3, − 13, − 23, …} = [7] 10 , and so on. When n = 2, the relation ≡ 2 has two distinct equivalence classes, the set of even integers and the set of odd integers. We can describe these sets as [0] 2 and [1] 2 , respectively. More generally, we will prove later that ≡ n has exactly n distinct equivalence classes, which can be named as [0] n , [1] n , …, [ n − 1] n . The set of these n equivalence classes is called the set of integers modulo n . We discuss the algebraic structure of the integers modulo n in § 6.6 .

例如,对于 Z 中的任何 k,[0] 5 = {0, ± 5, ± 10, ± 15, …} = [5] 5 = [10] 5 = [ − 15] 5 = [5k] 5 ;[1] 4 = {1, 5, 9, 13, …, − 3, − 7, − 11, …};[777] 10 = {7, 17, 27, 37, …, − 3, − 13, − 23, …} = [7] 10 ,依此类推。当 n = 2 时,关系 ≡ 2 有两个不同的等价类,即偶数集合和奇数集合。我们可以分别将这两个集合描述为 [0] 2 和 [1] 2 。更一般地,我们将在后面证明 ≡ n 恰好有 n 个不同的等价类,可以分别命名为 [0] n , [1] n , …, [n − 1] n 。这 n 个等价类的集合称为模 n 整数集合。我们将在 §6.6 中讨论模 n 整数的代数结构。

6.48 Example. Let X = R × R . For all ( a , b ), ( c , d ) ∈ X , define ( a , b ) R ( c , d ) iff a 2 + b 2 = c 2 + d 2 . R is the equivalence relation induced by the function f : X R given by f ( a , b ) = a 2 + b 2 . The equivalence class of (3, 4) ∈ X is [(3, 4)] R = {( x , y ) ∈ X : x 2 + y 2 = 3 2 + 4 2 = 25}, which is the circle of radius 5 centered at the origin. More generally, for any a 0 , b 0 R , [ ( a 0 , b 0 ) ] = { ( x , y ) X : x 2 + y 2 = a 0 2 + b 0 2 } is the circle of radius r 0 = a 0 2 + b 0 2 centered at the origin. A degenerate case occurs at the origin: [(0, 0)] R = {(0, 0)}. Note that every point in the plane belongs to exactly one circle centered at the origin. So the full set X has been decomposed into a union of equivalence classes (the circles), where any two unequal equivalence classes are disjoint (have empty intersection).

6.48 例。令 X=R×R。对于所有 (a, b), (c, d) ∈ X,定义 (a, b)R(c, d) 当且仅当 a 2 + b 2 = c 2 + d 2 。R 是由函数 f:X→R 导出的等价关系,其中 f(a, b) = a 2 + b 2 。点 (3, 4) ∈ X 的等价类为 [(3, 4)] R = {(x, y) ∈ X:x 2 + y 2 = 3 2 + 4 2 = 25},它是以原点为圆心、半径为 5 的圆。更一般地,对于任意 a0,b0∈R,[(a0,b0)]={(x,y)∈X:x2+y2=a02+b02} 是以原点为圆心、半径为 r0=a02+b02 的圆。在原点处存在退化情况:[(0, 0)] R = {(0, 0)}。注意,平面上的每个点都恰好属于一个以原点为圆心的圆。因此,整个集合 X 被分解为等价类(圆)的并集,其中任意两个不相等的等价类都是不相交的(交集为空)。

Now consider another relation S on X , defined by ( a , b ) S ( c , d ) iff there exists r R > 0 with c = ra and d = rb . We saw in the last section that S is an equivalence relation on X . The equivalence class of (3, 4) ∈ X determined by S is [ ( 3 , 4 ) ] S = { ( 3 r , 4 r ) : r R > 0 } , which is the ray from the origin through the point (3, 4) [excluding the origin itself]. More generally, each equivalence class [( a 0 , b 0 )] S is a ray starting at (but excluding) the origin and passing through ( a 0 , b 0 ). As before, the origin is a special case: [(0, 0)] S = {(0, 0)}. Once again, X has been decomposed into a union of equivalence classes (the rays, along with the origin), where any two equivalence classes are either disjoint or equal.

现在考虑 X 上的另一个关系 S,定义为 (a, b)S(c, d) 当且仅当存在 r∈R>0 使得 c = ra 且 d = rb。我们在上一节中看到,S 是 X 上的一个等价关系。由 S 确定的 (3, 4) ∈ X 的等价类是 [(3,4)]S={(3r,4r):r∈R>0},它是从原点出发经过点 (3, 4) 的射线(不包括原点本身)。更一般地,每个等价类 [(a 0 , b 0 )] S 都是从原点出发(但不包括原点)经过点 (a 0 , b 0 ) 的射线。与之前一样,原点是一个特例:[(0, 0)] S = {(0, 0)}。X 再次被分解为等价类的并集(射线以及原点),其中任意两个等价类要么不相交,要么相等。

6.49 Example. Let X be the set of subsets of {1, 2, 3}, and define a relation ∼ on X by letting A B iff A and B have the same number of elements. Then ∼ is an equivalence relation on X with the following equivalence classes: [ ] = { } ;

6.49 例。设 X 为集合 {1, 2, 3} 的子集集合,并在 X 上定义关系 ∼,使得 A ∼ B 当且仅当 A 和 B 的元素个数相同。则 ∼ 是 X 上的一个等价关系,其等价类如下:[∅]∼={∅};

[ { 1 } ] = { { 1 } , { 2 } , { 3 } } = [ { 2 } ] = [ { 3 } ] ;
[ { 1 , 2 } ] = { { 1 , 2 } , { 1 , 3 } , { 2 , 3 } } = [ { 1 , 3 } ] = [ { 2 , 3 } ] ;

and [{1, 2, 3}] = {{1, 2, 3}}. Here too, every member of X belongs to exactly one equivalence class, and two distinct equivalence classes do not share any common members.

并且 [{1, 2, 3}] = {{1, 2, 3}}。同样,X 的每个成员都属于恰好一个等价类,并且两个不同的等价类没有任何共同的成员。

6.50 Example. Let X be the set of all functions f : R R , and for all f , g X , define f g iff f (3) = g (3). The equivalence class of the squaring function s (given by s ( x ) = x 2 for x R ) is [ s ] = { f X : f (3) = s (3)} = { f X : f (3) = 9}. The equivalence class of the identity function is [ Id R ] = { f X : f ( 3 ) = 3 } . The equivalence class of a constant function f c with constant value c R is [ f c ] = { f X : f (3) = c }. In this example, each equivalence class contains exactly one constant function, which could be singled out as a particularly natural choice of a representative for that equivalence class. The constant tells us the value of f at input 3 for all functions f in the equivalence class.

6.50 例。设 X 为所有函数 f:R→R 的集合,对于任意 f, g ∈ X,定义 f ∼ g 当且仅当 f(3) = g(3)。平方函数 s(由 s(x) = x 2 给出,其中 x∈R)的等价类为 [s] = {f ∈ X:f(3) = s(3)} = {f ∈ X:f(3) = 9}。恒等函数的等价类为 [IdR]∼={f∈X:f(3)=3}。常数函数 f c 的等价类为 [f c ] = {f ∈ X:f(3) = c}。在这个例子中,每个等价类都恰好包含一个常数函数,这可以被视为该等价类的一个特别自然的代表函数。该常数告诉我们,对于等价类中的所有函数 f,输入 3 时 f 的值。

Properties of Equivalence Classes

等价类的性质

The following theorem lists some key features of equivalence classes illustrated by the preceding examples.

以下定理列出了前面例子所说明的等价类的一些关键特征。

6.51 Theorem on Equivalence Classes. Let R be an equivalence relation on a set X .

6.51 等价类定理。设 R 是集合 X 上的等价关系。

(a) When Equivalence Classes are Equal: For all a , b X , the following five conditions are equivalent: [ a ] R = [ b ] R ; b ∈ [ a ] R ; bRa ; aRb ; a ∈ [ b ] R .

#

(a)当等价类相等时:对于所有 a, b ∈ X,以下五个条件等价:[a] R = [b] R ; b ∈ [a] R ; bRa; aRb; a ∈ [b] R

(b) Disjointness of Equivalence Classes: For all equivalence classes S , T of R , if S T , then S T = . So two equivalence classes are either disjoint or equal .

(b)等价类的不相交性:对于 R 的所有等价类 S 和 T,如果 S ≠ T,则 S∩T=∅。因此,两个等价类要么不相交,要么相等。

(c) Partitioning Property: For all a X , there exists a unique equivalence class S of R with a S , namely S = [ a ] R . Every equivalence class S is nonempty.

(c)划分性质:对于所有 a ∈ X,存在 R 的唯一等价类 S,其中 a ∈ S,即 S = [a] R 。每个等价类 S 都是非空的。

Proof . (a) Fix a , b X . We give a circle proof that the indicated conditions are equivalent. First assume [ a ] R = [ b ] R . We know bRb since R is reflexive on X , so b ∈ [ b ] R by definition. As the sets [ a ] R and [ b ] R are equal, we see that b ∈ [ a ] R . Next, assume b ∈ [ a ] R . By definition of equivalence class, our assumption means that bRa . Next, assume bRa . Then aRb because R is symmetric. Next, assume aRb . Then a ∈ [ b ] R by definition. Finally, assume a ∈ [ b ] R , and prove [ a ] R = [ b ] R . Our assumption means aRb , and hence bRa by symmetry. Let us prove [ a ] R ⊆ [ b ] R . Fix c ; assume c ∈ [ a ] R ; prove c ∈ [ b ] R . We have assumed cRa and must prove cRb . Using cRa , aRb , and transitivity, we deduce cRb , as needed. Now we prove [ b ] R ⊆ [ a ] R . Fix d ; assume d ∈ [ b ] R ; prove d ∈ [ a ] R . We have assumed dRb and must prove dRa . Since dRb and bRa are known, transitivity gives dRa , as needed. We have now shown that [ a ] R = [ b ] R .

证明。(a) 固定 a, b ∈ X。我们给出循环证明,证明所给条件等价。首先假设 [a] R = [b] R 。由于 R 在 X 上自反,我们知道 bRb,因此根据定义,b ∈ [b] R 。由于集合 [a] R 和 [b] R 相等,我们看到 b ∈ [a] R 。接下来,假设 b ∈ [a] R 。根据等价类的定义,我们的假设意味着 bRa。接下来,假设 bRa。由于 R 是对称的,因此 aRb。接下来,假设 aRb。根据定义,因此 a ∈ [b] R 。最后,假设 a ∈ [b] R ,并证明 [a] R = [b] R 。我们的假设意味着 aRb,因此根据对称性,bRa 也成立。接下来,我们证明 [a] R ⊆ [b] R 。固定 c;假设 c ∈ [a] R ;证明 c ∈ [b] R 。我们已经假设 cRa,现在必须证明 cRb。利用 cRa、aRb 和传递性,我们推导出所需的 cRb。现在,我们证明 [b] R ⊆ [a] R 。固定 d;假设 d ∈ [b] R ;证明 d ∈ [a] R 。我们假设 dRb 成立,现在需要证明 dRa 成立。由于 dRb 和 bRa 已知,传递性使得 dRa 成立,满足要求。现在我们已经证明 [a] R = [b] R

(b) Fix two equivalence classes S , T of R . Using a contrapositive proof, we assume S T and prove S = T . By definition of equivalence class, there exist a , b X with S = [ a ] R and T = [ b ] R . Our assumption means that there exists c S T . Thus, c ∈ [ a ] R and c ∈ [ b ] R , so that cRa and cRb by definition. Now symmetry gives aRc , and transitivity gives aRb . By part (a), we conclude that [ a ] R = [ b ] R , which means that S = T .

(b) 固定 R 的两个等价类 S 和 T。利用逆否命题证明,我们假设 S∩T≠∅ 并证明 S = T。根据等价类的定义,存在 a, b ∈ X,使得 S = [a] R 且 T = [b] R 。我们的假设意味着存在 c∈S∩T。因此,c ∈ [a] R 且 c ∈ [b] R ,所以根据定义,cRa 和 cRb。现在,对称性给出 aRc,传递性给出 aRb。由 (a) 部分可知,[a] R = [b] R ,这意味着 S = T。

(c) Fix a X . By reflexivity, aRa , and hence a ∈ [ a ] R . Thus, S = [ a ] R is an equivalence class of R containing a . To prove uniqueness, suppose T and U are any two equivalence classes of R with a T and a U ; we must prove T = U . This follows from part (b), since a T U and hence T U . Finally, any equivalence class V has the form V = [ b ] R for some b X . Since b ∈ [ b ] R as remarked earlier, V must be nonempty. □

(c) 固定 a ∈ X。根据自反性,aRa,因此 a ∈ [a] R 。所以,S = [a] R 是 R 的一个包含 a 的等价类。为了证明唯一性,假设 T 和 U 是 R 的任意两个等价类,且 a ∈ T 和 a ∈ U;我们必须证明 T = U。这由 (b) 部分得出,因为 a∈T∩U,因此 T∩U≠∅。最后,任何等价类 V 都具有 V = [b] R 的形式,其中 b ∈ X。如前所述,由于 b ∈ [b] R ,因此 V 必定非空。□

Regarding part (c) of the theorem, note that each a X belongs to a unique equivalence class of R ; but the name of this equivalence class is usually not unique. Part (a) of the theorem tells us all possible names of a given equivalence class: [ b ] R is another name for the set [ a ] R iff [ b ] R = [ a ] R iff aRb iff b ∈ [ a ] R . This confirms our earlier remark that each representative (member) of a given equivalence class provides one of the possible names for this equivalence class.

关于定理的 (c) 部分,需要注意的是,每个 a ∈ X 都属于 R 的一个唯一等价类;但这个等价类的名称通常不是唯一的。定理的 (a) 部分告诉我们给定等价类的所有可能名称:[b] R 是集合 [a] R 的另一个名称,当且仅当 [b] R = [a] R 当且仅当 aRb 当且仅当 b ∈ [a] R 。这证实了我们之前的论断,即给定等价类的每个代表(成员)都提供了该等价类的一个可能名称。

Section Summary

章节概要

  1. Equivalence Classes. Given an equivalence relation R on a set X and given b X , the equivalence class of b determined by R is [ b ] R = { x X : xRb }. So x ∈ [ b ] R iff xRb iff bRx . A set S is an equivalence class of R iff b X , S = [ b ] R .
    等价类。给定集合 X 上的等价关系 R 和元素 b ∈ X,由 R 确定的 b 的等价类为 [b] R = {x ∈ X: xRb}。因此,x ∈ [b] R 当且仅当 xRb 当且仅当 bRx。集合 S 是 R 的一个等价类当且仅当存在 b∈X,S=[b]R。
  2. Fibers of a Function. Given f : X Y and y Y , the fiber of f over y is f −1 [{ y }]. For any a X , [ a ] f = { x X : f ( a ) = f ( x ) } = f 1 [ { f ( a ) } ] , which is the fiber of f over f ( a ).
    函数的纤维。给定 f:X → Y 和 y ∈ Y,f 在 y 上的纤维为 f −1 [{y}]。对于任意 a ∈ X,[a]∼f={x∈X:f(a)=f(x)}=f−1[{f(a)}],即 f 在 f(a) 上的纤维。
  3. Equivalence Classes of Congruence mod n . For all n ≥ 1 and a Z , [ a ] n = { x Z : x a ( mod n ) } = { a + k n : k Z } .
    模 n 同余的等价类。对于所有 n ≥ 1 和 a∈Z,[a] n = {x∈Z:x≡a(modn)}={a+kn:k∈Z}。
  4. Theorem on Equivalence Classes. Given an equivalence relation R on a set X and a , b X , [ a ] R = [ b ] R iff aRb iff bRa iff b ∈ [ a ] R iff a ∈ [ b ] R . For all equivalence classes S , T of R , if S T , then S T = . Equivalence classes are nonempty, and every a X belongs to exactly one equivalence class of R , namely [ a ] R .
    等价类定理。给定集合 X 上的等价关系 R 和 a, b ∈ X,[a] R = [b] R 当且仅当 aRb 当且仅当 bRa 当且仅当 b ∈ [a] R 当且仅当 a ∈ [b] R 。对于 R 的所有等价类 S, T,如果 S ≠ T,则 S∩T=∅。等价类非空,且每个 a ∈ X 都恰好属于 R 的一个等价类,即 [a] R

Exercises

练习

  1. For the equivalence relation R 3 shown in Figure 6.3 , explicitly list all elements in [ k ] R 3 for 1 ≤ k ≤ 7.
    对于图 6.3 所示的等价关系 R 3 ,明确列出 [k]R3 中 1 ≤ k ≤ 7 的所有元素。
  2. Let X = {1, 2, 3, 4, 5} and R = I X { ( 1 , 3 ) , ( 3 , 1 ) , ( 3 , 5 ) , ( 5 , 3 ) , ( 1 , 5 ) , ( 5 , 1 ) } . Find all equivalence classes of R .
    设 X = {1, 2, 3, 4, 5},R = IX∪{(1,3),(3,1),(3,5),(5,3),(1,5),(5,1)}。求 R 的所有等价类。
  3. Let X = {1, 2, 3, 4, 5, 6}. For each function f : X → { a , b , c , d , e }, find all equivalence classes of ∼ f . (a) f ( x ) = c for all x X . (b) f ( x ) = a for odd x X , f ( x ) = e for even x X . (c) f (1) = b , f (2) = e , f (3) = a , f (4) = b , f (5) = e , f (6) = a . (d) f (1) = a , f (2) = b , f (3) = c , f (4) = d , f (5) = e , f (6) = b .
    设 X = {1, 2, 3, 4, 5, 6}。对于每个函数 f:X → {a, b, c, d, e},求 ∼ f . 的所有等价类。(a) 对于所有 x ∈ X,f(x) = c。 (b) 对于奇数 x ∈ X,f(x) = a;对于偶数 x ∈ X,f(x) = e。 (c) f(1) = b,f(2) = e,f(3) = a,f(4) = b,f(5) = e,f(6) = a。 (d) f(1) = a,f(2) = b,f(3) = c,f(4) = d,f(5) = e,f(6) = b。
  4. For each function f : R R , describe the requested equivalence classes of ∼ f . (a) f ( x ) = x 4 ; find [ − 3], [2], and [0]. (b) f ( x ) = sin x ; find [0], [ π /2], and [4 π /3]. (c) f ( x ) = 3 x − 7; find [2], [5], and [ − 3]. (d) f ( x ) = 1 if x is rational, and 0 otherwise; find [3], [9/4], [ π ], and [ e ]. (e) f ( x ) = | x | if x is an integer, and 0 otherwise; find [4], [4.5], and [0].
    对于每个函数 f:R→R,描述 ∼ f . 的所请求的等价类:(a) f(x) = x 4 ;求 [ − 3]、[2] 和 [0]. (b) f(x) = sin x;求 [0]、[π/2] 和 [4π/3]. (c) f(x) = 3x − 7;求 [2]、[5] 和 [ − 3]. (d) f(x) = 1 当 x 为有理数时,否则为 0;求 [3]、[9/4]、[π] 和 [e]. (e) f(x) = |x| 当 x 为整数时,否则为 0;找到[4]、[4.5]和[0]。
  5. For each set X , equivalence relation R , and object b X , describe the equivalence class [ b ] R as explicitly as possible. Informally explain each answer. (a) X = {1, 2, 3, 4, 5, 6}, R = I X , b = 3. (b) X = {1, 2, 3, 4, 5, 6}, R = X × X , b = 3. (c) X = {1, 2, 3, 4, 5, 6}, R = ( { 1 , 3 , 5 } × { 1 , 3 , 5 } ) ( { 2 , 4 , 6 } × { 2 , 4 , 6 } ) , b = 3. (d) X = Z , R is congruence mod 2, b = 7. (e) X = Z , R is congruence mod 7, b = 7.
    对于每个集合 X、等价关系 R 和对象 b ∈ X,尽可能明确地描述等价类 [b] R 。请简要解释每个答案。 (a) X = {1, 2, 3, 4, 5, 6}, R = I X , b = 3. (b) X = {1, 2, 3, 4, 5, 6}, R = X × X, b = 3. (c) X = {1, 2, 3, 4, 5, 6}, R=({1,3,5}×{1,3,5})∪({2,4,6}×{2,4,6}), b = 3. (d) X=Z, R 模 2 全等, b = 7. (e) X=Z, R 模 7 全等, b = 7.
  6. For x R , let x be the least integer ≥ x . For instance, 5 = 5 , π = 4 , and 1 / 2 = 0 . For x , y R , define ( x , y ) ∈ R iff x = y . (a) State why R is an equivalence relation on R . (b) Describe the equivalence class [ π ] R . (c) Find three different real numbers x , y , z such that [ π ] R = [ x ] R = [ y ] R = [ z ] R . (d) Draw a picture of the real line showing all the distinct equivalence classes of R .
    对于 x∈R,令 ⌈x⌉ 为大于等于 x 的最小整数。例如,⌈5⌉=5,⌈π⌉=4,⌈−1/2⌉=0。对于 x,y∈R,定义 (x, y) ∈ R 当且仅当 ⌈x⌉=⌈y⌉。(a) 说明为什么 R 是 R 上的一个等价关系。(b) 描述等价类 [π] R 。(c) 找出三个不同的实数 x, y, z,使得 [π] R = [x] R = [y] R = [z] R 。(d) 画出实数轴,标出 R 的所有不同的等价类。
  7. For each equivalence relation R on X = R 2 , draw the equivalence classes [(1, 2)] R , [(3, 4)] R , [( − 4, − 3)] R , [(1, 1)] R , and [(0, 0)] R . Are any of these classes equal? (a) ( a , b ) R ( c , d ) means a = c . (b) ( a , b ) R ( c , d ) means b = d . (c) ( a , b ) R ( c , d ) means a b = c d . (d) ( a , b ) R ( c , d ) means a 2 + b 2 = c 2 + d 2 . (e) ( a , b ) R ( c , d ) means ( c , d ) = ( ra , rb ) for some r R 0 .
    对于 X=R2 上的每个等价关系 R,绘制等价类 [(1, 2)] R 、[(3, 4)] R 、[( − 4, − 3)] R 、[(1, 1)] R 和 [(0, 0)] R 。这些类中是否有相等的? (a) (a, b)R(c, d) 表示 a = c。 (b) (a, b)R(c, d) 表示 b = d。 (c) (a, b)R(c, d) 表示 a − b = c − d。 (d) (a, b)R(c, d) 表示 a 2 + b 2 = c 2 + d 2 (e) (a, b)R(c, d) 表示对于某些 r∈R≠0,(c, d) = (ra, rb)。
  8. Define an equivalence relation ∼ on the set of nonempty subsets of {1, 2, 3, 4} by letting A B mean that A and B have the same least element. Find all equivalence classes of ∼.
    在集合 {1, 2, 3, 4} 的非空子集上定义等价关系 ∼,令 A ∼ B 表示 A 和 B 具有相同的最小元素。求 ∼ 的所有等价类。
  9. Describe an equivalence relation on R 2 whose equivalence classes are squares centered at the origin.
    描述 R2 上的一个等价关系,其等价类是以原点为中心的正方形。
  10. Describe an equivalence relation R on Z such that for each k Z > 0 , R has an equivalence class of size k .
    描述 Z 上的等价关系 R,使得对于每个 k∈Z>0,R 都有一个大小为 k 的等价类。
  11. Find all positive integers n such that [ − 13] n = [37] n .
    找到所有正整数 n,使得 [ − 13] n = [37] n
  12. Find all equivalence relations R on X = {1, 2, 3, 4, 5} such that [1] R = [4] R and [2] R = [3] R .
    找出集合 X = {1, 2, 3, 4, 5} 上的所有等价关系 R,使得 [1] R = [4] R 且 [2] R = [3] R
  13. Prove: for all sets X , all equivalence relations R on X , and all b X , [ b ] R = R [{ b }].
    证明:对于所有集合 X、X 上的所有等价关系 R 以及所有 b ∈ X,[b] R = R[{b}]。
  14. Prove: for all sets X and all equivalence relations R on X , R = a X [ a ] R × [ a ] R .
    证明:对于所有集合 X 和 X 上的所有等价关系 R,R=⋃a∈X[a]R×[a]R。
  15. Prove: for all sets X and all equivalence relations R , S on X , R S iff for all a X , [ a ] R ⊆ [ a ] S .
    证明:对于所有集合 X 和 X 上的所有等价关系 R、S,R⊆ S 当且仅当对于所有 a ∈ X,[a] R ⊆ [a] S
  16. Prove: for all sets X , all equivalence relations R , S on X , and all a X , [ a ] R S = [ a ] R [ a ] S .
    证明:对于所有集合 X,所有 X 上的等价关系 R、S,以及所有 a ∈ X,[a]R∩S=[a]R∩[a]S。

6.4 Set Partitions

6.4 设置分区

This section introduces set partitions , which are ways of dividing a given set into collections of non-overlapping subsets. For example, the set of equivalence classes of a given equivalence relation on a set X is always a set partition of X . It turns out that this set partition contains enough information to reconstruct the equivalence relation that produced it. This observation leads us to the main theorem of this section, which states (intuitively) that equivalence relations and set partitions are two different ways of capturing the same underlying concept. More formally, for any set X , there is a natural one-to-one correspondence (bijection) between the set of all equivalence relations on X and the set of all set partitions of X .

本节介绍集合划分,即把给定的集合划分成若干互不重叠的子集的方法。例如,集合 X 上给定等价关系的等价类集合始终是 X 的一个集合划分。事实证明,这个集合划分包含足够的信息来重构产生它的等价关系。这一观察引出了本节的主要定理,该定理(直观地)指出,等价关系和集合划分是捕捉同一底层概念的两种不同方式。更正式地说,对于任意集合 X,X 上所有等价关系的集合与 X 的所有集合划分的集合之间存在自然的一一对应关系(双射)。

Definition and Examples of Set Partitions

集合划分的定义和示例

To motivate the introduction of set partitions, consider Figure 6.7 . The top row shows the digraphs of three equivalence relations on the set X = {1, 2, 3, 4, 5, 6}. The digraphs are rather cluttered because of all the arrows joining pairs of objects in the same equivalence class. The second row of the figure reduces the clutter by omitting all the arrows and circling elements that belong to the same equivalence class. Intuitively, the new diagrams contain exactly the same information as the old diagrams, since we can reconstruct the arrows by drawing all possible arrows between two objects in the same little oval.

为了说明引入集合划分的必要性,请参考图 6.7。第一行展示了集合 X = {1, 2, 3, 4, 5, 6} 上三个等价关系的有向图。由于连接同一等价类中对象对的箭头过多,这些有向图显得较为杂乱。图的第二行通过省略所有箭头并将属于同一等价类的元素圈起来,减少了这种杂乱感。直观地说,新图包含的信息与旧图完全相同,因为我们可以通过在同一个小椭圆内绘制两个对象之间所有可能的箭头来重构这些箭头。

fig6_7

Figure 6.7

图 6.7

Digraphs for equivalence relations and the associated set partitions.

等价关系的有向图及其相关的集合划分。

We can mathematically model each picture in the second row of Figure 6.7 as a set of subsets of X . For example, the first picture is encoded by P 1 = {{2, 3}, {1, 4}, {6, 5}}. The next two pictures correspond to the sets P 2 = {{1, 2}, {3}, {4, 5, 6}} and P 3 = {{2}, {1, 3, 4, 6}, {5}}. Note that the order in which we list the subsets, as well as the order in which the members of each subset are listed, does not matter. For instance, we could also write P 2 = {{3}, {5, 6, 4}, {2, 1}}. Also note that the subsets in each P i do not overlap each other, and every element of X belongs to one of the subsets. Each P i is an example of a set partition, which we formally define as follows.

我们可以将图 6.7 第二行中的每张图片用数学模型表示为 X 的子集集合。例如,第一张图片编码为 P 1 = {{2, 3}, {1, 4}, {6, 5}}。接下来的两张图片分别对应于集合 P 2 = {{1, 2}, {3}, {4, 5, 6}} 和 P 3 = {{2}, {1, 3, 4, 6}, {5}}。需要注意的是,子集的排列顺序以及每个子集中元素的排列顺序都无关紧要。例如,我们也可以写成 P 2 = {{3}, {5, 6, 4}, {2, 1}}。另请注意,每个 P i 中的子集彼此不重叠,并且 X 的每个元素都属于其中一个子集。每个 P i 都是一个集合划分的示例,我们将其正式定义如下。

6.52. Definition: Set Partitions.. For all sets X and P , P is a set partition of X iff:

6.52. 定义:集合划分。对于任意集合 X 和 P,P 是 X 的集合划分当且仅当:

(a) S P , S and S X ;

(a)∀S∈P,S≠∅且S⊆X;

(b) x X , S P , x S ; and

(b)∀x∈X,∃S∈P,x∈S;以及

(c) S , T P , S T S T = .

(c)∀S,T∈P,S≠T⇒S∩T=∅。

Each set S P is called a block of the set partition P .

每个集合 S ∈ P 称为集合划分 P 的一个块。

Condition (a) says that a set partition is a set of nonempty subsets of X . Condition (b) says that every element of X belongs to some subset in the set partition. Condition (c) says that any two blocks of the set partition are either equal or disjoint. When proving part (c), we often check the contrapositive: for all S , T P , if S T , then S = T . Conditions (a) and (b) together show that X = S P S , so that the whole set X is the union of all the blocks of the set partition. Conditions (b) and (c) are equivalent to the requirement that for all x X , there exists a unique block S P with x S (see Exercise 9).

条件 (a) 指出,集合划分是 X 的非空子集的集合。条件 (b) 指出,X 中的每个元素都属于集合划分中的某个子集。条件 (c) 指出,集合划分中的任意两个块要么相等,要么不相交。在证明条件 (c) 时,我们通常会验证逆否命题:对于所有 S, T ∈ P,如果 S∩T≠∅,则 S = T。条件 (a) 和 (b) 共同表明 X=⋃S∈PS,因此整个集合 X 是集合划分中所有块的并集。条件 (b) 和 (c) 等价于:对于所有 x ∈ X,存在唯一的块 S ∈ P,使得 x ∈ S(参见练习 9)。

6.53 Example. Which of the following are set partitions of X = {1, 2, 3, 4, 5}?

6.53 例题。下列哪些是集合 X = {1, 2, 3, 4, 5} 的划分?

P P 1 = { { 1 , 3 , 4 } , { 2 , 5 } } ; P P 2 = { 1 , 2 , 3 , 4 , 5 } ; P P 3 = { { 1 , 2 , 3 , 4 , 5 } } ; P P 4 = { { 1 , 5 } , { 2 , 4 } } ; P P 5 = { { 1 , 3 , 5 } , { 2 , 4 } , } ; P P 6 = { { 1 , 2 } , { 3 , 4 , 5 } , { 2 , 1 } } ; P P 7 = { { 1 , 2 , 3 } , { 3 , 4 , 5 } } ; P P 8 = { { 1 } , { 2 } , { 3 } , { 4 } , { 5 } } ; P P 9 = { { 1 , 2 , 3 } , { 4 , 5 , 6 } } .

Solution. P 1 is a set partition of X with blocks {1, 3, 4} and {2, 5}. P 2 is not a set partition of X because the elements of P 2 are members of X , rather than subsets of X . P 3 is a set partition of X whose only block is the full set X = {1, 2, 3, 4, 5}. At the other extreme, P 8 is a set partition with five blocks each containing a single element. P 4 is not a set partition of X because 3 ∈ X does not belong to any block of P 4 ; note P 4 is a set partition of the smaller set {1, 2, 4, 5}. P 5 is not a set partition because one of its members is the empty set, and all blocks are required to be nonempty. P 6 is a set partition with two distinct blocks {1, 2} and {3, 4, 5}; it does not matter that the block {1, 2} was listed twice. P 7 is not a set partition because the two unequal subsets {1, 2, 3} and {3, 4, 5} have nonempty intersection. Finally, P 9 is not a set partition of X since one of the members of P 9 , namely {4, 5, 6}, is not a subset of X . However, P 9 is a set partition of the larger set {1, 2, 3, 4, 5, 6}.

解:P 1 是 X 的一个集合划分,包含块 {1, 3, 4} 和 {2, 5}。P 2 不是 X 的一个集合划分,因为 P 2 中的元素是 X 的成员,而不是 X 的子集。P 3 是 X 的一个集合划分,其唯一块是整个集合 X = {1, 2, 3, 4, 5}。另一方面,P 8 是一个集合划分,包含五个块,每个块只包含一个元素。P 4 不是 X 的一个集合划分,因为 3 ∈ X 不属于 P 4 的任何块;注意 P 4 是较小集合 {1, 2, 4, 5} 的一个集合划分。 P 5 不是一个集合划分,因为它包含空集,而所有块都必须非空。P 6 是一个集合划分,包含两个不同的块 {1, 2} 和 {3, 4, 5};块 {1, 2} 出现两次并不影响结果。P 7 不是一个集合划分,因为两个不相等的子集 {1, 2, 3} 和 {3, 4, 5} 的交集非空。最后,P 9 不是 X 的一个集合划分,因为 P 9 的一个成员 {4, 5, 6} 不是 X 的子集。然而,P 9 是更大的集合 {1, 2, 3, 4, 5, 6} 的一个集合划分。

The Set Partition Induced by an Equivalence Relation

由等价关系诱导的集合划分

The major point of this section is that every equivalence relation on X determines a set partition of X , and vice versa. Intuitively, given an equivalence relation R on X , we obtain the associated set partition P by forming the set of all equivalence classes of R . The next definition gives a name to this set partition.

本节的主要观点是,集合 X 上的每个等价关系都确定了 X 的一个集合划分,反之亦然。直观地说,给定 X 上的一个等价关系 R,我们可以通过构造 R 的所有等价类的集合来得到与之相关的集合划分 P。下一个定义将为这个集合划分命名。

6.54. Definition: The Quotient Set X / R .. Given a set X and an equivalence relation R on X , the quotient set X / R is the set of all equivalence classes of R . Formally, X / R = {[ a ] R : a X }. So for all sets S , S X / R iff a X , S = [ a ] R . X / R is also called the set partition induced by R .

6.54. 定义:商集 X/R。给定一个集合 X 和 X 上的一个等价关系 R,商集 X/R 是 R 的所有等价类构成的集合。形式上,X/R = {[a] R :a ∈ X}。因此,对于所有集合 S,S∈X/R 当且仅当存在 a∈X,S=[a]R。X/R 也称为由 R 诱导的集合划分。

The fact that X / R really is a set partition of X follows from the Theorem on Equivalence Classes, parts (b) and (c).

X/R 确实是 X 的一个集合划分,这一事实可由等价类定理的 (b) 和 (c) 部分得出。

6.55 Example. Let X = {1, 2, 3, 4, 5, 6}. For any equivalence relation R on X , we always have X / R = {[1] R , [2] R , [3] R , [4] R , [5] R , [6] R }. But this presentation is misleading because a given equivalence class may be listed more than once (compare to the set partition P 6 in Example 6.53 ). We get a better understanding of the quotient set by choosing a single name for each distinct equivalence class. For example, consider the equivalence relations R 1 , R 2 , and R 3 from Figure 6.7 . We find that

6.55 例。设 X = {1, 2, 3, 4, 5, 6}。对于 X 上的任意等价关系 R,总有 X/R = {[1] R , [2] R , [3] R , [4] R , [5] R , [6] R 。但这种表示方式容易产生误导,因为一个给定的等价类可能会被多次列出(参见例 6.53 中的集合划分 P 6 )。通过为每个不同的等价类选择一个名称,我们可以更好地理解商集。例如,考虑图 6.7 中的等价关系 R 1 、R 2 和 R 3 。我们发现:

X X / R 1 = { { 1 , 4 } , { 2 , 3 } , { 5 , 6 } } = { [ 1 ] R 1 , [ 2 ] R 1 , [ 5 ] R 1 } = { [ 4 ] R 1 , [ 3 ] R 1 , [ 6 ] R 1 } ;
X X / R 2 = { { 1 , 2 } , { 3 } , { 4 , 5 , 6 } } = { [ 2 ] R 2 , [ 3 ] R 2 , [ 5 ] R 2 } ;
X X / R 3 = { { 1 , 3 , 4 , 6 } , { 2 } , { 5 } } = { [ 1 ] R 3 , [ 2 ] R 3 , [ 5 ] R 3 } .

We have X / I X = {{1}, {2}, {3}, {4}, {5}, {6}}. For R = X × X , X / R = {{1, 2, 3, 4, 5, 6}} = {[ k ] R } for any k X . For the function f from Example 6.44 ,

我们有 X/I X = {{1}, {2}, {3}, {4}, {5}, {6}}。对于 R = X × X,X/R = {{1, 2, 3, 4, 5, 6}} = {[k] R ,其中 k ∈ X。对于例 6.44 中的函数 f,

X X / f f = { { 1 , 2 , 3 } , { 4 , 6 } , { 5 } } = { [ 1 ] f f , [ 4 ] f f , [ 5 ] f f } = { f f 1 [ { s s } ] , f f 1 [ { t t } ] , f f 1 [ { u } ] } .

6.56 Example. For f : R R given by f ( x ) = | x |,

6.56 例。对于函数 f:R→R,定义为 f(x) = |x|,

R / f f = { [ x x ] f f : x x R } = { { x x , x x } : x x R 0 } .

For the sign function, R / sgn = { R < 0 , { 0 } , R > 0 } . For g : R 2 R given by g ( x , y ) = x 2 + y 2 , R 2 / g is the set of all circles centered at the origin (including the degenerate circle consisting of just the origin). For the relation S on R 2 from Example 6.48 , R 2 / S is the set of all rays starting at the origin, together with {0}.

对于符号函数,R/∼sgn={R<0,{0},R>0}。对于由g(x, y) = x 2 + y 2 给出的函数g:R2→R,R2/∼g是所有以原点为中心的圆的集合(包括仅由原点构成的退化圆)。对于例6.48中R2上的关系S,R2/S是所有以原点为起点的射线以及{0}的集合。

6.57 Example: Integers Modulo n . For fixed n ≥ 1, the quotient set Z / n is called the set of integers modulo n . We often write Z / n instead of Z / n . By definition, Z / n = { [ a ] n : a Z } . However, this presentation of the elements of Z / n contains a lot of redundancy. For instance, consider the case n = 5. By our calculation of equivalence classes in the previous section, we know that [ 0 ] 5 = { 5 k + 0 : k Z } , [ 1 ] 5 = { 5 k + 1 : k Z } , [ 2 ] 5 = { 5 k + 2 : k Z } , [ 3 ] 5 = { 5 k + 3 : k Z } , and [ 4 ] 5 = { 5 k + 4 : k Z } . These five subsets of Z already exhaust the entire set Z ; see Figure 6.8 . Note, for example, that [5] 5 = [0] 5 since 5 ∈ [0] 5 ; similarly [10] 5 = [15] 5 = [ − 5] 5 = [0] 5 , [7] 5 = [12] 5 = [17] 5 = [ − 3] 5 = [2] 5 , and so on. To summarize, Z / 5 is the five-element set Z / 5 = { [ 0 ] 5 , [ 1 ] 5 , [ 2 ] 5 , [ 3 ] 5 , [ 4 ] 5 } , where each element of Z / 5 is an infinite subset of Z . In § 6.6 , we prove the general fact that Z / n consists of the n distinct equivalence classes [0] n , [1] n , …, [ n − 1] n .

6.57 示例:模 n 的整数。对于固定的 n ≥ 1,商集 Z/≡n 称为模 n 的整数集。我们通常用 Z/n 代替 Z/≡n。根据定义,Z/n = {[a]n: a∈Z}。然而,这种 Z/n 元素的表示方式包含大量冗余。例如,考虑 n = 5 的情况。根据上一节中对等价类的计算,我们知道 [0]5 = {5k+0: k∈Z},[1]5 = {5k+1: k∈Z},[2]5 = {5k+2: k∈Z},[3]5 = {5k+3: k∈Z},以及 [4]5 = {5k+4: k∈Z}。这五个 Z 的子集已经穷尽了整个集合 Z;参见图 6.8。例如,注意 [5] 5 = [0] 5 ,因为 5 ∈ [0] 5 ;类似地,[10] 5 = [15] 5 = [ − 5] 5 = [0] 5 ,[7] 5 = [12] 5 = [17] 5 = [ − 3] 5 = [2] 5 ,依此类推。综上所述,Z/5 是包含五个元素的集合 Z/5={[0]5,[1]5,[2]5,[3]5,[4]5},其中 Z/5 的每个元素都是 Z 的无限子集。在 §6.6 中,我们证明了 Z/n 由 n 个不同的等价类 [0] n , [1] n , …, [n − 1] n 组成。

fig6_8

Figure 6.8

图 6.8

The quotient set Z / 5 , consisting of five equivalence classes.

商集 Z/5,由五个等价类组成。

The Equivalence Relation Induced by a Set Partition

由集合划分诱导的等价关系

So far, we have seen how an equivalence relation R on a set X generates an associated set partition X / R . Next, let us start with an arbitrary set partition P on a set X and see how to use P to build an equivalence relation ∼ P on X . The key idea is that two objects in X are related under ∼ P iff those objects belong to the same block of P ; compare to Figure 6.7 .

到目前为止,我们已经了解了集合 X 上的等价关系 R 如何生成一个相关的集合划分 X/R。接下来,让我们从集合 X 上的任意集合划分 P 开始,看看如何利用 P 在 X 上构建等价关系 ∼ P 。关键思想是,X 中的两个对象在 ∼ P 下相关当且仅当它们属于 P 的同一个块;参见图 6.7。

6.58. Definition: Equivalence Relation Induced by a Set Partition.. Given a set partition P of a set X , define a relation ∼ P on X as follows: for all a , b X , a P b iff S P , a S and b S . We call ∼ P the equivalence relation on X induced by P .

6.58. 定义:由集合划分诱导的等价关系。给定集合 X 的一个集合划分 P,定义 X 上的关系 ∼ P 如下:对于所有 a, b ∈ X,a∼Pb 当且仅当存在 S∈P,a∈S 且 b∈S。我们称 ∼ P 为由 P 诱导的 X 上的等价关系。

We check in a moment that ∼ P really is an equivalence relation on X . First, however, let us consider some examples.

我们稍后会验证 ∼ P 是否确实是 X 上的一个等价关系。不过,首先让我们考虑一些例子。

6.59 Example. Suppose X = {1, 2, 3, 4, 5} and consider the set partitions P 1 , P 3 , P 6 , and P 8 from Example 6.53 . Applying the definition to P 1 = {{1, 3, 4}, {2, 5}}, we find that

6.59 例。假设 X = {1, 2, 3, 4, 5},并考虑例 6.53 中的集合划分 P 1 、P 3 、P 6 和 P 8 。将定义应用于 P 1 = {{1, 3, 4}, {2, 5}},我们发现

P P 1 = { ( 1 , 1 ) , ( 3 , 3 ) , ( 4 , 4 ) , ( 1 , 3 ) , ( 3 , 1 ) , ( 1 , 4 ) , ( 4 , 1 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 2 , 2 ) , ( 5 , 5 ) , ( 2 , 5 ) , ( 5 , 2 ) } .

Since every x X belongs to the same block of P 3 , P 3 = { ( a , b ) : a , b X } = X × X . At the other extreme, P 8 = { ( a , a ) : a X } = I X . We can write P 6 as ( { 1 , 2 } × { 1 , 2 } ) ( { 3 , 4 , 5 } × { 3 , 4 , 5 } ) . If we try this construction on P 4 = {{1, 5}, {2, 4}} (which is not a set partition of X ), we get a relation

由于每个 x ∈ X 都属于 P 3 的同一个块,因此 ∼P3={(a,b):a,b∈X}=X×X。在另一个极端,∼P8={(a,a):a∈X}=IX。我们可以将 ∼P6 写成 ({1,2}×{1,2})∪({3,4,5}×{3,4,5})。如果我们尝试在 P 4 = {{1, 5}, {2, 4}}(它不是 X 的一个集合划分)上进行这种构造,我们会得到一个关系。

P P 4 = { ( 1 , 1 ) , ( 5 , 5 ) , ( 1 , 5 ) , ( 5 , 1 ) , ( 2 , 2 ) , ( 4 , 4 ) , ( 2 , 4 ) , ( 4 , 2 ) } ,

which is not reflexive on X since 3 P 4 3 . However, P 4 is an equivalence relation on the smaller set {1, 2, 4, 5}. Similarly, P 9 is not an equivalence relation on X since it is not a relation on X , but it is an equivalence relation on the larger set {1, 2, 3, 4, 5, 6}. Finally, if we apply the definition above to P 7 (which is not a set partition of any set), we get a non-transitive relation P 7 , since 1 P 7 3 and 3 P 7 5 but 1 P 7 5 .

由于 3≁P43,因此 ∼P4 在 X 上不具有自反性。然而,∼P4 在较小的集合 {1, 2, 4, 5} 上是一个等价关系。类似地,∼P9 在 X 上不是等价关系,因为它不是 X 上的关系,但它在较大的集合 {1, 2, 3, 4, 5, 6} 上是一个等价关系。最后,如果我们把上面的定义应用于 P 7 (它不是任何集合的划分),我们会得到一个非传递关系 ∼P7,因为 1∼P73 且 3∼P75 但 1≁P75。

6.60 Lemma. For any set partition P of any set X , ∼ P is an equivalence relation on X .

6.60 引理。对于任意集合 X 的任意集合划分 P,∼ P 是 X 上的一个等价关系。

Proof . Fix a set partition P of a set X . By definition, ∼ P is a relation on X ; we check ∼ P is reflexive on X , symmetric, and transitive. Fix a , b , c X . By (b) in the definition of set partition, there exists S P such that a S . So “ S P , a S a S ” is true, which means that a P a . Thus, ∼ P is reflexive on X . To prove symmetry, assume a P b and prove b P a . We have assumed S P , a S b S . By commutativity of AND, we see that S P , b S a S , so b P a holds. To prove transitivity, assume a P b and b P c ; prove a P c . We have assumed there exist blocks S , T P with a S , b S , b T , and c T . The critical observation is that b S T , so S T is nonempty. By (c) in the definition of set partition, S = T follows. So now a S and c S with S P , which means a P c . □

证明。固定集合 X 的一个集合划分 P。根据定义,∼ P 是 X 上的一个关系;我们验证 ∼ P 在 X 上是自反的、对称的和传递的。固定 a, b, c ∈ X。根据集合划分的定义 (b),存在 S ∈ P 使得 a ∈ S。因此,“∃S∈P,a∈S∧a∈S”为真,这意味着 a ∼ P a。因此,∼ P 在 X 上是自反的。为了证明对称性,假设 a ∼ P b,并证明 b ∼ P a。我们已经假设了 ∃S∈P,a∈S∧b∈S。由与运算的交换律可知,存在 S∈P, b∈S ∧ a∈S,因此 b ∼ a 成立。为了证明传递性,假设 a ∼ b 且 b ∼ c;证明 a ∼ c。我们假设存在块 S, T ∈ P,其中 a ∈ S, b ∈ S, b ∈ T, c ∈ T。关键在于 b∈S∩T,因此 S∩T 非空。根据集合划分的定义 (c),S = T。因此,现在 a ∈ S 且 c ∈ S,其中 S ∈ P,这意味着 a ∼ c。□

6.61 Example. Let X = R and P = { [ n , n + 1 ) : n Z } . By drawing a picture of the real number line, it is intuitively evident that P is a set partition of X (see Exercise 11 in § 8.5 for a formal proof). The associated equivalence relation ∼ P can be described as follows: for x , y R , x P y iff we obtain the same answer when we round x and y down to the next closest integer. Letting f ( x ) = x denote x rounded down to the next integer, we see that ∼ P is the equivalence relation ∼ f induced by the floor function f : R Z .

6.61 例。设 X=R 且 P={[n,n+1):n∈Z}。通过绘制实数轴,可以直观地看出 P 是 X 的一个集合划分(形式证明见 §8.5 中的练习 11)。相关的等价关系 ∼ P 可以描述如下:对于 x,y∈R,x ∼ P y 当且仅当将 x 和 y 向下取整到下一个最接近的整数时,结果相同。令 f(x)=⌊x⌋ 表示 x 向下取整到下一个整数,则 ∼ P 是由向下取整函数 f:R→Z 诱导的等价关系 ∼ f

The Bijection Between Equivalence Relations and Set Partitions

等价关系与集合划分之间的双射

Fix a set X . We have now discussed a construction that maps any equivalence relation R on X to an associated set partition X / R on X , along with another construction that maps any set partition P on X to an associated equivalence relation ∼ P on X . We now prove the key point that these constructions are inverses of each other. Intuitively, this means that equivalence relations and set partitions are really just two different formalizations of the same intuitive concept of dividing the objects in X into non-overlapping clumps, where all objects in a given clump share some common property.

设 X 为一个固定的集合。我们已经讨论了两种构造:一种构造将 X 上的任意等价关系 R 映射到 X 上的一个关联集合划分 X/R;另一种构造将 X 上的任意集合划分 P 映射到 X 上的一个关联等价关系 ∼ P 。现在,我们将证明关键的一点:这两种构造互为逆。直观地说,这意味着等价关系和集合划分实际上只是对同一个直观概念的两种不同形式化描述,即把 X 中的对象划分成互不重叠的簇,其中给定簇中的所有对象都具有某些共同的属性。

6.62 Theorem on Equivalence Relations and Set Partitions. Fix a set X , let E Q be the set of all equivalence relations on X , and let S P be the set of all set partitions of X . The map f : E Q S P given by f ( R ) = X / R for R E Q is a bijection with inverse g : S P E Q given by g ( P ) = ∼ P for P S P .

6.62 关于等价关系和集合划分的定理。固定一个集合 X,令 EQ 为 X 上所有等价关系的集合,令 SP 为 X 的所有集合划分的集合。映射 f:EQ→SP,定义为 f(R) = X/R,其中 R∈EQ,是一个双射,其逆映射 g:SP→EQ,定义为 g(P) = ∼ P ,其中 P∈SP。

Proof . [This proof may be considered optional.] We have already seen that f maps E Q into the codomain S P (by the Theorem on Equivalence Classes), and g maps S P into the codomain E Q (by Lemma 6.60 ). It suffices to check that g f = Id E Q and f g = Id S P . For g f , fix an equivalence relation R E Q , let P = f ( R ) = X / R , and let R ′ = g ( f ( R )) = g ( P ) = ∼ P . We must prove R = R ′. To do so, we fix a , b X , and we prove aRb iff aR b . First assume aRb , and prove aR b . Since aRb and bRb , we know that a ∈ [ b ] R and b ∈ [ b ] R . Now, [ b ] R is one of the blocks in the set partition P . Since a and b both belong to this block of P , we have a P b by definition of ∼ P , and hence aR b . Conversely, assume aR b , and prove aRb . We have assumed a P b , which means S P , a S b S . Now, every block S in P = X / R is an equivalence class of R . So S must have the form [ c ] R for some c X . Since a , b ∈ [ c ] R , we know aRc and bRc . Using symmetry and transitivity of R , we conclude aRb .

证明。[此证明可视为可选。] 我们已经看到,根据等价类定理,f 将 EQ 映射到值域 SP,而根据引理 6.60,g 将 SP 映射到值域 EQ。只需验证 g∘f=IdEQ 且 f∘g=IdSP。对于 g ○ f,固定一个等价关系 R∈EQ,令 P = f(R) = X/R,并令 R′ = g(f(R)) = g(P) = ∼ P 。我们必须证明 R = R′。为此,我们固定 a, b ∈ X,并且证明 aRb 当且仅当 aR′b。首先假设 aRb,并证明 aR′b。由于 aRb 且 bRb,我们知道 a ∈ [b] R 且 b ∈ [b] R 。现在,[b] R 是集合划分 P 中的一个块。由于 a 和 b 都属于 P 的这个块,根据 ∼ P 的定义,我们有 a ∼ P b,因此 aR′b 成立。反之,假设 aR′b 成立,并证明 aRb 成立。我们已经假设 a ∼ P b 成立,这意味着存在 S∈P, a∈S ∧ b∈S。现在,P = X/R 中的每个块 S 都是 R 的一个等价类。因此,S 必须具有 [c] R 的形式,其中 c ∈ X。由于 a, b ∈ [c] R ,我们知道 aRc 和 bRc 成立。利用 R 的对称性和传递性,我们得出 aRb 成立。

Next we check f g is the identity function on S P . Fix a set partition P S P , let R = g ( P ) = P E Q , and let P ′ = f ( g ( P )) = f ( R ) = X / R . We must prove P = P ′. First, fix S , assume S P , and prove S P ′. Since P is a set partition of X , there exists a X with a S . The equivalence class [ a ] R is one of the blocks in P ′, so it suffices to show S = [ a ] R . On one hand, fix b S , and prove b ∈ [ a ] R . Since b and a both belong to the block S of P , we have b P a , hence bRa , hence b ∈ [ a ] R . On the other hand, fix d ∈ [ a ] R , and prove d S . Here dRa , so d P a , so T P , d T a T . Now, S is the unique block of P containing a , so we must have T = S . Then d S , as needed.

接下来,我们检验 f ○ g 是否为 SP 上的恒等函数。固定一个集合划分 P∈SP,令 R=g(P)=∼P∈EQ,并令 P′ = f(g(P)) = f(R) = X/R。我们需要证明 P = P′。首先,固定 S,假设 S ∈ P,并证明 S ∈ P′。由于 P 是 X 的一个集合划分,因此存在 a ∈ X 使得 a ∈ S。等价类 [a] R 是 P′ 中的一个块,因此只需证明 S = [a] R 。一方面,固定 b ∈ S,并证明 b ∈ [a] R 。由于 b 和 a 都属于 P 的块 S,因此 b ∼ P a,从而 bRa,进而 b ∈ [a] R 。另一方面,固定 d ∈ [a] R ,并证明 d ∈ S。这里 dRa,所以 d ∼ P a,因此存在 T∈P,d∈T∧a∈T。现在,S 是 P 中包含 a 的唯一块,所以我们必须有 T = S。那么 d ∈ S,满足要求。

Conversely, fix a new S , assume S P ′, and prove S P . Here we know that S is an equivalence class of ∼ P , say S = [ a ] P for some a X . Let T be the unique block in P containing a ; it suffices to show S = T . On one hand, fix b S , and prove b T . Here b [ a ] P , so b P a , so there exists a block U of P containing b and a . By uniqueness, the only block containing a is T ; so U = T , and b T as needed. On the other hand, fix c T , and prove c S . Here c and a are in the same block T of the set partition P , so c P a , so c [ a ] P = S , as needed.      □

反之,固定一个新的集合 S,假设 S ∈ P′,并证明 S ∈ P。这里我们知道 S 是 ∼ P 的一个等价类,例如,对于某个 a ∈ X,S=[a]∼P。设 T 是 P 中包含 a 的唯一块;只需证明 S = T。一方面,固定 b ∈ S,并证明 b ∈ T。这里 b∈[a]∼P,所以 b ∼ P a,因此 P 中存在一个包含 b 和 a 的块 U。由唯一性可知,包含 a 的唯一块是 T;因此 U = T,b ∈ T,满足要求。另一方面,固定 c ∈ T,并证明 c ∈ S。这里 c 和 a 位于集合划分 P 的同一个块 T 中,所以 c ∼ P a,因此 c∈[a]∼P=S,满足要求。□

Section Summary

章节概要

  1. Set Partitions. For all sets X and P , P is a set partition of X iff every S P is a nonempty subset of X ; every x X belongs to some S P ; and for all S , T P , if S T , then S T = . Given a set partition P of X , every a X belongs to a unique block of P .
    集合划分。对于任意集合 X 和 P,P 是 X 的一个集合划分当且仅当每个 S ∈ P 都是 X 的非空子集;每个 x ∈ X 都属于某个 S ∈ P;并且对于任意 S, T ∈ P,如果 S ≠ T,则 S∩T=∅。给定 X 的一个集合划分 P,每个 a ∈ X 都属于 P 的一个唯一块。
  2. Quotient Sets. Given an equivalence relation R on a set X , the quotient set X / R is the set of all equivalence classes of R , namely X / R = {[ a ] R : a X }. So S X / R iff a X , S = [ a ] R . X / R is always a set partition of X , called the set partition induced by the equivalence relation R .
    商集。给定集合 X 上的等价关系 R,商集 X/R 是 R 的所有等价类的集合,即 X/R = {[a] R :a ∈ X}。因此,S ∈ X/R 当且仅当存在 a∈X,S=[a]R。X/R 始终是 X 的一个集合划分,称为由等价关系 R 诱导的集合划分。
  3. Equivalence Relation Induced by a Set Partition. Given a set partition P of a set X , the relation ∼ P , defined by a P b iff S P , a S b S (where a , b X ), is an equivalence relation on X . Informally, this equivalence relation relates each pair of objects that belong to the same block of P .
    由集合划分诱导的等价关系。给定集合 X 的一个划分 P,关系 ∼ P ,定义为 a ∼ P b 当且仅当 ∃S∈P,a∈S∧b∈S(其中 a, b ∈ X),是 X 上的一个等价关系。通俗地说,这个等价关系关联了属于 P 中同一划分块的每对对象。
  4. Correspondence between Equivalence Relations and Set Partitions. For any set X , there is a bijection from the set of equivalence relations on X to the set of set partitions of X . This bijection sends an equivalence relation R to the quotient set X / R . The inverse bijection sends a set partition P to the equivalence relation ∼ P . Figure 6.7 gives visual intuition for how this correspondence works.
    等价关系与集合划分之间的对应关系。对于任意集合 X,存在一条从 X 上的等价关系集合到 X 的集合划分集合的双射。这条双射将等价关系 R 映射到商集 X/R。逆双射将集合划分 P 映射到等价关系 ∼ P 。图 6.7 直观地展示了这种对应关系是如何运作的。

Exercises

练习

  1. Which of the following are set partitions of X = { v , w , x , y , z }? Explain. (a) {{ v , x , z }, { w , y }} (b) {{ v }, { w , z }, { x , y }, { z , w }, { v }} (c) {{ v , w }, { w , x }, { y , z }} (d) {{ v , x , y }, { z }} (e) { v , w , x , y , z } (f) {{ v , w , x , y , z }} (g) {{{ v , w , x , y , z }}}
    下列哪些是集合 X = {v, w, x, y, z} 的划分?请解释。 (a) {{v, x, z}, {w, y}} (b) {{v}, {w, z}, {x, y}, {z, w}, {v}} (c) {{v, w}, {w, x}, {y, z}} (d) {{v, x, y}, {z}} (e) {v, w, x, y, z} (f) {{v, w, x, y, z}} (g) {{{v, w, x, y, z}}}
  2. Which of the following are set partitions of R ? Explain. (a) { Z , R Z } (b) { R } (c) { { x } : x R } (d) { ( n 1 , n ) : n Z } (e) { ( n 1 , n ] : n Z } (f) { [ n 1 , n ] : n Z } (g) { ( n , n ) : n Z } .
    下列哪些是 R 的集合划分?请解释。 (a) {Z,R−Z} (b) {R} (c) {{x}:x∈R} (d) {(n−1,n):n∈Z} (e) {(n−1,n]:n∈Z} (f) {[n−1,n]:n∈Z} (g) {(−n,n):n∈Z}.
  3. Which of the following are set partitions of some set X ? For those that are, say what X is. (a) {1, 2, {3, 4}} (b) {{1, 2}, {3, 4}} (c) {{{1, 2}}, {3, 4}} (d) {{1, 3}, {4, 6}, {2, 8}} (e) { , { 1 } , { 2 , 3 } } (f) { { } , { { 1 } } , { { 1 , 2 } } } .
    下列哪些是集合 X 的集合划分?对于是的划分,请说明 X 是什么。(a) {1, 2, {3, 4}} (b) {{1, 2}, {3, 4}} (c) {{{1, 2}}, {3, 4}} (d) {{1, 3}, {4, 6}, {2, 8}} (e) {∅,{1},{2,3}} (f) {{∅},{{1}},{{1,2}}}.
  4. (a) List all possible set partitions of X = {1, 2, 3, 4}. (b) Use (a) to count the number of equivalence relations on X .
    (a)列出集合 X = {1, 2, 3, 4} 的所有可能的划分。(b)利用(a)计算集合 X 上的等价关系的数量。
  5. For each set X and equivalence relation R on X , explicitly describe the associated set partition X / R . (a) X = { a , b , c , d , e }, R = I X { ( a , c ) , ( c , a ) , ( c , e ) , ( e , c ) , ( a , e ) , ( e , a ) } . (b) X = { w , x , y , z }, R = X × X . (c) X = Z , R is ≡ 7 (congruence mod 7). (d ) X = R , xRy means y x Z . (e) X = P ( { 1 , 2 , 3 } ) , ( S , T ) ∈ R iff S and T have the same size. (f) X = R 2 , ( a , b ) R ( c , d ) means b a 2 = d c 2 .
    对于每个集合 X 和 X 上的等价关系 R,显式地描述其关联的集合划分 X/R。(a) X = {a, b, c, d, e},R = IX∪{(a,c),(c,a),(c,e),(e,c),(a,e),(e,a)}。(b) X = {w, x, y, z},R = X × X。(c) X = Z,R ≡ 7 (模 7 同余)。(d) X = R,xRy 表示 y−x∈Z。(e) X = P({1,2,3}),(S, T) ∈ R 当且仅当 S 和 T 的大小相同。(f) X = R²,(a, b)R(c, d) 表示 b − a 2 = d − c 2
  6. For each function f in Exercise 2 of § 6.2 , explicitly describe the set partition associated to the equivalence relation ∼ f .
    对于第 6.2 节练习 2 中的每个函数 f,明确描述与等价关系 ∼ f 相关的集合划分。
  7. For each function f : X Y , explicitly describe the set partition associated to the equivalence relation ∼ f . (a) f : [ 1 , 1 ] R given by f ( x ) = 1 x 2 for x ∈ [ − 1, 1]. (b) f : Z Z given by f ( n ) = −7 for all n Z . (c) f : Z 0 Z 0 given by f (0) = 0 and f ( n ) = n − 1 for all n > 0. (d) f : Z Z given by f ( n ) = n /2 for n even, and f ( n ) = ( n − 1)/2 for n odd. (e) f : R R given by f ( x ) = x x for x R .
    对于每个函数 f:X → Y,显式描述与等价关系 ∼ f . 相关的集合划分:(a) f:[−1,1]→R,其中 f(x)=1−x2,x ∈ [ − 1, 1]。 (b) f:Z→Z,其中 f(n) = −7,n∈Z。 (c) f:Z≥0→Z≥0,其中 f(0) = 0 且 f(n) = n − 1,n > 0。 (d) f:Z→Z,其中 f(n) = n/2,n 为偶数;f(n) = (n − 1)/2,n 为奇数。 (e) f:R→R,其中f(x)=x−⌊x⌋,其中 x∈R。
  8. For each set partition P of each set X , describe the induced equivalence relation ∼ P on X . (a) X = { a , b , c , d }, P = {{ a , c }, { b , d }}. (b) X = {1, 2, 3, 4, 5, 6}, P = {{1, 3, 6}, {2, 5}, {4}}. (c) X = Z , P = { Z 0 , Z > 3 , { 1 , 2 , 3 } } . (d) X = R 2 , P = { { a } × R : a R } . (e) X = R 2 , P = { { ( a , a + c ) : a R } : c R } .
    对于每个集合 X 的每个集合划分 P,描述 X 上的导出等价关系 ∼ P (a) X = {a, b, c, d}, P = {{a, c}, {b, d}}。 (b) X = {1, 2, 3, 4, 5, 6}, P = {{1, 3, 6}, {2, 5}, {4}}。 (c) X=Z, P={Z≤0,Z>3,{1,2,3}}。 (d) X=R2, P={{a}×R:a∈R}。 (e) X=R2, P={{(a,a+c):a∈R}:c∈R}。
  9. Let P be a collection of nonempty subsets of a set X . Prove: P is a set partition of X iff x X , ! S P , x S .
    设 P 是集合 X 的非空子集的集合。证明:P 是 X 的一个集合划分当且仅当 ∀x∈X,∃!S∈P,x∈S。
  10. Let A , B , C , D be pairwise disjoint nonempty sets. (a) Show that R = ( A × A ) ( B × B ) ( C × C ) ( D × D ) is an equivalence relation on X = A B C D . (b) What is the set partition X / R ?
    设 A、B、C、D 为两两不相交的非空集合。(a) 证明 R=(A×A)∪(B×B)∪(C×C)∪(D×D) 是 X=A∪B∪C∪D 上的等价关系。(b) 求 X/R 的集合划分。
  11. (a) Prove: for all sets X , Y , P , Q , if P is a set partition of X and Q is a set partition of Y and X Y = , then P Q is a set partition of X Y . (b) Show that (a) can be false without the hypothesis X Y = .
    (a)证明:对于所有集合 X、Y、P、Q,如果 P 是 X 的一个集合划分,Q 是 Y 的一个集合划分,且 X∩Y=∅,则 P∪Q 是 X∪Y 的一个集合划分。(b)证明在没有假设 X∩Y=∅ 的情况下,(a)可能不成立。
  12. Prove or disprove: for all sets X , Y , P , Q , if P is a set partition of X and Q is a set partition of Y , then P Q is a set partition of X Y .
    证明或反证:对于所有集合 X、Y、P、Q,如果 P 是 X 的一个集合划分,Q 是 Y 的一个集合划分,那么 P∩Q 是 X∩Y 的一个集合划分。
  13. Give an example of a set X and set partitions P , Q of X such that P is a four-element set, Q is an eight-element set, and every member of Q is a subset of some member of P . Is Q P here?
    举例说明集合 X 和 X 的划分 P、Q,其中 P 是一个包含四个元素的集合,Q 是一个包含八个元素的集合,并且 Q 中的每个元素都是 P 中某个元素的子集。这里 Q⊆ P 吗?
  14. Prove or disprove: for all sets X , P , Q , if P and Q are set partitions of X and P Q , then P = Q .
    证明或反证:对于所有集合 X、P、Q,如果 P 和 Q 是 X 的集合划分且 P⊆ Q,则 P = Q。
  15. Let R be an equivalence relation on a set X , and define p : X X / R by p ( a ) = [ a ] R for all a X . (a) Prove p is surjective. (b) Prove that R equals ∼ p . [This means that every equivalence relation has the form ∼ f for an appropriate choice of f .] (c) Prove p is injective iff R = I X .
    设 R 是集合 X 上的等价关系,定义 p:X → X/R 为 p(a) = [a] R ,其中 a ∈ X。(a)证明 p 是满射。(b)证明 R 等于 ∼ p 。[这意味着对于适当的 f,每个等价关系都具有 ∼ f 的形式。](c)证明 p 是单射当且仅当 R = I X

6.5 Partially Ordered Sets

6.5 部分有序集

Recall that the abstract idea of an equivalence relation came from the concrete concept of equality by isolating three key properties of equality: reflexivity, symmetry, and transitivity. In this section, we formulate the abstract concept of ordering by singling out appropriate properties of the concrete ordering relation ≤ on real numbers. This relation has the following properties: for all x , y , z R , (1) x x (reflexivity); (2) if x y and y z , then x z (transitivity); (3) if x y and y x , then x = y (antisymmetry); and (4) x y or y x (comparability). It turns out that many mathematical relations, like subset inclusion and divisibility, satisfy the first three of these properties. Thus we use these properties to define partial order relations. Partial orderings that also possess property (4) are called total orders. We also discuss well-ordered sets, in which every nonempty subset of the set has a least element.

回想一下,等价关系的抽象概念源于等式的具体概念,它提取了等式的三个关键性质:自反性、对称性和传递性。在本节中,我们通过选取实数上的具体排序关系 ≤ 的适当性质来构建排序的抽象概念。该关系具有以下性质:对于所有 x, y, z∈R,(1) x ≤ x(自反性);(2) 如果 x ≤ y 且 y ≤ z,则 x ≤ z(传递性);(3) 如果 x ≤ y 且 y ≤ x,则 x = y(反对称性);以及 (4) x ≤ y 或 y ≤ x(可比性)。许多数学关系,例如子集包含和可除性,都满足前三个性质。因此,我们利用这些性质来定义偏序关系。同时具有性质 (4) 的偏序关系称为全序。我们还讨论了良序集,其中集合的每个非空子集都有一个最小元素。

Antisymmetric Relations and Partial Orders

反对称关系和偏序关系

For all real numbers x and y , if x y and y x , then x = y . The generalization of this property to arbitrary relations is given in the next definition.

对于任意实数 x 和 y,如果 x ≤ y 且 y ≤ x,则 x = y。该性质推广到任意关系将在下一个定义中给出。

6.63. Definition: Antisymmetric Relations.. For all relations R :

6.63. 定义:反对称关系。对于所有关系 R:

R is antisymmetric 反对称 iff 当……时 a 一个 , b b , if 如果 ( a 一个 , b b ) R and ( b b , a 一个 ) R , then 然后 a 一个 = b b .

Using infix notation, R is antisymmetric iff for all a , b , if aRb and bRa , then a = b .

使用中缀表示法,R 是反对称的,当且仅当对于所有 a、b,如果 aRb 且 bRa,则 a = b。

In terms of the digraph for R , antisymmetry means that for any two distinct vertices a and b , the arrows from a to b and from b to a cannot both appear in the digraph. In Figure 6.1 of § 6.1 , every relation shown is antisymmetric except for R 4 . Relation R 1 is both symmetric and antisymmetric.

就关系 R 的有向图而言,反对称性意味着对于任意两个不同的顶点 a 和 b,从 a 到 b 的箭头和从 b 到 a 的箭头不能同时出现在有向图中。在 §6.1 的图 6.1 中,除关系 R 4 外,所示的每个关系都是反对称的。关系 R 1 既是对称的又是反对称的。

6.64. Definition: Partial Orders and Posets.. For all sets X and all relations R on X :

6.64. 定义:偏序与偏序集。对于所有集合 X 和 X 上的所有关系 R:

R is a 一个 partial 部分的 order 命令 on X X iff 当……时 R is reflexive 反身 on X X , antisymmetric 反对称 , and transitive 及物动词 .

The ordered pair ( X , R ) is called a partially ordered set or poset .

有序对 (X, R) 称为偏序集或偏序集。

We often use the ≤ symbol, or related symbols such as , to denote a partial order relation.

我们经常使用 ≤ 符号或相关符号(如 ⪯)来表示偏序关系。

6.65 Example: Subset Ordering Relation. Let X be any collection of sets. We can define a relation R on X as follows: for all A , B X , ( A , B ) ∈ R iff A B . We know that for all A , B , C X , A A (reflexivity); if A B and B A , then A = B (antisymmetry); and if A B and B C , then A C (transitivity). Thus, R is a partial ordering relation on X , and ( X , ⊆) is a partially ordered set.

6.65 示例:子集排序关系。设 X 为任意集合族。我们可以定义 X 上的关系 R 如下:对于任意 A, B ∈ X,(A, B) ∈ R 当且仅当 A ⊆ B。我们知道,对于任意 A, B, C ∈ X,A ⊆ A(自反性);如果 A ⊆ B 且 B ⊆ A,则 A = B(反对称性);并且如果 A ⊆ B 且 B ⊆ C,则 A ⊆ C(传递性)。因此,R 是 X 上的一个偏序关系,且 (X, ⊆) 是一个偏序集。

6.66 Example: Divisibility Ordering Relation. Let X be any set of positive integers. Define a relation R on X as follows: for all a , b X , ( a , b ) ∈ R iff a divides b . The notation a | b means that a divides b . We know that for all a , b , c Z > 0 , a | a (reflexivity); if a | b and b | a , then a = b (antisymmetry); and if a | b and b | c , then a | c (transitivity). Thus, ( X , |) is a partially ordered set. In contrast, ( Z , | ) is not a poset, because antisymmetry fails: 2|( − 2), and ( − 2)|2, yet 2 ≠ −2. To see why antisymmetry works in Z > 0 , fix positive integers a , b , and assume a | b and b | a . Because a and b are positive, we can deduce a b and b a (see Example 2.36). Then a = b because the standard ordering on integers is antisymmetric.

6.66 示例:整除性关系。设 X 为任意正整数集合。定义 X 上的关系 R 如下:对于任意 a, b ∈ X,(a, b) ∈ R 当且仅当 a 整除 b。记号 a|b 表示 a 整除 b。我们知道,对于任意 a, b, c ∈ Z>0,a|a(自反性);如果 a|b 且 b|a,则 a = b(反对称性);如果 a|b 且 b|c,则 a|c(传递性)。因此,(X, |) 是一个偏序集。相反,(Z, |) 不是一个偏序集,因为反对称性不成立:2|(-2),且 (-2)|2,但 2 ≠ -2。为了理解反对称性在 Z>0 中成立的原因,固定正整数 a 和 b,并假设 a|b 且 b|a。因为 a 和 b 都是正数,我们可以推导出 a ≤ b 且 b ≤ a(参见例 2.36)。那么 a = b,因为整数的标准排序是反对称的。

When X is finite, we can visualize a poset ( X , ≤ ) using a picture called a Haase diagram. This picture is obtained from the digraph of the relation ≤ by omitting all arrows that can be deduced from reflexivity or transitivity. It is customary to position the elements so that each relation arrow goes from a lower element to a higher element; this is possible by antisymmetry. For example, Figure 6.9 shows the Haase diagrams of four posets: the set P ( { 1 , 2 , 3 } ) ordered by set inclusion, the set of positive divisors of 30 ordered by divisibility, the set of positive divisors of 12 ordered by divisibility, and the set of positive divisors of 12 ordered by the standard ordering of integers.

当 X 为有限集时,我们可以用称为哈斯图的图像来可视化一个偏序集 (X, ≤)。该图像是通过省略关系 ≤ 的所有可由自反性或传递性推导出的箭头而得到的。通常,元素的排列方式使得每个关系箭头都从较低的元素指向较高的元素;这可以通过反对称性来实现。例如,图 6.9 展示了四个偏序集的哈斯图:按集合包含关系排序的集合 P({1,2,3})、按整除性排序的 30 的正因子集合、按整除性排序的 12 的正因子集合,以及按整数的标准排序的 12 的正因子集合。

fig6_9

Figure 6.9

图 6.9

Haase diagrams for four posets.

四个偏序集的哈斯图。

The last two examples show that there can be more than one partial order relation on a given set. It also frequently happens in a poset ( X , ) that there are incomparable pairs of elements a , b where neither a b nor b a is true. For instance, in the poset ( Z > 0 , | ) , 3|4 and 4|3 are both false. In contrast, for any two real numbers x and y , x y or y x must be true. Generalizing this latter property, we arrive at the definition of total ordering relations.

最后两个例子表明,给定集合上可以存在多个偏序关系。在偏序集 (X, ⪯) 中,经常会出现元素对 a 和 b 不可比较的情况,其中 a⪯b 和 b⪯a 都不成立。例如,在偏序集 (Z>0, |) 中,3|4 和 4|3 都为假。相反,对于任意两个实数 x 和 y,x ≤ y 或 y ≤ x 必然成立。推广后一个性质,我们便得到了全序关系的定义。

6.67. Definition: Total Orders.. A relation on a set X is a total order on X iff is a partial order on X such that a , b X , a b or b a . The pair ( X , ) is called a totally ordered set .

6.67. 定义:全序关系。集合 X 上的关系 ⪯ 是 X 上的全序关系,当且仅当 ⪯ 是 X 上的偏序关系,使得对于任意 a, b∈X,a⪯ 或 b⪯ a。二元组 (X, ⪯) 称为全序集。

We have seen above that ( R , ) is a totally ordered set, but ( Z > 0 , | ) is not totally ordered. Similarly, a poset ( X , ⊆) is not totally ordered if we can find subsets A , B X such that A B and B A . For example, if X = P ( { 1 , 2 , 3 , 4 } ) , we may take A = {1, 2} and B = {2, 3} to see that ( X , ⊆) is not totally ordered. On the other hand, if S = { , { 1 } , { 1 , 3 } , { 1 , 3 , 4 } , { 1 , 2 , 3 , 4 } } , then ( S , ⊆) is a totally ordered set.

前面我们已经看到,(R,≤) 是一个全序集,但 (Z>0,|) 不是全序集。类似地,如果存在子集 A, B ∈ X 使得 A⊈B 且 B⊈A,则偏序集 (X, ⊆) 不是全序集。例如,如果 X=P({1,2,3,4}),我们可以取 A = {1, 2} 和 B = {2, 3},从而看出 (X, ⊆) 不是全序集。另一方面,如果 S={∅,{1},{1,3},{1,3,4},{1,2,3,4}},则 (S, ⊆) 是一个全序集。

Constructing New Posets

构建新的姿态集

We now give more examples of posets by introducing methods of constructing new posets from old ones. The first construction lets us turn a subset of a poset into a partially ordered set.

现在,我们将通过介绍从旧偏序集构造新偏序集的方法,给出更多偏序集的例子。第一种构造方法可以将偏序集的子集转化为偏序集。

6.68. Definition: Subposets.. Given a poset ( X , ≤ ) and a subset S of X , define a relation ≤′ on S as follows: for all a , b S , let a ≤ ′ b iff a b . The relation ≤′ is called the restriction of ≤ to S , and ( S , ≤ ′) is called a subposet of ( X , ≤ ).

6.68. 定义:子偏序集.. 给定一个偏序集 (X, ≤ ) 和 X 的一个子集 S,定义 S 上的关系 ≤′ 如下:对于所有 a, b ∈ S,令 a ≤ ′ b 当且仅当 a ≤ b。关系 ≤′ 称为 ≤ 在 S 上的限制,(S, ≤ ′) 称为 (X, ≤ ) 的一个子偏序集。

The poset axioms for ( S , ≤ ′) follow immediately from the poset axioms for ( X , ≤ ). For example, to check antisymmetry of ≤′, suppose a , b S satisfy a ≤ ′ b and b ≤ ′ a . Then a b and b a , so a = b because ≤ is antisymmetric. Every subposet of a totally ordered set is also totally ordered. On the other hand, it is quite possible for ( S , ≤ ′) to be totally ordered even when ( X , ≤ ) is not totally ordered. We saw an example of this above when ( X , ≤ ) is ( P ( { 1 , 2 , 3 , 4 } ) , ) . For another example, taking S = { 2 k : k Z 0 } gives a totally ordered subposet of the divisibility poset ( Z > 0 , | ) . In general, given a poset ( X , ≤ ) and a subset S of X , S is called a chain in X iff the subposet ( S , ≤ ′) is totally ordered.

偏序集 (S, ≤ ′) 的公理可直接由偏序集 (X, ≤) 的公理推出。例如,为了检验 ≤′ 的反对称性,假设 a, b ∈ S 满足 a ≤ ′ b 且 b ≤ ′ a。那么 a ≤ b 且 b ≤ a,所以 a = b,因为 ≤ 是反对称的。全序集的每个子偏序集也是全序的。另一方面,即使 (X, ≤) 不是全序的,(S, ≤ ′) 也完全有可能是全序的。我们上面已经看到了一个例子,当 (X, ≤) 是 (P({1,2,3,4}),⊆) 时。再举一个例子,取 S={2k:k∈Z≥0} 则给出了整除性偏序集 (Z>0,|) 的一个全序子偏序集。一般而言,给定一个偏序集 (X, ≤ ) 和 X 的一个子集 S,当且仅当子偏序集 (S, ≤ ′) 是全序的时,称 S 为 X 中的链。

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We can concatenate a list of non-overlapping posets to obtain a new poset, as follows.

我们可以将一系列不重叠的偏序集连接起来,得到一个新的偏序集,如下所示。

6.69. Definition: Concatenation Posets.. Let ( X 1 , ≤ 1 ), ( X 2 , ≤ 2 ), …, ( X n , ≤ n ) be a list of posets such that the sets X 1 , X 2 , …, X n are pairwise disjoint (i.e., X i X j = for all i j ). Define a relation ≤ on X = X 1 X 2 X n as follows: for a , b X with a X i and b X j , let a b iff i < j , or i = j and a i b . The pair ( X , ≤ ) is a poset called the concatenation of X 1 , …, X n .

6.69. 定义:连接偏序集.. 令 (X 1 , ≤ 1 ), (X 2 , ≤ 2 ), …, (X n , ≤ n ) 为偏序集列表,使得集合 X 1 , X 2 , …, X n 两两不相交(即,对于所有 i ≠ j,Xi∩Xj=∅)。在 X=X1∪X2∪⋯∪Xn 上定义关系 ≤ 如下:对于 a, b ∈ X,其中 a ∈ X i 且 b ∈ X j ,令 a ≤ b 当且仅当 i < j,或者 i = j 且 a ≤ i b。二元组 (X, ≤ ) 是一个偏序集,称为 X 1 , …, X n 的连接。

Let us check that ≤ is transitive. Fix a , b , c X , assume a b and b c , and prove a c . We know there are unique i , j , k ∈ {1, 2, …, n } with a X i , b X j , and c X k . Consider various cases. Case 1. Assume i < j < k . Then i < k (by transitivity of < on Z ), so a c . Case 2. Assume i < j = k . Here i < k , so a c . Case 3. Assume i = j < k . Here i < k , so a c . Case 4. Assume i = j = k . Here we know a i b and b i c , so a i c by transitivity of ≤ i . Thus, a c follows by definition of ≤. You can check that ≤ is reflexive on X and antisymmetric. Moreover, if every ( X i , ≤ i ) is totally ordered, then ( X , ≤ ) is also totally ordered.

让我们验证 ≤ 的传递性。固定 a, b, c ∈ X,假设 a ≤ b 且 b ≤ c,并证明 a ≤ c。我们知道存在唯一的 i, j, k ∈ {1, 2, …, n},使得 a ∈ X i ,b ∈ X j ,c ∈ X k 。考虑以下几种情况。情况 1:假设 i < j < k。则 i < k(由 < 在 Z 上的传递性可知),所以 a ≤ c。情况 2:假设 i < j = k。这里 i < k,所以 a ≤ c。情况 3:假设 i = j < k。这里 i < k,所以 a ≤ c。情况 4:假设 i = j = k。这里我们知道 a ≤ b 且 b ≤ c,所以根据 ≤ 的传递性,a ≤ c。因此,根据 ≤ 的定义,a ≤ c 成立。你可以验证 ≤ 在 X 上是自反的且反对称的。此外,如果每个 (X, ≤) 都是全序的,那么 (X, ≤) 也是全序的。

Our next construction gives a way to turn the product of posets into a poset.

我们接下来的构造提供了一种将偏序集的乘积转化为偏序集的方法。

6.70. Definition: Product Posets.. Given posets ( X 1 , ≤ 1 ), ( X 2 , ≤ 2 ), …, ( X n , ≤ n ), define a relation on X = X 1 × X 2 × · · · × X n as follows: for all a = ( a 1 , …, a n ) and b = ( b 1 , …, b n ) in X , define a b iff a i b i for all i ∈ {1, 2, …, n }. The pair ( X , ) is called the product poset with factors ( X i , ≤ i ).

6.70。定义:产品偏序集。给定偏序集 (X 1 , ≤ 1 ), (X 2 , ≤ 2 ), …, (X n , ≤ n ),在 X = X 1 × X 2 × · · · × X n 上定义关系 ⪯,如下所示:对于 X 中的所有 a = (a 1 , …, a n ) 和 b = (b 1 , …, b n ),定义 a⪯b 当且仅当对于所有 i ∈ {1, 2, …, n}。二元组 (X,⪯) 称为因子为 (X i , ≤ i ) 的乘积偏序集。

We check that ( X , ) is transitive, letting the reader verify reflexivity and antisymmetry. Fix a , b , c X , and write a = ( a 1 , …, a n ), b = ( b 1 , …, b n ), c = ( c 1 , …, c n ) for some a i , b i , c i X i . Assume a b and b c , which means a i b i and b i c i for all i ∈ {1, 2, …, n }. Since each relation ≤ i is transitive, we deduce a i c i for all i , so a c holds.

我们验证 (X,⪯) 是传递的,让读者验证自反性和反对称性。固定 a, b, c ∈ X,并令 a = (a 1 , …, a n ), b = (b 1 , …, b n ), c = (c 1 , …, c n ),其中 a i , b i , c i ∈ X i 。假设 a⪯b 且 b⪯c,这意味着对于所有 i ∈ {1, 2, …, n},都有 a i ≤ b i 且 b i ≤ c i 。由于每个 ≤ i 关系都是传递的,因此我们推断对于所有 i,都有 a i ≤ c i ,所以 a⪯c 成立。

The product of totally ordered posets is usually not totally ordered. For example, let X = {1, 2, 3} and Y = {1, 2, 3, 4} considered as subposets of ( Z , ) . In the product poset ( X × Y , ) , we have ( 1 , 4 ) ⪯̸ ( 2 , 3 ) and ( 2 , 3 ) ⪯̸ ( 1 , 4 ) , so this poset is not totally ordered. In the exercises, we describe another ordering on product sets called lexicographic ordering, which is a total ordering if all the factors are totally ordered.

全序偏序集的乘积通常不是全序的。例如,设 X = {1, 2, 3} 和 Y = {1, 2, 3, 4} 是 (Z, ≤) 的子集。在乘积偏序集 (X×Y, ≤) 中,有 (1, 4) ≤ (2, 3) 且 (2, 3) ≤ (1, 4),因此该偏序集不是全序的。在练习中,我们将介绍乘积集上的另一种排序方式,称为字典序,如果所有因子都是全序的,则字典序是全序的。

Greatest and Least Elements

最大元素和最小元素

Given a subset S of a partially ordered set, it is often useful to single out the largest and smallest elements in S (when such elements exist).

给定偏序集的子集 S,通常需要找出 S 中的最大元素和最小元素(如果存在这样的元素)。

6.71. Definition: Greatest and Least Elements.. Let ( X , ≤ ) be a poset and S X .

6.71. 定义:最大元素和最小元素.. 设 (X, ≤ ) 为一个偏序集,且 S⊆ X。

For all z , z is the greatest element of S iff z S and y S , y z .

对于所有 z,zis 是 S 的最大特征当且仅当 z∈S 且 ∀y∈S,y≤z。

For all x , x is the least element of S iff x S and y S , x y .

对于所有 x,x 是 S 的最小元素当且仅当 x∈S 且对于所有 y∈S,x≤y。

We sometimes write z = max S to mean that z is the greatest (maximum) element of S , and x = min S to mean that x is the least (minimum) element of S .

我们有时写 z=maxS 表示 z 是 S 中的最大元素,写 x=minS 表示 x 是 S 中最小的元素。

The next examples show that not all subsets have greatest and least elements. But when these elements exist, they are unique. For suppose z 1 and z 2 are both greatest elements of S . Since z 1 is a greatest element and z 2 S , z 2 z 1 . Since z 2 is also a greatest element and z 1 S , z 1 z 2 . Then z 1 = z 2 follows from antisymmetry. Similarly, least elements are unique when they exist.

接下来的例子表明,并非所有子集都存在最大元素和最小元素。但当这些元素存在时,它们是唯一的。假设 z 1 和 z 2 都是子集 S 的最大元素。由于 z 1 也是最大元素且 z 2 ∈ S,所以 z 2 ≤ z 1 。由于 z 2 也是最大元素且 z 1 ∈ S,所以 z 1 ≤ z 2 。那么,根据反对称性,z 1 = z 2 。类似地,当最小元素存在时,它们也是唯一的。

6.72 Example. In the poset ( Z , ) , the subset Z > 0 has least element 1 but no greatest element. The subset Z 5 has greatest element 5 and no least element. The subset has no least element and no greatest element. Similarly, Z itself has no least element and no greatest element. Let us prove carefully that the set Z odd of all odd integers has no greatest element. Assume, to get a contradiction, that z Z odd is the greatest element of this subset. Then z + 2 is also odd, and z + 2 ≤ z is not true. This contradiction proves that no greatest element exists.

6.72 例。在偏序集 (Z,≤) 中,子集 Z>0 的最小元素为 1,但没有最大元素。子集 Z≤5 的最大元素为 5,但没有最小元素。子集 ∅ 没有最小元素也没有最大元素。类似地,Z 本身也没有最小元素也没有最大元素。让我们仔细证明所有奇数构成的集合 Zodd 没有最大元素。为了得出矛盾,假设 z∈Zodd 是该子集的最大元素。那么 z + 2 也是奇数,因此 z + 2 ≤ z 不成立。这个矛盾证明了不存在最大元素。

6.73 Example. In the poset ( R , ) , the half-open interval ( 0 , 1 ] = { x R : 0 < x 1 } has greatest element 1 but no least element. Note that 0 is not the least element of (0, 1] because it does not belong to this subset. For any x 0 ∈ (0, 1], x 0 cannot be the smallest element of (0, 1] since x 0 /2 is in this subset and x 0 /2 < x 0 . Similarly, for all real a < b , we see that [ a , b ) has least element a but no greatest element, and ( a , b ) has no least element and no greatest element.

6.73 例。在偏序集 (R,≤) 中,半开区间 (0,1]={x∈R:0<x≤1} 的最大元素为 1,但没有最小元素。注意,0 不是 (0,1] 的最小元素,因为它不属于这个子集。对于任意 x 0 ∈ (0,1],x 0 不可能是 (0,1] 的最小元素,因为 x 0 /2 属于这个子集,且 x 0 /2 < x 0 。类似地,对于所有实数 a < b,我们看到 [a, b) 的最小元素为 a,但没有最大元素,而 (a, b) 既没有最小元素也没有最大元素。

6.74 Example. For any set X , the poset ( P ( X ) , ) consists of all subsets of X ordered by set inclusion. This poset has least element and greatest element X . However, you can show that the subset S of P ( X ) consisting of all nonempty proper subsets of X has no least element and no greatest element. For example, consider X = {1, 2, 3}, so S = {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}. No one of these six sets is a subset of all other members of S , so S has no least element. Similarly, no one of these six sets contains all the other sets as subsets, so S has no greatest element.

6.74 例。对于任意集合 X,偏序集 (P(X),⊆) 由 X 的所有子集按集合包含关系排序构成。该偏序集的最小元素为 ∅,最大元素为 X。然而,你可以证明 P(X) 的子集 S(由 X 的所有非空真子集构成)既没有最小元素也没有最大元素。例如,考虑 X = {1, 2, 3},则 S = {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}。这六个集合中没有一个是 S 中所有其他集合的子集,因此 S 没有最小元素。类似地,这六个集合中也没有一个是 S 中所有其他集合的子集,因此 S 没有最大元素。

6.75 Example. Let X be the set of positive divisors of 36, ordered by divisibility. X has least element 1 and greatest element 36. Removing these two elements, we get the subset S = {2, 3, 4, 6, 9, 12, 18} ordered by divisibility. You may check that S has no least element and no greatest element with respect to this partial order . Of course, with respect to the standard ordering ≤, S has least element 2 and greatest element 18. On the other hand, now take X to be the set of positive divisors of 32, ordered by divisibility. Here X = {1, 2, 4, 8, 16, 32} is totally ordered by |. You may verify that every nonempty subset of X has a least element and a greatest element with respect to |.

6.75 例。设 X 为 36 的所有正因子的集合,并按整除性排序。X 的最小元素为 1,最大元素为 36。移除这两个元素后,我们得到子集 S = {2, 3, 4, 6, 9, 12, 18},并按整除性排序。你可以验证,对于这个偏序关系,S 没有最小元素也没有最大元素。当然,对于标准排序 ≤,S 的最小元素为 2,最大元素为 18。另一方面,现在设 X 为 32 的所有正因子的集合,并按整除性排序。这里 X = {1, 2, 4, 8, 16, 32} 是按 | 整除的全序关系。你可以验证,对于 | 整除,X 的每个非空子集都有最小元素和最大元素。

6.76 Example. Suppose ( X 1 , ≤ 1 ), …, ( X n , ≤ n ) are nonempty posets. The concatenation poset ( X , ≤ ) has a least element iff ( X 1 , ≤ 1 ) has a least element, and these elements are the same. Similarly, the greatest element of ( X , ≤ ) is the greatest element of ( X n , ≤ n ) if the latter exists. On the other hand, a = ( a 1 , …, a n ) is the least element of the product poset ( X 1 × × X n , ) iff a i is the least element of ( X i , ≤ i ) for all i ; likewise for greatest elements.

6.76 示例。假设 (X 1 , ≤ 1 ), …, (X n , ≤ n ) 是非空偏序集。连接偏序集 (X, ≤ ) 存在最小元素当且仅当 (X 1 , ≤ 1 ) 存在最小元素,且这两个元素相同。类似地,如果 (X n , ≤ n ) 存在最大元素,则 (X, ≤ ) 的最大元素也是 (X n , ≤ n ) 的最大元素。另一方面,a = (a 1 , …, a n ) 是乘积偏序集 (X1×⋯×Xn,⪯) 的最小元素,当且仅当对于所有 i,a i 是 (X i , ≤ i ) 的最小元素;最大元素也是如此。

Well-Ordered Posets

有序姿势集

In the examples above, we have seen many posets ( X , ≤ ) with no least element. We have also seen that X itself may have a least element, but there could exist nonempty subsets S of X where S has no least element. For example, this happened when X = R and S is an open interval ( a , b ). We now give a special name to partially ordered sets where every subset (other than ) has a least element.

在上面的例子中,我们看到了许多没有最小元素的偏序集 (X, ≤)。我们也看到 X 本身可能存在最小元素,但也可能存在 X 的非空子集 S,其中 S 没有最小元素。例如,当 X=R 且 S 是开区间 (a, b) 时,就会出现这种情况。现在,我们给每个子集(∅ 除外)都存在最小元素的偏序集赋予一个特殊的名称。

6.77 Definition: Well-Ordered Posets. A poset ( X , ≤ ) is well-ordered iff every nonempty subset of X has a least element .

6.77 定义:良序偏序集。偏序集 (X, ≤ ) 是良序的,当且仅当 X 的每个非空子集都有最小元素。

A well-ordered poset must be totally ordered. For if ( X , ≤ ) is well-ordered and a , b X , then S = { a , b } is a nonempty subset of X . So this subset has a least element. If a is the least element of S , then a b . If b is the least element of S , then b a . Thus, a b or b a holds.

良序偏序集必定是全序的。因为如果 (X, ≤) 是良序的,且 a, b ∈ X,则 S = {a, b} 是 X 的一个非空子集。所以这个子集存在最小元素。如果 a 是 S 的最小元素,则 a ≤ b。如果 b 是 S 的最小元素,则 b ≤ a。因此,a ≤ b 或 b ≤ a 成立。

It is intuitively plausible that every totally ordered finite set is well-ordered. We delay the formal proof until § 7.1 , where finite sets are officially defined. On the other hand, ( Z , ) and ( R , ) are not well-ordered sets, since they have no least element. Although ( R 0 , ) has a least element (zero), this poset is not well-ordered either, since its subset R > 0 has no least element. Intuitively, ( Z 0 , ) and ( Z > 0 , ) are well-ordered sets, although the proof of this fact is not obvious. In some formal developments of the integers, the well-ordering of Z > 0 is postulated as an axiom.

直观上,每个全序有限集都是良序集是合理的。我们将正式证明推迟到第 7.1 节,该节正式定义了有限集。另一方面,(Z,≤) 和 (R,≤) 不是良序集,因为它们没有最小元素。虽然 (R≥0,≤) 有最小元素(零),但这个偏序集也不是良序集,因为它的子集 R>0 没有最小元素。直观上,(Z≥0,≤) 和 (Z>0,≤) 是良序集,尽管这个事实的证明并不明显。在一些整数的正式发展中,Z>0 的良序性被假定为一个公理。

6.78 Well-Ordering Property of Z > 0 . The poset ( Z > 0 , ) is well-ordered.

6.78 Z>0 的良序性质。偏序集 (Z>0,≤) 是良序的。

We now show that the Well-Ordering Property is logically equivalent to the Induction Axiom from § 4.1 . [This proof may be considered optional.] First, we assume the Induction Axiom holds and prove the Well-Ordering Property. Fix a subset S of Z > 0 that has no least element; we must prove that S = . To do so, we use strong induction to prove the statement n Z > 0 , n S . Fix n 0 Z > 0 . By strong induction, we may assume that for all integers m in the range 0 < m < n 0 , m S . Using this, we must prove that n 0 S . Assume, to get a contradiction, that n 0 S . We derive a contradiction to our original assumption on S by showing that n 0 is the least element of S . Fix k S , so k is a positive integer. We cannot have k < n 0 , since otherwise the induction hypothesis would give k S . Because ( Z > 0 , ) is totally ordered, we conclude that n 0 k . We have now shown that n 0 is the least element of S . This contradiction forces n 0 S , completing the proof by induction.

现在我们证明良序性质在逻辑上等价于§4.1中的归纳公理。[此证明可视为可选。] 首先,我们假设归纳公理成立,并证明良序性质。固定Z>0的一个子集S,使其没有最小元素;我们必须证明S=∅。为此,我们使用强归纳法证明命题∀n∈Z>0,n∉S。固定n0∈Z>0。根据强归纳法,我们可以假设对于所有在0 < m < n范围内的整数m,m∉S。利用这一点,我们必须证明n0∉S。为了得出矛盾,假设 n 0 ∈ S。我们通过证明 n 0 是 S 的最小元素,从而推导出与我们对 S 的最初假设相矛盾。固定 k ∈ S,使得 k 为正整数。我们不能有 k < n 0 ,否则归纳假设将得出 k∉S。由于 (Z>0,≤) 是全序集,我们得出 n 0 ≤ k。现在我们已经证明了 n 0 是 S 的最小元素。这个矛盾迫使 n0∉S,从而完成了归纳证明。

Next, we assume the Well-Ordering Property holds and prove the Induction Axiom. Fix an open sentence P ( n ). As discussed in § 4.1 , it is enough to prove that the statement Q : “ P ( 1 ) n Z > 0 , ( P ( n ) P ( n + 1 ) ) ” implies the statement R : “ n Z > 0 , P ( n ) .” We use proof by contradiction, assuming that Q is true and R is false. To use the Well-Ordering Property, we need a nonempty subset of the positive integers. We define this subset to be S = { n Z > 0 : ¬ P ( n ) } , the set of positive integers for which P ( n ) is false. A useful denial of R is n Z > 0 , ¬ P ( n ) , which means that S is nonempty. The Well-Ordering Property states that there is a least element of S , say n 0 . Either n 0 = 1 or n 0 > 1. In the case n 0 = 1, we see that P (1) is false because n 0 belongs to S , but also P (1) is true by statement Q . This is a contradiction. In the case n 0 > 1, then n 0 − 1 is a positive integer less than n 0 . On one hand, n 0 is the least element of S , so n 0 1 S and n 0 S . This means that P ( n 0 − 1) is true and P ( n 0 ) is false. On the other hand, taking n = n 0 − 1 in statement Q , we see that P ( n 0 − 1) ↠ P ( n 0 ) is true. This contradicts the truth table for IF, and the proof is complete. [The preceding proof tacitly uses certain additional facts about integers. For example, we needed to know that for any integer n 0 > 1, n 0 − 1 is also a positive integer. For a careful development of the integers where all necessary properties are derived from axioms, see Chapter 8 . In particular, compare the discussion here to Lemma 8.41 and the proof of Theorem 8.51.]

接下来,我们假设良序性成立,并证明归纳公理。固定一个开语句 P(n)。如 §4.1 所述,只需证明命题 Q:“P(1)∧∀n∈Z>0,(P(n)⇒P(n+1))”蕴含命题 R:“∀n∈Z>0,P(n)”。我们使用反证法,假设 Q 为真且 R 为假。为了使用良序性,我们需要一个非空的正整数子集。我们将这个子集定义为 S={n∈Z>0:¬P(n)},即 P(n) 为假的正整数集合。R 的一个有用否定是 ∃n∈Z>0,¬P(n),这意味着 S 非空。良序性指出 S 中存在最小元素,例如 n 0 。要么 n 0 = 1,要么 n 0 > 1。当 n 0 = 1 时,P(1) 为假,因为 n 0 属于 S,但根据命题 Q,P(1) 也为真。这自相矛盾。当 n 0 > 1 时,n 0 − 1 是一个小于 n 0 的正整数。一方面,n 0 是 S 中的最小元素,所以 n0−1∉S 且 n 0 ∈ S。这意味着 P(n 0 − 1) 为真,而 P(n 0 ) 为假。另一方面,在命题 Q 中取 n = n 0 − 1,我们发现 P(n 0 − 1) ↠ P(n 0 ) 为真。这与 IF 的真值表矛盾,证明完毕。[前面的证明隐含地使用了关于整数的一些附加事实。例如,我们需要知道对于任何大于 1 的整数 n 0 ,n 0 − 1 也是一个正整数。关于整数的详细推导(其中所有必要的性质均由公理导出),请参见第 8 章。特别是,请将此处的讨论与引理 8.41 和定理 8.51 的证明进行比较。]

Section Summary

章节概要

  1. Partial Orders and Total Orders. A relation ≤ on a set X is a partial order iff ≤ is reflexive on X , transitive, and antisymmetric. This means that for all x , y , z X , x x ; if x y and y z , then x z ; and if x y and y x , then x = y . A total order is a partial order such that for all x , y X , x y or y x .
    偏序和全序。集合 X 上的关系 ≤ 是偏序关系,当且仅当 ≤ 在 X 上自反、传递且反对称。这意味着对于所有 x, y, z ∈ X,x ≤ x;如果 x ≤ y 且 y ≤ z,则 x ≤ z;如果 x ≤ y 且 y ≤ x,则 x = y。全序关系是指对于所有 x, y ∈ X,x ≤ y 或 y ≤ x 的偏序关系。
  2. Examples of Posets. Any set of sets is partially ordered by set inclusion. Any set of positive integers is partially ordered by divisibility. Any subset of a poset is a poset with the restricted ordering. The concatenation of a list of pairwise disjoint posets is a poset. The product of posets is a poset if we let ( a 1 , …, a n ) ≤ ( b 1 , …, b n ) mean a i i b i for all i .
    偏序集示例。任何集合的集合都可以通过集合包含关系进行偏序。任何正整数集合都可以通过整除关系进行偏序。任何偏序集的子集都是具有受限排序的偏序集。两两不相交的偏序集列表的连接仍然是一个偏序集。如果令 (a 1 , …, a n ) ≤ (b 1 , …, b n ) 表示对于所有 i,a i i b i ,则偏序集的乘积仍然是一个偏序集。
  3. Greatest and Least Elements. If S is any subset of a poset ( X , ≤ ), z is the greatest element of S iff z S and y S , y z ; x is the least element of S iff x S and y S , x y . Greatest and least elements do not always exist, but they are unique when they do exist.
    最大元素和最小元素。如果 S 是偏序集 (X, ≤) 的任意子集,则 z 是 S 的最大元素当且仅当 z ∈ S 且对于所有 y∈S,y≤z;x 是 S 的最小元素当且仅当 x ∈ S 且对于所有 y∈S,x≤y。最大元素和最小元素并非总是存在,但当它们存在时,它们是唯一的。
  4. Well-Ordered Sets. A poset ( X , ≤ ) is well-ordered iff every nonempty subset of X has a least element. The Well-Ordering Property states that Z > 0 with the standard ordering is well-ordered. This property is equivalent to the Induction Axiom.
    良序集。偏序集 (X, ≤) 是良序的,当且仅当 X 的每个非空子集都存在最小元素。良序性质指出,具有标准序的 Z>0 是良序的。该性质等价于归纳公理。

Exercises

练习

  1. Which relations shown in Figure 6.3 are antisymmetric?
    图 6.3 中所示的哪些关系是反对称的?
  2. Give a specific example of a relation R that is not symmetric and not antisymmetric.
    请举出一个既不对称也不反对称的关系 R 的具体例子。
  3. Find all total order relations on the set X = {1, 2, 3}.
    找出集合 X = {1, 2, 3} 上的所有全序关系。
  4. Draw Haase diagrams for each poset. (a) the set of positive divisors of 6 ordered by divisibility. (b) the set of positive divisors of 8 ordered by divisibility. (c) the set of positive divisors of 100 ordered by divisibility. (d) the set of all subsets of { a , b , c , d } ordered by set inclusion.
    为每个偏序集绘制哈斯图。 (a) 6 的正因子集合,按整除性排序。 (b) 8 的正因子集合,按整除性排序。 (c) 100 的正因子集合,按整除性排序。 (d) {a, b, c, d} 的所有子集集合,按集合包含性排序。
  5. Let X = {1, 2, 3, 4, 5}. A Haase diagram for the poset ( X , ≤ ) has arrows from 1 to 2 to 4 to 5 and arrows from 1 to 3 to 5. (a) Draw this diagram. (b) Describe the relation ≤ as a set of ordered pairs. (c) Confirm that ≤ really is a partial order on X .
    设 X = {1, 2, 3, 4, 5}。偏序集 (X, ≤) 的哈斯图包含从 1 到 2 到 4 到 5 的箭头,以及从 1 到 3 到 5 的箭头。(a) 画出这个哈斯图。(b) 将关系 ≤ 描述为一个有序对集合。(c) 验证 ≤ 确实是 X 上的一个偏序关系。
  6. (a) Find two different chains of maximum size in the poset ( P ( { 1 , 2 , 3 , 4 } ) , ) . (b) Find all chains of maximum size in the poset of positive divisors of 75 ordered by divisibility.
    (a)在偏序集(P({1,2,3,4}),⊆). 中找出两条不同的最大链。(b)在按整除性排序的 75 的正因子偏序集中找出所有最大链。
  7. Suppose ( X , ≤ ) is a poset. Define a new relation ≥ on X as follows: for all a , b X , define a b iff b a . Prove that ≥ is a partial order on X , and ≥ is a total ordering iff ≤ is a total ordering.
    假设 (X, ≤) 是一个偏序集。在 X 上定义一个新的关系 ≥ 如下:对于所有 a, b ∈ X,定义 a ≥ b 当且仅当 b ≤ a。证明 ≥ 是 X 上的一个偏序关系,并且 ≥ 是一个全序关系当且仅当 ≤ 是一个全序关系。
  8. (a) Suppose X is a fixed set, and for each i in an index set I , R i is a partial order on X . Show that R = i I R i is also a partial order on X . (b) If every R i is a total order, must R be a total ordering on X ? Prove your answer. (c) Is S = i I R i a partial order on X ? Prove your answer.
    (a) 假设 X 是一个固定的集合,对于索引集 I 中的每个 i,R i 是 X 上的一个偏序关系。证明 R=⋂i∈IRi 也是 X 上的一个偏序关系。(b) 如果每个 R i 都是一个全序关系,那么 R 是否一定是 X 上的一个全序关系?证明你的答案。(c) S=⋃i∈IRi 是否是 X 上的一个偏序关系?证明你的答案。
  9. (a) Let ( X , ≤ ) be the concatenation of n posets ( X i , ≤ i ). Prove ≤ satisfies reflexivity and antisymmetry. (b) Prove ≤ is a total order on X iff every ≤ i is a total order on X i .
    (a) 令 (X, ≤) 为 n 个偏序集 (X i , ≤ i ) 的串联。证明 ≤ 满足自反性和反对称性。(b) 证明 ≤ 是 X 上的全序当且仅当每个 ≤ i 都是 X i 上的全序。
  10. Let ( X , ) be the product of n posets ( X i , ≤ i ). Prove satisfies reflexivity and antisymmetry.
    设 (X,⪯) 为 n 个偏序集 (X i , ≤ i ) 的乘积。证明 ⪯ 满足自反性和反对称性。
  11. Suppose ( I , ) is a partially ordered index set, and for each i I , ( X i , ≤ i ) is a poset. Assume the sets X i are pairwise disjoint. Define a relation ≤ on X = i I X i as in Definition 6.69 . Prove that ( X , ≤ ) is a poset.
    假设 (I, ≤) 是一个偏序索引集,对于每个 i ∈ I,(X i , ≤ i ) 是一个偏序集。假设集合 X i 两两不相交。定义 X=⋃i∈IXi 上的关系 ≤,如定义 6.69 所示。证明 (X, ≤) 是一个偏序集。
  12. Suppose ( X i , ≤ i ) is a poset for 1 ≤ i n . Let X = X 1 × X 2 × · · · × X n . Define the lexicographic order lex on X as follows: for a = ( a 1 , …, a n ) and b = ( b 1 , …, b n ) in X , let a lex b iff a = b or there exists i ∈ {1, 2, …, n } with a j = b j for all j < i , a i b i , and a i i b i . (a) Prove the poset axioms for ≤ lex . (b) Prove: if every ≤ i is a total ordering, then ≤ lex is a total ordering.
    假设 (X i , ≤ i ) 是 1 ≤ i ≤ n 的一个偏序集。令 X = X 1 × X 2 × · · · × X n 。在 X 上定义字典序 ≤ lex 如下:对于 X 中的 a = (a 1 , …, a n ) 和 b = (b 1 , …, b n ),令 a ≤ lex b 当且仅当 a = b 或存在 i ∈ {1, 2, …, n} 使得对于所有 j < i,a j = b j ,a i ≠ b i ,并且 a i i b i 。(a)证明 ≤ lex 的偏序集公理。 (b)证明:如果每个 ≤ i 都是一个全序,那么 ≤ lex 也是一个全序。
  13. (a) Show that the concatenation of n well-ordered sets is well-ordered. (b) Show that the product of n well-ordered sets, ordered lexicographically (see the previous exercise), is well-ordered.
    (a) 证明 n 个良序集的连接仍然是良序集。(b) 证明 n 个良序集的乘积(按字典序排列,参见前一个练习)仍然是良序集。
  14. Let Y = P ( { 1 , 2 } ) , Z = {3, 4, 5, 6}, and R = I Z { ( 3 , 4 ) , ( 3 , 5 ) , ( 3 , 6 ) } . (a) Draw Haase diagrams for ( Y , ⊆) and ( Z , R ). (b) Draw a Haase diagram for the concatenation of Y and Z (in this order). (c) Draw a Haase diagram for the concatenation of Z and Y (in this order).
    设 Y=P({1,2}),Z={3, 4, 5, 6},R=IZ∪{(3,4),(3,5),(3,6)}。(a)绘制 (Y, ⊆) 和 (Z, R) 的哈斯图。(b)绘制 Y 和 Z(按此顺序)连接的哈斯图。(c)绘制 Z 和 Y(按此顺序)连接的哈斯图。
  15. Let X = {1, 2, 3} and Y = {0, 2, 4, 6}. Draw a Haase diagram for the product poset ( X , ≤ ) × ( Y , ≤ ), where ≤ is the standard ordering on Z .
    设 X = {1, 2, 3} 且 Y = {0, 2, 4, 6}。绘制乘积偏序集 (X, ≤ ) × (Y, ≤ ) 的哈斯图,其中 ≤ 是 Z 上的标准序。
  16. Determine (with explanation) whether each poset below is well-ordered. (a) Z > 0 ordered by divisibility. (b) Z a ordered by ≤, for fixed a Z . (c) the set of positive divisors of 80 ordered by divisibility. (d) the set of positive divisors of 81 ordered by divisibility. (e) [1, 5] ordered by ≤. (f) {1, 2, 3, 4, 5} ordered by ≤.
    判断(并解释)以下每个偏序集是否良序。 (a) Z>0 按整除性排序。 (b) Z≤a 按 ≤ 排序,其中 a∈Z 为固定值。 (c) 80 的正因子集合按整除性排序。 (d) 81 的正因子集合按整除性排序。 (e) [1, 5] 按 ≤ 排序。 (f) {1, 2, 3, 4, 5} 按 ≤ 排序。
  17. Use the Well-Ordering Property for Z > 0 to prove that for each a Z , ( Z a , ) is a well-ordered set.
    利用 Z>0 的良序性质证明,对于每个 a∈Z,(Z≥a,≤) 是一个良序集。
  18. Prove that every nonempty subset of ( Z < 0 , ) has a greatest element.
    证明 (Z<0,≤) 的每个非空子集都有最大元素。
  19. Define a relation R on Z by setting, for all a , b Z ,
    a R b ( ( 0 a b ) ( b a < 0 ) ( a 0 b < 0 ) ) .
    Prove that ( Z , R ) is a poset, and this poset is well-ordered.
    在 Z 上定义关系 R,使得对于所有 a,b∈Z, aRb⇔((0≤a≤b)∨(b≤a<0)∨(a≥0∧b<0))。 证明 (Z,R) 是一个偏序集,并且该偏序集是良序的。
  20. Carefully prove the assertions in the last sentence of Example 6.73 .
    仔细证明例 6.73 最后一句中的断言。
  21. Prove the assertions made in Example 6.76 .
    证明例 6.76 中提出的断言。
  22. Show that a relation R on a set X is both symmetric and antisymmetric iff R I X .
    证明集合 X 上的关系 R 既是对称的又是反对称的,当且仅当 R⊆ I X
  23. A relation < on a set X is called a strict partial order on X iff < is transitive and for all a X , a < a is false. (The last condition is called irreflexivity .) (a) Given a poset ( X , ≤ ), define a < b to mean a b and a b for a , b X . Show that < is a strict partial order on X . (b) Conversely, given a strict partial order < on a set X , define a b to mean a < b or a = b for a , b X . Show that ≤ is a partial order relation on X .
    集合 X 上的关系 < 被称为 X 上的严格偏序关系,当且仅当 < 是传递的,且对于所有 a ∈ X,a < a 为假。(最后一个条件称为反自反性。) (a) 给定一个偏序集 (X, ≤ ),定义 a < b 表示对于 a, b ∈ X,a ≤ b 且 a ≠ b。证明 < 是 X 上的严格偏序关系。 (b) 反之,给定集合 X 上的严格偏序关系 <,定义 a ≤ b 表示对于 a, b ∈ X,a < b 或 a = b。证明 ≤ 是 X 上的偏序关系。
  24. Count the number of partial order relations on X = {1, 2, 3}.
    计算集合 X = {1, 2, 3} 上的偏序关系的数量。
  25. For a fixed set X , let S P be the set of all set partitions of X . We introduce a partial ordering on S P as follows: given P , Q S P , let P Q mean that every block of P is contained in a block of Q ; i.e., S P , T Q , S T . (a) Prove that is a partial order on S P . (b) Draw the Haase diagram for ( S P , ) when X = {1, 2, 3}. (c) Draw the Haase diagram for ( S P , ) when X = { a , b , c , d }.
    对于固定的集合 X,令 SP 为 X 的所有集合划分的集合。我们在 SP 上引入如下偏序关系:给定 P, Q∈SP,令 P⪯Q 表示 P 的每个块都包含在 Q 的一个块中;即,∀S∈P, ∃T∈Q, S⊆T。(a)证明 ⪯ 是 SP 上的一个偏序关系。(b)当 X = {1, 2, 3} 时,绘制 (SP, ⪯) 的哈斯图。(c)当 X = {a, b, c, d} 时,绘制 (SP, ⪯) 的哈斯图。

6.6 Equivalence Relations and Algebraic Structures (Optional)

6.6 等价关系和代数结构(可选)

In this chapter, we have defined equivalence relations, equivalence classes, and quotient sets. These ideas are very abstract, and you may be wondering how these concepts could be at all useful. It turns out that quotient sets can be used to define new algebraic structures generalizing familiar number systems such as the integers. In particular, starting with Z and the arithmetic operations of integer addition and multiplication, we use quotient sets to build new number systems: the integers modulo n (denoted Z / n ) and the rational numbers (denoted Q ). In this section, we investigate how the known arithmetic operations on Z let us introduce new arithmetic operations on Z / n and Q . The notion of a well-defined (or single-valued ) operation plays a key role here. We will see that facts in Z , such as the commutative law or the distributive law, can be used to deduce corresponding facts about these new number systems.

在本章中,我们定义了等价关系、等价类和商集。这些概念非常抽象,你可能想知道它们究竟有什么用处。事实上,商集可以用来定义新的代数结构,从而推广诸如整数之类的熟悉数系。具体来说,从整数集 Z 以及整数加法和乘法运算出发,我们利用商集构建新的数系:模 n 的整数(记为 Z/n)和有理数(记为 Q)。在本节中,我们将探讨如何利用 Z 上已知的算术运算,在 Z/n 和 Q 上引入新的算术运算。定义明确(或单值)的运算在这里起着关键作用。我们将看到,Z 中的一些性质,例如交换律或分配律,可以用来推导出关于这些新数系的相应性质。

The Integers Modulo n

整数模 n

We have already introduced the integers modulo n in some earlier examples. For convenience, we review the relevant definitions here. Fix an integer n ≥ 1. For all a , b Z , recall a n b means n divides a b . The relation ≡ n is an equivalence relation on Z , and the equivalence class of a Z is [ a ] n = { a + k n : k Z } by Example 6.47 . We abbreviate the equivalence class notation to [ a ] n or [ a ], when n is understood.

我们在前面的例子中已经引入了模 n 的整数。为了方便起见,我们在这里回顾一下相关的定义。固定一个整数 n ≥ 1。对于所有 a, b∈Z,回顾一下,a ≡ n b 表示 n 整除 a − b。关系 ≡ n 是 Z 上的一个等价关系,根据例 6.47,a∈Z 的等价类为 [a]≡n={a+kn:k∈Z}。当 n 不显式时,我们将等价类简写为 [a] n 或 [a]。

6.79. Definition: Integers Modulo n .. For each fixed integer n ≥ 1, Z / n is the quotient set Z / n . Thus, Z / n = { [ a ] n : a Z } is the set of all equivalence classes of congruence modulo n .

6.79. 定义:模 n 的整数。对于每个固定的整数 n ≥ 1,Z/n 是商集 Z/≡n。因此,Z/n={[a]n:a∈Z} 是模 n 同余的所有等价类的集合。

The next theorem shows that Z / n consists of n distinct equivalence classes.

下一个定理表明 Z/n 由 n 个不同的等价类组成。

6.80 Theorem on the Elements of Z / n . For all n ≥ 1, Z / n = { [ 0 ] n , [ 1 ] n , , [ n 1 ] n } . Furthermore, the n equivalence classes [0] n , [1] n , …, [ n − 1] n are all distinct.

6.80 关于 Z/n 的元素的定理。对于所有 n ≥ 1,Z/n={[0]n,[1]n,…,[n−1]n}。此外,n 个等价类 [0] n , [1] n , …, [n − 1] n 都是不同的。

Proof . Fix n ≥ 1. Since Z / n = { [ a ] n : a Z } , the set {[0] n , [1] n , …, [ n − 1] n } is a subset of Z / n . To establish the reverse set inclusion, fix [ a ] n Z / n , where a Z . We must prove there exists r ∈ {0, 1, …, n − 1} with [ a ] n = [ r ] n . By the Division Theorem, we know a = qn + r for some q , r Z with 0 ≤ r < n . Then n divides a r = qn , so a n r , so [ a ] n = [ r ] n by part (a) of the Theorem on Equivalence Classes (§ 6.3 ). To see that [0] n , …, [ n − 1] n are all distinct, we assume s , t ∈ {0, 1, …, n − 1} satisfy [ s ] n = [ t ] n , and we prove s = t . Using the Theorem on Equivalence Classes again, we see that s n t , so n divides s t . Since 0 ≤ t < n , we see that − n < − t ≤ 0. Adding this inequality to the inequality 0 ≤ s < n , we conclude − n < s t < n . The only integer multiple of n strictly between − n and n is zero, so s t = 0, giving s = t . □

证明。设 n ≥ 1。由于 Z/n = {[a]n: a∈Z},集合 {[0] n , [1] n , …, [n − 1] n 是 Z/n 的子集。为了证明反向集合包含关系,设 [a]n∈Z/n,其中 a∈Z。我们需要证明存在 r ∈ {0, 1, …, n − 1} 使得 [a] n = [r] n 。根据除法定理,我们知道对于某个 q, r∈Z,有 a = qn + r,其中 0 ≤ r < n。那么,n 整除 a − r = qn,所以 a ≡ n r,因此根据等价类定理(§6.3)的 (a) 部分,[a] n = [r] n 。为了证明 [0] n , …, [n − 1] n 彼此不同,我们假设 s, t ∈ {0, 1, …, n − 1} 满足 [s] n = [t] n ,并证明 s = t。再次利用等价类定理,我们看到 s ≡ n t,因此 n 整除 s − t。由于 0 ≤ t < n,所以 −n < −t ≤ 0。将此不等式加到不等式 0 ≤ s < n 上,我们得出 −n < s − t < n。n 的唯一整数倍严格介于 −n 和 n 之间,且该整数倍为零,因此 s − t = 0,从而 s = t。□

Addition and Multiplication Modulo n

模 n 的加法和乘法

Now we introduce versions of addition and multiplication defined on integers modulo n .

现在我们引入定义在模 n 的整数上的加法和乘法的版本。

6.81. Definition: Addition and Multiplication on Z / n .. Fix n ≥ 1. For all a , b Z , we define addition mod n by [ a ] n n [ b ] n = [ a + b ] n and multiplication mod n by [ a ] n n [ b ] n = [ a · b ] n .

6.81. 定义:Z/n 上的加法和乘法。固定 n ≥ 1。对于所有 a,b∈Z,我们定义模 n 的加法为 [a]n⊕n[b]n=[a+b]n,定义模 n 的乘法为 [a] n n [b] n = [a · b] n

6.82 Example. Take n = 5, so Z / 5 = { [ 0 ] , [ 1 ] , [ 2 ] , [ 3 ] , [ 4 ] } . We can make an addition table and a multiplication table for arithmetic modulo 5. In these tables, the entry in the row labeled [ a ] n and the column labeled [ b ] n is [ a ] n n [ b ] n (left table) or [ a ] n n [ b ] n (right table).

5 [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 0 ] [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 1 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 2 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 3 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 4 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 8 ] 5 [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 1 ] [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 2 ] [ 0 ] [ 2 ] [ 4 ] [ 6 ] [ 8 ] [ 3 ] [ 0 ] [ 3 ] [ 6 ] [ 9 ] [ 12 ] [ 4 ] [ 0 ] [ 4 ] [ 8 ] [ 12 ] [ 16 ]
We can get more informative tables by changing the names of the outputs so that every entry in the table is one of the standard names [0], [1], [2], [3], or [4]. For example, [ 3 ] 5 [ 4 ] = [ 7 ] = [ 2 ] and [3] ⊗ 5 [4] = [12] = [2]. Making these changes produces the following tables:
5 [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 0 ] [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 1 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 0 ] [ 2 ] [ 2 ] [ 3 ] [ 4 ] [ 0 ] [ 1 ] [ 3 ] [ 3 ] [ 4 ] [ 0 ] [ 1 ] [ 2 ] [ 4 ] [ 4 ] [ 0 ] [ 1 ] [ 2 ] [ 3 ] 5 [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 1 ] [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 2 ] [ 0 ] [ 2 ] [ 4 ] [ 1 ] [ 3 ] [ 3 ] [ 0 ] [ 3 ] [ 1 ] [ 4 ] [ 2 ] [ 4 ] [ 0 ] [ 4 ] [ 3 ] [ 2 ] [ 1 ]
6.82 例。取 n = 5,则 Z/5={[0],[1],[2],[3],[4]}。我们可以为模 5 的算术运算制作一个加法表和一个乘法表。在这些表中,标记为 [a] n 的行和标记为 [b] n 的列中的条目是 [a]n⊕n[b]n(左表)或 [a] n n [b] n (右表)。 ⊕5[0][1][2][3][4][0][0][1][2][3][4][1][1][2][3][4][5][2][2][3][4][5][6][3][3][4][5][6][7][4][4][5][6][7][8]⊗5[0][1][2][3][4][0][0][0][0][0][0][1][0][1][2][3][4][2][0][2][4][6][8][3][0][3][6][9][12][4][0][4][8][12][16] 我们可以通过更改输出的名称来获得信息更丰富的表格,以便表格中的每个条目都是标准名称 [0]、[1]、[2]、[3] 或 [4] 之一。例如,[3]⊕5[4]=[7]=[2] 且 [3] ⊗ 5 [4] = [12] = [2]。进行这些更改后,将生成以下表格:⊕5[0][1][2][3][4][0][0][1][2][3][4][1][1][2][3][4][0][2][2][3][4][0][1][3][3][4][0][1][2][4][4][0][1][2][3]⊗5[0][1][2][3][4][0][0][0][0][0][0][1][0][1][2][3][4][2][0][2][4][1][3][3][0][3][1][4][2][4][0][4][3][2][1]

We must now discuss a crucial technical point about the definitions of n and ⊗ n . Just as the outputs of these operations have many names, so also do the inputs . For instance, when we computed [3] ⊗ 5 [4] = [12], we chose the particular names [3] and [4] for the equivalence classes { 3 + 5 k : k Z } and { 4 + 5 k : k Z } that are being combined by the new addition operation. What happens if we use different names for these two inputs? For instance, we know [3] = [8] and [4] = [ − 6] (mod 5) by the Theorem on Equivalence Classes. Using the new names, we would compute [8] ⊗ 5 [ − 6] = [ − 48]. This may at first seem to disagree with our previous answer of [12], until we realize that [ − 48] = [2] = [12]. However, it is not yet clear that we must always get the same output (possibly with a new name) whenever we change the names of the two inputs to the operations n and ⊗ n .

现在我们必须讨论一个关于 ⊕n 和 ⊗ n 定义的一个关键技术点。正如这些运算的输出有很多名称一样,它们的输入也有很多名称。例如,当我们计算 [3] ⊗ 5 [4] = [12] 时,我们为新的加法运算组合的等价类 {3+5k:k∈Z} 和 {4+5k:k∈Z} 选择了特定的名称 [3] 和 [4]。如果我们为这两个输入使用不同的名称会发生什么?例如,根据等价类定理,我们知道 [3] = [8] 且 [4] = [ − 6] (mod 5)。使用新的名称,我们将计算 [8] ⊗ 5 [ − 6] = [ − 48]。这乍一看似乎与我们之前的答案 [12] 相矛盾,直到我们意识到 [ − 48] = [2] = [12]。然而,当我们更改操作 ⊕n 和 ⊗ n 的两个输入的名称时,是否总是能得到相同的输出(可能名称不同),这一点尚不明确。

To resolve this issue in general, we need to address the following question. Suppose n ≥ 1 is fixed, and a , a ′, b , b ′ are integers such that [ a ] n = [ a ′] n and [ b ] n = [ b ′] n . Using the original names [ a ] n and [ b ] n , we would compute [ a ] n n [ b ] n = [ a + b ] n . Using the new names [ a ′] n and [ b ′] n , we would instead compute [ a ] n n [ b ] n = [ a + b ] n . Is it always true that [ a + b ] n = [ a ′ + b ′] n ? The answer is yes, thanks to some previous theorems. By the Theorem on Equivalence Classes (§ 6.3 ), our assumptions on a , a ′, b , b ′ mean that a n a ′ and b n b ′. Next, by part (b) of the Theorem on Congruence Modulo n 6.2 ), we see that a + b n a ′ + b ′. So the Theorem on Equivalence Classes tells us that [ a + b ] n = [ a ′ + b ′] n . We have now proved that changing the names of the inputs to n does not change the output of the operation (although the name of the output may very well change). More briefly, we say that n is a well-defined operation. Similarly, to say that ⊗ n is well-defined means that

为了解决这个问题,我们需要解决以下问题。假设 n ≥ 1 是固定的,a、a′、b、b′ 是整数,且 [a] n = [a′] n 且 [b] n = [b′] n 。使用原始名称 [a] n 和 [b] n ,我们可以计算 [a]n⊕n[b]n=[a+b]n。使用新名称 [a′] n 和 [b′] n ,我们可以计算 [a′]n⊕n[b′]n=[a′+b′]n。 [a + b] n = [a′ + b′] n 总是成立吗?答案是肯定的,这要归功于前面的一些定理。根据等价类定理(§6.3),我们对 a、a′、b、b′ 的假设意味着 a ≡ n a′ 且 b ≡ n b′。接下来,根据模 n 同余定理的 (b) 部分(§6.2),我们看到 a + b ≡ n a′ + b′。因此,等价类定理告诉我们 [a + b] n = [a′ + b′] n 。我们现在已经证明,将输入名称更改为 ⊕n 不会改变运算的输出(尽管输出名称很可能会改变)。简而言之,我们称 ⊕n 是一个定义良好的运算。类似地,说 ⊗ n 是定义良好的,意味着

a 一个 , b b , a 一个 , b b Z Z , ( [ a 一个 ] n n = [ a 一个 ] n n [ b b ] n n = [ b b ] n n ) [ a 一个 b b ] n n = [ a 一个 b b ] n n .

The proof we gave above works here, this time using part (c) of the Theorem on Congruence Modulo n .

我们上面给出的证明在这里也适用,这次使用模 n 同余定理的 (c) 部分。

We can rephrase our discussion in a more formal way that makes contact with our definition of a function. Addition mod n is really a function S : Z / n × Z / n Z / n such that S ( [ a ] n , [ b ] n ) = [ a ] n n [ b ] n = [ a + b ] n for all a , b Z . The graph of this addition function is, by definition, the set of ordered pairs G = { ( ( [ a ] n , [ b ] n ) , [ a + b ] n ) : a , b Z } . To confirm that S really is a function, we must check three things (see § 5.4 ): (a) G ( Z / n × Z / n ) × Z / n ; (b) for all x Z / n × Z / n , there exists y Z / n with ( x , y ) ∈ G ; and (c) for all x , x Z / n × Z / n and all y , y Z / n , if ( x , y ) ∈ G and ( x ′, y ′) ∈ G and x = x ′, then y = y ′. Part (c) tells us that the proposed function S is single-valued : putting equal inputs x and x ′ into the function must produce equal outputs y and y ′. This is exactly what we checked above, taking x to be the ordered pair ([ a ] n , [ b ] n ), x ′ to be the ordered pair ([ a ′] n , [ b ′] n ), y = [ a + b ] n , and y ′ = [ a ′ + b ′] n . Conditions (a) and (b) are immediate in this example, though in other situations these conditions might need to be checked explicitly.

我们可以用更正式的方式重新表述我们的讨论,使其与函数的定义相联系。模 n 加法实际上是一个函数 S:Z/n×Z/n→Z/n,使得对于所有 a,b∈Z,都有 S([a]n,[b]n)=[a]n⊕n[b]n=[a+b]n。根据定义,这个加法函数的图像是有序对的集合 G={(([a]n,[b]n),[a+b]n):a,b∈Z}。为了确认 S 确实是一个函数,我们必须检查三件事(参见 §5.4):(a) G⊆(Z/n×Z/n)×Z/n;(b) 对于所有 x∈Z/n×Z/n,存在 y∈Z/n 使得 (x, y) ∈ G; (c) 对于所有 x,x′∈Z/n×Z/n 和所有 y,y′∈Z/n,如果 (x, y) ∈ G 且 (x′, y′) ∈ G 且 x = x′,则 y = y′。(c) 部分告诉我们,所提出的函数 S 是单值的:将相等的输入 x 和 x′ 代入该函数必须产生相等的输出 y 和 y′。这正是我们上面所验证的,其中 x 为有序对 ([a] n , [b] n ),x′ 为有序对 ([a′] n , [b′] n ),y = [a + b] n ,y′ = [a′ + b′] n 。在本例中,条件 (a) 和 (b) 是立即满足的,但在其他情况下,这些条件可能需要明确地进行检查。

Algebraic Properties of Operations Modulo n

模 n 运算的代数性质

In § 2.1 , we listed some algebraic properties of the addition and multiplication operations on the set of integers. Almost all of these properties extend to the new operations of addition and multiplication modulo n on Z / n . Furthermore, now that we know the latter operations are well-defined, we can quickly deduce properties for Z / n from the corresponding properties for Z .

在第 2.1 节中,我们列举了整数集上加法和乘法运算的一些代数性质。几乎所有这些性质都可以推广到 Z/n 上的模 n 加法和乘法运算。此外,既然我们知道后者的运算是良定义的,我们就可以迅速地从 Z 的相应性质推导出 Z/n 的性质。

6.83 Theorem on Addition and Multiplication Modulo n . For all n ≥ 1, Axioms 2.4(a) through (j) in § 2.1 (closure, commutativity, associativity, identity, additive inverses, and the distributive law) hold with Z replaced by Z / n , + replaced by n , · replaced by ⊗ n , 0 replaced by [0] n , and 1 replaced by [1] n .

Proof . We prove a few axioms to illustrate, leaving the others as exercises. To show that addition modulo n is commutative, fix two elements of Z / n , say [ a ] n and [ b ] n for some a , b Z . We compute

[ a ] n n [ b ] n = [ a + b ] n (by definition of n ) = [ b + a ] n (since integer addition is commutative) = [ b ] n n [ a ] n (by definition of n ) .
To prove the distributive law, fix three elements of Z / n , say [ a ] n , [ b ] n , [ c ] n where a , b , c Z . We compute
[ a ] n n ( [ b ] n n [ c ] n ) = [ a ] n n [ b + c ] n (by definition of n ) = [ a ( b + c ) ] n (by definition of n ) = [ ( a b ) + ( a c ) ] n (by the distributive law in Z ) = [ a b ] n n [ a c ] n (by definition of n ) = ( [ a ] n n [ b ] n ) n ( [ a ] n n [ c ] n ) (by definition of n ) .
Assume we have already proved that [0] n is the identity element for n . To prove that additive inverses exist, fix [ a ] n Z / n , where a Z . Note that a Z , so [ − a ] n is an element of Z n , and
[ a ] n n [ a ] n = [ a + ( a ) ] n = [ 0 ] n .

Thus, [ − a ] n is an additive inverse for [ a ] n in Z / n . □

6.83 模 n 加法和乘法定理。对于所有 n ≥ 1,§2.1 中的公理 2.4(a) 至 (j)(封闭性、交换律、结合律、单位元、加法逆元和分配律)成立,其中 Z 替换为 Z/n,+ 替换为 ⊕n,· 替换为 ⊗ n ,0 替换为 [0] n ,1 替换为 [1] n 。证明。我们证明几个公理以作说明,其余公理留作练习。为了证明模 n 加法满足交换律,固定 Z/n 中的两个元素,例如 [a] n 和 [b] n ,其中 a, b∈Z。我们计算 [a]n⊕n[b]n=[a+b]n(根据⊕n的定义)=[b+a]n(因为整数加法满足交换律)=[b]n⊕n[a]n(根据⊕n的定义)。为了证明分配律,固定 Z/n 中的三个元素,例如 [a] n 、[b] n 、[c] n ,其中 a、b、c∈Z。我们计算 [a]n⊗n([b]n⊕n[c]n)=[a]n⊗n[b+c]n(根据⊕n的定义)=[a⋅(b+c)]n(根据⊗n的定义)=[(a⋅b)+(a⋅c)]n(根据Z中的分配律)=[a⋅b]n⊕n[a⋅c]n(根据⊕n的定义)=([a]n⊗n[b]n)⊕n([a]n⊗n[c]n)(根据⊗n的定义)。假设我们已经证明 [0] n 是⊕n的单位元。为了证明加法逆元存在,固定 [a]n∈Z/n,其中 a∈Z。注意 −a∈Z,所以 [ − a] n 是 Zn 的一个元素,且 [a]n⊕n[−a]n=[a+(−a)]n=[0]n。因此,[ − a] n 是 Z/n 中 [a] n 的加法逆元。□

The question of whether Z / n has zero divisors or multiplicative inverses for nonzero elements is more subtle. Some of the exercises ask you to show that when n is prime, every nonzero element of Z / n has a multiplicative inverse modulo n , and there are no zero divisors. On the other hand, when n > 1 is not prime, Z / n has zero divisors and some nonzero elements of Z / n do not have multiplicative inverses.

矩阵 Z/n 是否存在零因子或非零元素是否存在乘法逆元这个问题比较微妙。有些练习要求证明,当 n 为素数时,Z/n 中的每个非零元素都存在模 n 的乘法逆元,且不存在零因子。另一方面,当 n > 1 且 n 不是素数时,Z/n 存在零因子,且 Z/n 中的某些非零元素不存在乘法逆元。

Constructing Rational Numbers from Integers

利用整数构造有理数

In most of this book, we have taken the set R of real numbers as our starting point, and we have defined the set Q of rational numbers as a particular subset of R (namely, the subset of real numbers of the form a / b , where a Z and b Z 0 ). However, it is also possible to build the set of rational numbers from the more basic set of integers. We can then introduce addition, multiplication, and an ordering relation on Q using the corresponding concepts for Z , and prove that these new entities obey the expected axioms. These constructions provide an excellent example of an algebraic application of equivalence relations and quotient sets.

本书大部分内容都以实数集 R 为出发点,并将有理数集 Q 定义为 R 的一个特例(即形如 a/b 的实数集,其中 a∈Z 且 b∈Z≠0)。然而,我们也可以从更基本的整数集构造有理数集。然后,我们可以利用整数集 Z 的相应概念,在 Q 上引入加法、乘法和排序关系,并证明这些新实体满足预期的公理。这些构造为等价关系和商集在代数中的应用提供了一个极佳的例子。

Intuitively, we are going to build Q from Z using the notion of fractions, which we have already mentioned above. We want a rational number to be a ratio a / b of two integers a and b , where b cannot be zero. The main difficulty is that the division operation for real numbers is no longer available, so we cannot interpret a / b as “ a divided by b .” We could try to get around this by introducing the ordered pair ( a , b ) as a formal model for the intuitive fraction a / b .

直观地讲,我们将利用前面提到的分数概念,从 Z 构建 Q。我们希望有理数是两个整数 a 和 b 的比值 a/b,其中 b 不能为零。主要难点在于实数的除法运算不再适用,因此我们不能将 a/b 解释为“a 除以 b”。我们可以尝试引入有序对 (a, b) 作为直观分数 a/b 的形式模型来解决这个问题。

But now a new difficulty arises: there are many ways to represent a particular rational number in the form a / b . For example, 3/5 = 6/10 = 9/15 = −3/ − 5 = · · ·. However, the ordered pairs (3, 5), (6, 10), etc., are not equal! Now the key idea presents itself: we would like to regard all of these ordered pairs as equivalent for the purposes of representing fractions. This suggests introducing an equivalence relation on the set of ordered pairs, and using the equivalence class of ( a , b ) as the formal model for the fraction a / b . We want ( a , b ) to be equivalent to ( c , d ) iff a / b = c / d , but the latter condition involves the forbidden notion of division. So we rewrite the condition as ad = bc , which is a statement that only involves integers and multiplication of integers. At last, we are ready for our new definition of rational numbers.

但现在出现了一个新的难题:用 a/b 的形式表示一个特定的有理数有很多种方法。例如,3/5 = 6/10 = 9/15 = −3/ − 5 = …。然而,有序数对 (3, 5)、(6, 10) 等等并不相等!关键思想由此显现:为了表示分数,我们希望将所有这些有序数对视为等价的。这提示我们在有序数对集合上引入一个等价关系,并将 (a, b) 的等价类作为分数 a/b 的形式模型。我们希望 (a, b) 等价于 (c, d) 当且仅当 a/b = c/d,但后一个条件涉及除法运算,而除法运算是被禁止的。因此,我们将条件重写为 ad = bc,这是一个只涉及整数和整数乘法的语句。最终,我们得到了有理数的新定义。

6.84. Definition: Rational Numbers.. Let X = Z × Z 0 be the set of ordered pairs ( a , b ) where a , b are integers and b ≠ 0. Define an equivalence relation ∼ on X by setting ( a , b ) ∼ ( c , d ) iff ad = bc . The equivalence class [( a , b )] is denoted a / b or a b . The set X / ∼ of all equivalence classes is denoted Q . Elements of Q are called rational numbers .

6.84. 定义:有理数。设 X=Z×Z≠0 为有序对 (a, b) 的集合,其中 a, b 为整数且 b ≠ 0。定义 X 上的等价关系 ∼,使得 (a, b) ∼ (c, d) 当且仅当 ad = bc。等价类 [(a, b)] 记为 a/b 或 ab。所有等价类构成的集合 X/∼ 记为 Q。Q 中的元素称为有理数。

You checked that ∼ is an equivalence relation on X in Exercise 14 of § 6.2 . Look at your solution to that exercise and make sure your proof did not use division! For instance, here is the verification that ∼ is transitive. Fix a , b , c , d , e , f Z with b , d , f ≠ 0; assume ( a , b ) ∼ ( c , d ) and ( c , d ) ∼ ( e , f ); prove ( a , b ) ∼ ( e , f ). We have assumed ad = bc and cf = de ; we must prove af = be . Multiplying the assumptions, we get ( ad )( cf ) = ( bc )( de ), or ( afc ) d = ( bec ) d since integer multiplication is commutative and associative. We can rewrite this as ( afc bec ) d = 0 by the distributive law. Since d ≠ 0 and Z has no zero divisors, we conclude afc bec = 0, or ( af ) c = ( be ) c . If c is nonzero, we can similarly deduce that af = be , as needed. If c = 0, we have ad = bc = 0 = cf = de . Since d is nonzero, we see that a = e = 0, so af = 0 = be , as needed.

你在§6.2的练习14中验证了∼是X上的等价关系。回顾一下你对该练习的解答,确保你的证明过程中没有使用除法!例如,以下是∼传递性的验证。固定a,b,c,d,e,f∈Z,其中b,d,f≠0;假设(a,b)∼(c,d)且(c,d)∼(e,f);证明(a,b)∼(e,f)。我们假设ad=bc且cf=de;我们需要证明af=be。将这两个假设相乘,我们得到(ad)(cf)=(bc)(de),或者(afc)d=(bec)d,因为整数乘法满足交换律和结合律。根据分配律,我们可以将其改写为(afc-bec)d=0。由于 d ≠ 0 且 Z 没有零因子,我们得出 afc − bec = 0,即 (af)c = (be)c。如果 c 非零,我们可以类似地推导出 af = be,满足要求。如果 c = 0,则有 ad = bc = 0 = cf = de。由于 d 非零,我们看到 a = e = 0,因此 af = 0 = be,满足要求。

Let a , b , c , d Z with b , d ≠ 0. By the Theorem on Equivalence Classes and the definition of ∼, [( a , b )] = [( c , d )] iff ( a , b ) ∼ ( c , d ) iff ad = bc . We can now officially say that a / b = c / d iff a d = b c , provided we remember that the expressions a / b and c / d are abbreviations for the equivalence classes [( a , b )] and [( c , d )] .

设 a,b,c,d∈Z,其中 b, d ≠ 0。根据等价类定理和 ∼ 的定义,[(a, b)] = [(c, d)] 当且仅当 (a, b) ∼ (c, d) 当且仅当 ad = bc。现在我们可以正式地说 a/b=c/d 当且仅当 ad=bc,前提是我们要记住表达式 a/b 和 c/d 分别是等价类 [(a, b)] 和 [(c, d)] 的缩写。

Addition and Multiplication of Rational Numbers

有理数的加法和乘法

The next step in the formal development of Q is to introduce the operations of addition and multiplication for rational numbers. As in the case of Z / n , the key initial point is to verify that these operations are well-defined , not depending on which names we use for the inputs to the operations.

Q 形式化发展的下一步是引入有理数的加法和乘法运算。与 Z/n 的情况类似,关键的初始点是验证这些运算是否定义明确,而不依赖于我们对运算输入使用的名称。

6.85 Definition: Addition and Multiplication in Q . Given rational numbers a / b and c / d , define a b + c d = a d + b c b d and a b c d = a c b d .

6.85 定义:Q 中的加法和乘法。给定有理数 a/b 和 c/d,定义 ab+cd=ad+bcbd 和 ab⋅cd=acbd。

Using the more cumbersome equivalence class notation for fractions, this definition says that [( a , b )] + [( c , d )] = [( ad + bc , bd )] and [( a , b )] · [( c , d )] = [( ac , bd )].

使用分数的等价类表示法,该定义表示 [(a, b)] + [(c, d)] = [(ad + bc, bd)] 且 [(a, b)] · [(c, d)] = [(ac, bd)]。

6.86 Lemma. The operations + and · on Q are well-defined.

6.86 引理。Q 上的运算 + 和 · 是定义良好的。

Proof . Fix a , b , c , d , a , b , c , d Z with b , d , b ′, d ′ ≠ 0. Assume a / b = a ′/ b ′ and c / d = c ′/ d ′, which means ab ′ = ba ′ and cd ′ = dc ′. To show that addition is well-defined, we prove that a / b + c / d = a ′/ b ′ + c ′/ d ′. Expanding the definition, we must show that ( ad + bc )/( bd ) = ( a d ′ + b c ′)/( b d ′), which means ( ad + bc ) b d ′ = bd ( a d ′ + b c ′). Working in Z , we compute

证明。固定 a, b, c, d, a′, b′, c′, d′∈Z,其中 b, d, b′, d′ ≠ 0。假设 a/b = a′/b′ 且 c/d = c′/d′,这意味着 ab′ = ba′ 且 cd′ = dc′。为了证明加法定义良好,我们证明 a/b + c/d = a′/b′ + c′/d′。展开定义,我们必须证明 (ad + bc)/(bd) = (a′d′ + b′c′)/(b′d′),这意味着 (ad + bc)b′d′ = bd(a′d′ + b′c′)。在 Z 中,我们计算

( a 一个 d d + b b c c ) b b d d = a 一个 d d b b d d + b b c c b b d d = a 一个 b b d d d d + c c d d b b b b = b b a 一个 d d d d + d d c c b b b b = b b d d ( a 一个 d d + b b c c ) ,

as needed. To show that multiplication is well-defined, we prove that ( a / b ) · ( c / d ) = ( a ′/ b ′) · ( c ′/ d ′), which means ( ac )/( bd ) = ( a c ′)/( b d ′). Here we must prove acb d ′ = bda c ′. We compute acb d ′ = ( ab ′)( cd ′) = ( ba ′)( dc ′) = bda c ′, as needed. □

根据需要。为了证明乘法定义良好,我们证明 (a/b) · (c/d) = (a′/b′) · (c′/d′),这意味着 (ac)/(bd) = (a′c′)/(b′d′)。这里我们必须证明 acb′d′ = bda′c′。我们根据需要计算 acb′d′ = (ab′)(cd′) = (ba′)(dc′) = bda′c′。□

Now we can prove that addition and multiplication of rational numbers satisfies axioms similar to those in Item 2.4, and moreover every nonzero rational number has a multiplicative inverse.

现在我们可以证明有理数的加法和乘法满足与第 2.4 项中类似的公理,而且每个非零有理数都有乘法逆元。

6.87 Theorem on Algebraic Operations in Q .

6.87 Q 中的代数运算定理。

(a) Closure: For all x , y Q , x + y Q and x y Q .

(a)封闭性:对于所有 x,y∈Q,x+y∈Q 和 x⋅y∈Q。

(b) Commutativity: For all x , y Q , x + y = y + x and x · y = y · x .

(b)交换律:对于所有 x,y∈Q,x + y = y + x 且 x · y = y · x。

(c) Associativity: For all x , y , z Q , ( x + y ) + z = x + ( y + z ) and ( x · y ) · z = x · ( y · z ).

(c)结合律:对于所有 x,y,z∈Q,(x + y)+ z = x +(y + z)且(x · y)· z = x ·(y · z)。

(d) Identity: For all x Q , x + (0/1) = x and x · (1/1) = x .

(d)恒等式:对于所有 x∈Q,x + (0/1) = x 且 x · (1/1) = x。

(e) Additive Inverses: x Q , y Q , x + y = 0 / 1 .

(e)加法逆元:∀x∈Q,∃y∈Q,x+y=0/1。

(f) Multiplicative Inverses: x Q , if x ≠ 0/1, then y Q , x · y = 1/1.

(f)乘法逆元:∀x∈Q,如果x≠0/1,则存在y∈Q,x·y=1/1。

(g) No Zero Divisors: For all x , y Q , if x · y = 0/1, then x = 0/1 or y = 0/1.

(g)无零因子:对于所有 x,y∈Q,如果 x · y = 0/1,则 x = 0/1 或 y = 0/1。

Proof . We prove a few parts as illustrations, leaving the others as exercises. First we prove that addition is commutative. Fix x , y Q , and write x = a / b and y = c / d for some a , b , c , d Z with b , d ≠ 0. Note that x + y = ( a / b ) + ( c / d ) = ( ad + bc )/( bd ), whereas y + x = ( c / d ) + ( a / b ) = ( cb + da )/( db ). Since addition and multiplication in Z are commutative, ad + bc = cb + da and bd = db , so x + y = y + x . Next we prove that multiplication is associative. Fix x , y , z Q , and write x = a / b , y = c / d , z = e / f for some a , b , c , d , e , f Z with b , d , f ≠ 0. On one hand, ( x y ) z = a c b d e f = ( a c ) e ( b d ) f . On the other hand, x ( y z ) = a b c e d f = a ( c e ) b ( d f ) . Since multiplication in Z is associative, ( x · y ) · z = x · ( y · z ) follows. To check the additive identity axiom, fix x Q , and write x = a / b for some a , b Z with b ≠ 0. Compute x + 0/1 = a / b + 0/1 = ( a · 1 + b · 0)/( b · 1) = a / b = x . To check the multiplicative inverse axiom, fix x Q with x ≠ 0/1. Write x = a / b with a , b Z and b ≠ 0. Since a / b ≠ 0/1, a · 1 ≠ b · 0, and hence a ≠ 0. Thus, y = b / a is a member of Q , and we compute x · y = ( a / b ) · ( b / a ) = ( ab )/( ba ). The fraction ( ab )/( ba ) equals 1/1 because ( ab )1 = ( ba )1. Thus, xy = 1/1, as needed. □

证明。我们证明其中几个部分作为示例,其余部分留作练习。首先,我们证明加法满足交换律。固定 x, y∈Q,并令 x = a/b 和 y = c/d,其中 a, b, c, d∈Z,且 b, d ≠ 0。注意到 x + y = (a/b) + (c/d) = (ad + bc)/(bd),而 y + x = (c/d) + (a/b) = (cb + da)/(db)。由于 Z 中的加法和乘法满足交换律,所以 ad + bc = cb + da 且 bd = db,因此 x + y = y + x。接下来,我们证明乘法满足结合律。固定 x, y, z∈Q,令 x = a/b,y = c/d,z = e/f,其中 a, b, c, d, e, f∈Z,且 b, d, f ≠ 0。一方面,(x⋅y)⋅z=acbd⋅ef=(ac)e(bd)f。另一方面,x⋅(y⋅z)=ab⋅cedf=a(ce)b(df)。由于 Z 中的乘法满足结合律,因此 (x · y) · z = x · (y · z)。为了验证加法恒等式,固定 x∈Q,令 x = a/b,其中 a, b∈Z,且 b ≠ 0。计算 x + 0/1 = a/b + 0/1 = (a · 1 + b · 0)/(b · 1) = a/b = x。为了检验乘法逆元公理,固定 x∈Q 且 x ≠ 0/1。令 x = a/b,其中 a, b∈Z 且 b ≠ 0。由于 a/b ≠ 0/1,所以 a · 1 ≠ b · 0,进而 a ≠ 0。因此,y = b/a 属于 Q,我们计算 x · y = (a/b) · (b/a) = (ab)/(ba)。分数 (ab)/(ba) 等于 1/1,因为 (ab)¹ = (ba)¹。因此,xy = 1/1,符合要求。□

We usually write 0 instead of 0/1 and 1 instead of 1/1. More generally, for any integer n , we often identify n with the rational number n /1. You can check that for all n , m Z , n /1 + m /1 = ( n + m )/1; ( n /1) · ( m /1) = ( nm )/1; and if n m , then n /1 ≠ m /1. So the identification of n with n /1 preserves equality, addition, and multiplication of integers. This identification also allows us to view Z as a subset of Q .

我们通常用 0 代替 0/1,用 1 代替 1/1。更一般地,对于任意整数 n,我们通常将 n 与有理数 n/1 等同起来。你可以验证,对于所有 n,m∈Z,n/1 + m/1 = (n + m)/1;(n/1) · (m/1) = (nm)/1;并且如果 n ≠ m,则 n/1 ≠ m/1。因此,将 n 与 n/1 等同起来,保持了整数的相等性、加法和乘法。这种等同也使得我们可以将 Z 视为 Q 的子集。

Section Summary

章节概要

  1. Integers Modulo n . For n Z > 0 , Z / n consists of the n distinct equivalence classes [0] n , [1] n , …, [ n − 1] n of the equivalence relation ≡ n . For all a , b Z , [ a ] n = [ b ] n iff a n b iff n divides a b . The operations [ a ] n n [ b ] n = [ a + b ] n and [ a ] n n [ b ] n = [ ab ] n are well-defined and satisfy closure, commutativity, associativity, identity, distributive laws, and existence of additive inverses.
    整数模 n。对于 n∈Z>0,Z/n 由等价关系 ≡ n 的 n 个不同的等价类 [0] n , [1] n , …, [n − 1] n 组成。对于所有 a,b∈Z,[a] n = [b] n 当且仅当 a ≡ n b 当且仅当 n 整除 a − b。运算 [a]n⊕n[b]n=[a+b]n 和 [a] n n [b] n = [ab] n 是定义良好的,并且满足封闭性、交换律、结合律、单位元、分配律和加法逆元的存在性。
  2. Well-Defined Operations. To see that an operation defined on equivalence classes is well-defined, one must check that changing the names of the inputs to the operations does not change the value of the output (although the name of the output may change). For example, saying that n is well-defined means a , b , a , b Z , if [ a ] n = [ a ′] n and [ b ] n = [ b ′] n , then [ a + b ] n = [ a ′ + b ′] n .
    定义良好的运算。要判断定义在等价类上的运算是否定义良好,必须检查改变运算的输入名称是否不会改变输出值(尽管输出的名称可能会改变)。例如,说 ⊕n 是定义良好的,意味着对于任意 a, b, a′, b′∈Z,如果 [a] n = [a′] n 且 [b] n = [b′] n ,则 [a + b] n = [a′ + b′] n
  3. Rational Numbers. Formally, Q = X / , where X = Z × Z 0 and ( a , b ) ∼ ( c , d ) iff ad = bc (where a , b , c , d Z and b , d ≠ 0). The equivalence class [( a , b )] is denoted a / b or a b . The operations a / b + c / d + ( ad + bc )/( bd ) and ( a / b ) · ( c / d ) = ( ac )/( bd ) are well-defined and satisfy closure, commutativity, associativity, identity, inverse, and distributive laws.
    有理数。形式上,Q=X/∼,其中X=Z×Z≠0,且(a, b)∼(c, d)当且仅当ad=bc(其中a,b,c,d∈Z且b,d≠0)。等价类[(a, b)] 记为a/b或ab。运算a/b + c/d + (ad + bc)/(bd)和(a/b)·(c/d)=(ac)/(bd)是良定义的,并且满足封闭性、交换律、结合律、单位元、逆元和分配律。

Exercises

练习

  1. Make an addition table and multiplication table for Z / 6 . Does Z / 6 have zero divisors?
    为 Z/6 制作加法表和乘法表。Z/6 是否有零因数?
  2. Make an addition table and multiplication table for Z / 7 . Does Z / 7 have zero divisors?
    为 Z/7 制作加法表和乘法表。Z/7 是否有零因数?
  3. Fix n Z > 0 , and define f : Z / n Z / n by f ([ a ] n ) = [ − a ] n for all a Z . Prove that f is well-defined.
    固定 n∈Z>0,并定义 f:Z/n→Z/n 为 f([a] n ) = [ − a] n ,其中 a∈Z。证明 f 是良定义的。
  4. Prove the remaining axioms for addition in Theorem 6.83 .
    证明定理 6.83 中关于加法的其余公理。
  5. Prove the remaining axioms for multiplication in Theorem 6.83 .
    证明定理 6.83 中乘法的其余公理。
  6. (a) Find all x Z / 12 for which there exist y Z / 12 with x 12 y = 1. (b) Repeat (a) for Z / 13 . (c) Repeat (a) for Z / 16 .
    (a)找出所有满足 x ⊗ 12 y = 1 的 x∈Z/12。(b)对 Z/13 重复(a)。(c)对 Z/16 重复(a)。
  7. Suppose p is prime. (a) Prove: for all x Z / p , if x ≠ [0] p then there exists y Z / p with x p y = [1] p . [ Hint: What is gcd( x , p )?] (b) Prove Z / p has no zero divisors, i.e., for all x , y Z / p , if x p y = [0] p , then x = [0] p or y = [0] p .
    假设 p 是素数。(a) 证明:对于所有 x∈Z/p,如果 x ≠ [0] p ,则存在 y∈Z/p,使得 x ⊗ p y = [1] p 。[提示:gcd(x, p) 是什么?] (b) 证明 Z/p 没有零因子,即对于所有 x,y∈Z/p,如果 x ⊗ p y = [0] p ,则 x = [0] p 或 y = [0] p
  8. Suppose n > 1 is not prime. (a) Prove there exist nonzero x , y Z / n with x n y = [0] n . (b) Prove that some nonzero elements of Z / n do not have multiplicative inverses.
    假设 n > 1 不是素数。(a) 证明存在非零元素 x,y∈Z/n,使得 x ⊗ n y = [0] n 。(b) 证明 Z/n 中存在一些非零元素没有乘法逆元。
  9. Prove the remaining parts of Theorem 6.87 .
    证明定理 6.87 的其余部分。
  10. Define i : Z Q by i ( m ) = m /1 for m Z . Prove that i is one-to-one and for all m , n Z , i ( m + n ) = i ( m ) + i ( n ) and i ( mn ) = i ( m ) i ( n ). [The function i lets us identify Z with a subset of Q .]
    定义 i:Z→Q,其中 i(m) = m/1,m∈Z。证明 i 是单射,且对于所有 m,n∈Z,i(m + n) = i(m) + i(n) 且 i(mn) = i(m)i(n)。[函数 i 使我们能够将 Z 与 Q 的一个子集等同起来。]
  11. Which of the following functions are well-defined ? Prove your answers. (a) f : Q Z given by f ( m / n ) = m for m , n Z with n ≠ 0. (b) g : Q 0 Q given by g ( m / n ) = n / m for m , n Z 0 . (c) h : Q Q given by h ( m / n ) = 3 m / n for m , n Z with n ≠ 0. (d) k : Q Q given by k ( m / n ) = ( m + 5)/ n for m , n Z with n ≠ 0.
    下列哪些函数定义良好?请证明你的答案。 (a) f:Q→Z,定义 f(m/n) = m,其中 m,n∈Z,n ≠ 0。 (b) g:Q≠0→Q,定义 g(m/n) = n/m,其中 m,n∈Z≠0。 (c) h:Q→Q,定义 h(m/n) = 3m/n,其中 m,n∈Z,n ≠ 0。 (d) k:Q→Q,定义 k(m/n) = (m + 5)/n,其中 m,n∈Z,n ≠ 0。
  12. (a) Prove: for all a , b , c Z with b , c ≠ 0, a / b = ( ac )/( bc ). (b) Show that for every x Q , there exists a unique pair ( a , b ) Z 2 with b > 0, gcd( a , b ) = 1, and x = a / b . (Informally, the particular name a / b for x is called the fraction in lowest terms representing x .)
    (a) 证明:对于任意 a, b, c∈Z,且 b, c ≠ 0,有 a/b = (ac)/(bc)。(b) 证明:对于任意 x∈Q,存在唯一的 (a, b)∈Z2,使得 b > 0,gcd(a, b) = 1,且 x = a/b。(通俗地说,x 的特指 a/b 称为表示 x 的最简分数。)
  13. Ordering on Q . Given x , y Q , we know x = a / b and y = c / d for some a , b , c , d Z with b , d Z > 0 (see the previous exercise). Define a relation ≤ on Q by setting x y iff ad bc . (a) Prove that this relation is well-defined. (b) Prove that ≤ is a total ordering of Q . Also prove that for all m , n Z , m n (in Z ) iff m /1 ≤ n /1 (in Q ).
    在有理数域 Q 上定义一个关系 ≤。给定 x, y∈Q,我们知道对于任意 a, b, c, d∈Z,有 x = a/b 且 y = c/d,其中 b, d∈Z > 0(参见前面的练习)。在 Q 上定义一个关系 ≤,使得 x ≤ y 当且仅当 ad ≤ bc。(a)证明该关系是良定义的。(b)证明 ≤ 是 Q 的一个全序关系。此外,证明对于任意 m, n∈Z,m ≤ n(在 Z 中)当且仅当 m/1 ≤ n/1(在 Q 中)。
  14. Fix n Z > 1 . Suppose we try to define an ordering relation ≤ on Z / n as follows: for all a , b Z , let [ a ] n ≤ [ b ] n iff a b . Give a precise explanation of why this construction fails.
    固定 n∈Z>1。假设我们尝试在 Z/n 上定义一个排序关系 ≤ 如下:对于所有 a,b∈Z,令 [a] n ≤ [b] n 当且仅当 a ≤ b。请给出该构造失败的精确解释。
  15. Let X be a fixed set. Define a relation ∼ on X × X as follows: for a , b , c , d X , let ( a , b ) ∼ ( c , d ) iff ( a = c b = d ) ( a = d b = c ) . (a) Prove ∼ is an equivalence relation on X × X . (b) Let Y be the set of unordered pairs { a , b } with a , b X . Define f : X / ∼ → Y by f ([( a , b )] ) = { a , b }. Prove that f is a well-defined bijection.
    设 X 为一个固定的集合。在 X × X 上定义关系 ∼ 如下:对于 a, b, c, d ∈ X,令 (a, b) ∼ (c, d) 当且仅当 (a=c∧b=d)∨(a=d∧b=c)。(a) 证明 ∼ 是 X × X 上的等价关系。(b) 设 Y 为所有无序对 {a, b} 的集合,其中 a, b ∈ X。定义 f:X/∼ →Y 为 f([(a, b)] ) = {a, b}。证明 f 是一个良定义的双射。
  16. Let R be a reflexive, transitive relation on a set X . (a) Define a relation ∼ on X as follows: for all a , b X , let a b iff aRb and bRa . Prove that ∼ is an equivalence relation. (b) Define a relation ≤ on the quotient set X / ∼ as follows: for all a , b X , [ a ] ≤ [ b ] iff aRb . Prove that ≤ is well-defined and ( X / ∼ , ≤ ) is a poset. (c) Discuss the effect of the preceding construction when X = Z and R = { ( a , b ) Z 2 : a | b } .
    设 R 是集合 X 上的自反传递关系。(a) 在 X 上定义关系 ∼ 如下:对于任意 a, b ∈ X,令 a ∼ b 当且仅当 aRb 且 bRa。证明 ∼ 是一个等价关系。(b) 在商集 X/∼ 上定义关系 ≤ 如下:对于任意 a, b ∈ X,[a] ≤ [b] 当且仅当 aRb。证明 ≤ 是良定义的,并且 (X/∼, ≤) 是一个偏序集。(c) 讨论当 X=Z 且 R={(a,b)∈Z2:a|b} 时,上述构造的效果。
  17. This problem outlines a formal construction of Z (the set of all integers) from Z 0 (the set of nonnegative integers). When solving this problem, take care not to use subtraction or additive inverses. Assume we know: x , y Z 0 , x + 1 = y + 1 x + y . (a) Prove by induction on n : x , y , n Z 0 , x + n = y + n x = y . (b) Let X = Z 0 × Z 0 , and define a relation ∼ on X by ( a , b ) ∼ ( c , d ) iff a + d = b + c for a , b , c , d Z 0 . Prove that ∼ is an equivalence relation on X . Denote the equivalence class [( a , b )] by [ a b ], and let Z be the set of all such equivalence classes. (c) For a , b , c , d Z 0 , define [ a b ] + [ c d ] = [( a + c ) − ( b + d )] and [ a b ] · [ c d ] = [( ac + bd ) − ( ad + bc )]. Prove that these operations are well-defined. (d) Check the axioms in 2.4 , assuming the corresponding facts for Z 0 are known. (e) Show: for every x Z , there exists a unique a Z 0 such that x = [ a − 0] or x = [0 − a ].
    本题概述了从 Z≥0(非负整数集合)构造 Z(所有整数的集合)的正式方法。在解决此问题时,请注意不要使用减法或加法逆元。假设我们已知:∀x,y∈Z≥0,x+1=y+1⇒x+y。(a)用数学归纳法证明:∀x,y,n∈Z≥0,x+n=y+n⇒x=y。(b)令 X=Z≥0×Z≥0,并在 X 上定义关系 ∼:(a, b) ∼ (c, d) 当且仅当对于 a,b,c,d∈Z≥0,a + d = b + c。证明 ∼ 是 X 上的等价关系。将等价类 [(a, b)] 记为 [a − b],并令 Z 为所有此类等价类的集合。 (c) 对于 a, b, c, d∈Z≥0,定义 [a − b] + [c − d] = [(a + c) − (b + d)] 和 [a − b] · [c − d] = [(ac + bd) − (ad + bc)]。证明这些运算是良定义的。(d) 假设 Z≥0 的相应事实已知,检验 2.4 中的公理。(e) 证明:对于任意 x∈Z,存在唯一的 a∈Z≥0,使得 x = [a − 0] 或 x = [0 − a]。
  18. Let f : X Y be a function, let Z = f [ X ] be the image of f , and let R be the equivalence relation ∼ f . Show that g : X / R Z , given by g ([ a ] R ) = f ( a ) for a X , is well-defined and bijective.
    设 f:X → Y 为一个函数,设 Z = f[X] 为 f 的像,设 R 为等价关系 ∼ f 。证明由 g([a] R ) = f(a) 给出的 g:X/R → Z,其中 a ∈ X,是良定义的且是双射的。
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7

Cardinality

基数

7.1  Finite Sets

7.1 有限集

The theory of cardinality gives us a rigorous way to measure and compare the size of various sets (finite or infinite). The key idea is that two sets X and Y have the same size (cardinality) iff there exists a bijection from X to Y . This definition has many surprising consequences for infinite sets that may challenge your intuition. For example, not all infinite sets have the same size. We will see that, in a sense to be made precise later, R is a strictly larger infinite set than Q , and P ( R ) (the set of all subsets of R ) is larger than R . On the other hand, Z > 0 and Z and Q all have the same cardinality, even though we have strict set inclusions Z > 0 Z Q . We begin our study of cardinality by defining finite sets and proving their basic properties. Our theorems about finite sets should agree with your intuition and may even seem to be obvious, but some of these results are surprisingly tricky to prove carefully.

基数理论为我们提供了一种严谨的方法来衡量和比较各种集合(有限集或无限集)的大小。其核心思想是,两个集合 X 和 Y 具有相同的大小(基数)当且仅当存在从 X 到 Y 的双射。这个定义对无限集有许多令人惊讶的推论,可能会挑战你的直觉。例如,并非所有无限集都具有相同的大小。我们将看到,在某种意义上(稍后会详细阐述),R 是一个严格大于 Q 的无限集,并且 P(R)(R 的所有子集构成的集合)大于 R。另一方面,Z>0 以及 Z 和 Q 都具有相同的基数,即使存在严格的集合包含关系 Z>0⊊Z⊊Q。我们首先定义有限集并证明其基本性质,以此作为基数研究的起点。我们关于有限集的定理应该与你的直觉相符,甚至可能看起来显而易见,但其中一些结果的证明却出乎意料地复杂。

Definition of Finite Sets

有限集的定义

In ordinary life, we count a set of objects by pointing to each object in the set and saying “one, two, three, ….” This verbal procedure is essentially setting up a one-to-one correspondence (bijection) from some set of integers {1, 2, …, n } onto the given set, as shown in Figure 7.1 . We formalize this idea in the next definition. Here and below, we use the notation {1, 2, …, n } to denote the set { i Z : 1 i n } .

在日常生活中,我们通过指着集合中的每个物体并说“一、二、三……”来计数。这种口头计数的方式本质上是在给定的集合上建立从某个整数集合{1, 2, …, n}到该集合的一一对应关系(双射),如图7.1所示。我们将在下一个定义中形式化这一概念。在下文中,我们使用符号{1, 2, …, n}来表示集合{i∈Z:1≤i≤n}。

7.1. Definition: n -Element Sets. Let X be a set. For all n Z > 0 , X has n elements iff there exists a bijection f : {1, 2, …, n } → X . We say X has 0 elements iff X = 0 .

7.1. 定义:n 元集合。设 X 为一个集合。对于所有 n∈Z>0,X 有 n 个元素当且仅当存在双射 f : {1, 2, …, n} → X。我们说 X 有 0 个元素当且仅当 X=0。

For n Z 0 , | X | = n means that X has n elements.

对于 n∈Z≥0,|X| = n 表示 X 有 n 个元素。

fig7_1

Figure 7.1

图 7.1

Counting a set of objects.

统计一组物体的数量。

Since a function f is a bijection iff f −1 exists and is a bijection (see §5.9 ), we can also say that X has n elements iff there is a bijection g : X → {1, 2, …, n }.

由于函数 f 是双射当且仅当 f −1 存在且是双射(见 §5.9),我们也可以说 X 有 n 个元素当且仅当存在双射 g : X → {1, 2, …, n}。

7.2 Definition: Finite and Infinite Sets. For all sets X , X is finite iff n Z 0 , | X | = n . X is infinite iff X is not finite .

7.2 定义:有限集和无限集。对于所有集合 X,X 是有限的当且仅当存在 n∈Z≥0,|X|=n。X 是无限的当且仅当 X 不是有限的。

Properties of Finite Sets

有限集的性质

The next theorem lists the main properties of finite sets.

下一个定理列出了有限集的主要性质。

7.3 Theorem on Finite Sets For all sets X , Y , X 1 , …, X s (where s Z > 0 ):

7.3 有限集定理 对于所有集合 X, Y, X 1 , …, X s (其中 s∈Z>0):

(a) Cardinality is Well-Defined: For all n , m Z 0 , if | X | = n and | X | = m , then n = m .

(a)基数定义良好:对于所有 n,m∈Z≥0,如果 |X| = n 且 |X| = m,则 n = m。

(b) Subset Property: If | X | = n and Y X , then m Z 0 , | Y | = m and m n .

(b)子集性质:如果 |X| = n 且 Y⊆ X,则存在 m∈Z≥0,|Y| = m 且 m ≤ n。

(c) Sum Rule: If | X | = n and | Y | = k and X Y = 0 , then | X Y | = n + k .

(c)求和规则:如果 |X| = n 且 |Y| = k 且 X∩Y=0,则 |X∪Y|=n+k。

(d) Sum Rule for s Sets: If | X i | = n i for 1 ≤ i s and X i X j = 0 for all i j , then | X 1 X 2 X s | = n 1 + n 2 + + n s .

(d)s 集的求和规则:如果对于 1 ≤ i ≤ s,|X i | = n i ,并且对于所有 i ≠ j,Xi∩Xj=0,则 |X1∪X2∪⋯∪Xs|=n1+n2+⋯+ns。

(e) General Sum Rule: If | X | = n , | Y | = k , and | X Y | = i , then | X Y | = n + k i .

(e)一般求和规则:如果 |X| = n,|Y| = k,且 |X∩Y|=i,则 |X∪Y|=n+k−i。

(f) Product Rule: If | X | = n and | Y | = k , then | X × Y | = nk .

(f)乘法法则:如果 |X| = n 且 |Y| = k,则 |X × Y| = nk。

(g) Product Rule for s Sets: If | X i | = n i for 1 ≤ i s , then | X 1 × X 2 × ⋯ × X s | = n 1 n 2 n s .

(g)s 集的乘积规则:如果 |X i | = n i 对于 1 ≤ i ≤ s,则 |X 1 × X 2 × ⋯ × X s | = n 1 n 2 … n s

(h) Power Set Rule: If | X | = n , then | P ( X ) | = 2 n .

(h)幂集规则:如果 |X| = n,则 |P(X)|=2n。

(i) Finite Set Operations Preserve Finiteness: If X 1 , …, X s are all finite, then i = 1 s X i , i = 1 s X i , X 1 × ⋯ × X s , and P ( X 1 ) are also finite. Every subset of a finite set is finite.

(i) 有限集运算保持有限性:如果 X 1 , …, X s 都是有限的,那么 ⋃i=1sXi, ⋂i=1sXi, X 1 × ⋯ × X s 和 P(X1) 也是有限的。有限集的每个子集都是有限的。

The first two parts are examples of intuitively plausible properties that are not so straightforward to prove rigorously. We give proofs of these results in an optional subsection later. The remaining items are proved below. Throughout this chapter, we frequently use the following terminology. Sets X and Y are called disjoint iff X Y = 0 ; and a list of sets X 1 , …, X s is pairwise disjoint iff X i X j = 0 for all i j with 1 ≤ i , j s .

前两部分是一些直观上合理的性质的例子,但要严格证明它们并非易事。我们将在后面的可选小节中给出这些结果的证明。其余部分将在下文中证明。在本章中,我们将频繁使用以下术语。集合 X 和 Y 被称为不相交,当且仅当 X∩Y=0;集合列表 X 1 , …, X s 两两不相交,当且仅当对于所有 i ≠ j 且 1 ≤ i, j ≤ s,Xi∩Xj=0。

The Sum Rule

求和法则

Let us now prove the Sum Rule (part (c) of Theorem 7.3 ). Assume X and Y are disjoint sets with | X | = n and | Y | = k ; we must prove | X Y | = n + k . Our assumption means there exist bijections f : {1, 2, …, n } → X and g : {1, 2, …, k } → Y ; we need to build a bijection h : { 1 , 2 , , n + k } X Y . To do this, define h ( i ) = f ( i ) for 1 ≤ i n , and define h ( i ) = g ( i n ) for n + 1 ≤ i n + k (where i denotes an integer). Figure 7.2 illustrates the construction of h when n = 3 and k = 2. We prove h is a bijection by exhibiting a two-sided inverse h : X Y { 1 , 2 , , n + k } . Each z X Y is in exactly one of the sets X or Y , since X Y = 0 . If z X , define h ′( z ) = f −1 ( z ). If z Y , define h ′( z ) = n + g −1 ( z ). You can now prove that h and h ′ are well-defined functions such that h h = I d X Y and h h = I d { 1 , 2 , , n + k } . As an example of what needs to be checked, let us see that h ( h ′( z )) = z when z Y . By definition of h ′, we know h ′( z ) = n + g −1 ( z ). Since g −1 maps Y into the set {1, 2, …, k }, we see that n + 1 ≤ h ′( z ) ≤ n + k . Hence, applying the definition of h , we get h ( h ( z ) ) = g ( h ( z ) n ) = g ( g 1 ( z ) ) = z = I d X Y ( z ) = z , as needed.

现在我们来证明求和法则(定理 7.3 的 (c) 部分)。假设 X 和 Y 是不相交的集合,且 |X| = n 且 |Y| = k;我们需要证明 |X∪Y|=n+k。我们的假设意味着存在双射 f : {1, 2, …, n} → X 和 g : {1, 2, …, k} → Y;我们需要构造一个双射 h:{1,2,…,n+k}→X∪Y。为此,定义 h(i) = f(i),其中 1 ≤ i ≤ n;定义 h(i) = g(i − n),其中 n + 1 ≤ i ≤ n + k(其中 i 表示整数)。图 7.2 展示了当 n = 3 且 k = 2 时 h 的构造。我们通过展示一个双侧逆函数 h′:X∪Y→{1,2,…,n+k} 来证明 h 是一个双射。每个 z∈X∪Y 都恰好属于集合 X 或 Y 中的一个,因为 X∩Y=0。如果 z ∈ X,则定义 h′(z) = f −1 (z)。如果 z ∈ Y,则定义 h′(z) = n + g −1 (z)。现在你可以证明 h 和 h′ 是定义良好的函数,使得 h∘h′=IdX∪Y 且 h′∘h=Id{1,2,…,n+k}。举例来说,我们来验证当 z ∈ Y 时 h(h′(z)) = z。根据 h′ 的定义,我们知道 h′(z) = n + g −1 (z)。由于 g −1 将 Y 映射到集合 {1, 2, …, k},因此 n + 1 ≤ h′(z) ≤ n + k。由此,应用 h 的定义,我们得到 h(h′(z))=g(h′(z)−n)=g(g−1(z))=z=IdX∪Y(z)=z,符合要求。

fig7_2

Figure 7.2

图 7.2

Building h from f and g .

由 f 和 g 构建 h。

We can now prove the Sum Rule for s Sets (part (d) of the theorem) by induction on s . The result is immediate if s = 1, and it holds by what we just proved if s = 2. For the induction step, fix s Z 2 , assume the Sum Rule is known for any union of s pairwise disjoint sets, and prove this rule for a union of s + 1 pairwise disjoint sets. Fix arbitrary sets X 1 , …, X s , X s +1 with | X i | = n i for all i and X i X j = 0 for all i j . The idea is to apply the Sum Rule already proved (part (c) of the theorem) to the sets X = X 1 X s and Y = X s +1 . The key observation is that

现在我们可以用归纳法证明 s 个集合的求和规则(定理的 (d) 部分)。当 s = 1 时,结果显然成立;当 s = 2 时,结果由我们刚才证明的结论得出。归纳步骤如下:固定 s∈Z≥2,假设对于任意 s 个两两不相交的集合的并集,求和规则已知,并证明对于 s + 1 个两两不相交的集合的并集,求和规则成立。固定任意集合 X 1 , …, X s , X s +1 ,使得对于所有 i,|X i | = n i ,并且对于所有 i ≠ j,Xi∩Xj=0。其思路是将已证明的求和法则(定理的 (c) 部分)应用于集合 X=X1∪⋯∪Xs 和 Y = X s +1 。关键观察结果是:

X X Y = ( i = 1 s s X X i ) X X s s + 1 = i = 1 s s ( X X i X X s s + 1 ) = i = 1 s s 0 = 0 .

So we can invoke the Sum Rule to conclude that | X Y | = | X | + | Y | . Now, since the sets X 1 , …, X s are pairwise disjoint, the induction hypothesis applies to show that | X | = n 1 + ⋯ + n s . Since | Y | = | X s +1 | = n s +1 , we conclude that | X Y | = ( n 1 + + n s ) + n s + 1 , completing the induction step.

因此,我们可以运用求和规则得出 |X∪Y|=|X|+|Y|。现在,由于集合 X 1 , …, X s 两两不相交,归纳假设适用,证明 |X| = n 1 + ⋯ + n s 。由于 |Y| = |X s +1 | = n s +1 ,我们得出 |X∪Y|=(n1+⋯+ns)+ns+1,归纳步骤完成。

Finally, we prove the General Sum Rule (part (e) of the theorem) Fix sets X and Y with | X | = n , | Y | = k , and | X Y | = i . On one hand, X = ( X Y ) ( X Y ) where the two sets X Y and X Y have empty intersection. By the Sum Rule already proved, | X | = | X Y | + | X Y | . On the other hand, X Y = ( X Y ) Y where the two sets X Y and Y have empty intersection. Using the Sum Rule again, we get | X Y | = | X Y | + | Y | . Inserting the previous equation into this one, we find that | X Y | = | X | + | Y | | X Y | = n + k i , as needed. This proof tacitly used the fact that X Y is a finite set (being a subset of the finite set X ; see part (b) of the theorem), so that | X Y | = j for some integer j . Similarly, all of our counting formulas make implicit use of the fact that the size of a finite set is well-defined (part (a) in Theorem 7.3 ).

最后,我们证明一般求和法则(定理的 (e) 部分)。固定集合 X 和 Y,使得 |X| = n,|Y| = k,且 |X∩Y| = i。一方面,X = (X − Y)∪(X∩Y),其中集合 X − Y 和 X∩Y 的交集为空。根据已证明的求和法则,|X| = |X − Y| + |X∩Y|。另一方面,X∪Y = (X − Y)∪Y,其中集合 X − Y 和 Y 的交集为空。再次利用求和法则,我们得到 |X∪Y| = |X − Y| + |Y|。将前一个等式代入此等式,我们发现 |X∪Y| = |X| + |Y| − |X∩Y| = n + k − i,符合要求。该证明隐含地利用了 X − Y 是一个有限集(是有限集 X 的子集;参见定理的 (b) 部分)这一事实,因此对于某个整数 j,|X − Y| = j。类似地,我们所有的计数公式都隐含地利用了有限集的大小是明确定义的这一事实(定理 7.3 的 (a) 部分)。

The Product Rule and Power Set Rule

乘积法则和幂集法则

Before proving the Product Rule (parts (f) and (g) of Theorem 7.3 ), we need a preliminary lemma. We prove that for any n -element set X and any object b , | X × { b }| = n . Fix X and b with | X | = n . We know there exists a bijection f : {1, 2, …, n } → X , and we must show there is a bijection g : {1, 2, …, n } → X × { b }. To do this, define a function h : X X × { b } by h ( x ) = ( x , b ) for all x X . We see at once that h is a bijection with two-sided inverse given by h −1 ( x , b ) = x for ( x , b ) ∈ X × { b }. We may now choose g = h f , which is a bijection since it is the composition of two bijections.

在证明乘法法则(定理 7.3 的 (f) 和 (g) 部分)之前,我们需要一个预备引理。我们证明对于任意 n 元集合 X 和任意对象 b,|X × {b}| = n。固定 X 和 b,使得 |X| = n。我们知道存在一个双射 f : {1, 2, …, n} → X,因此我们必须证明存在一个双射 g : {1, 2, …, n} → X × {b}。为此,定义函数 h : X → X × {b},使得对于所有 x ∈ X,h(x) = (x, b)。我们立即看出 h 是一个双射,其双侧逆为 h −1 (x, b) = x,其中 (x, b) ∈ X × {b}。现在我们可以选择 g = h ○ f,由于它是两个双射的复合,因此 g 也是一个双射。

Now we prove the Product Rule for two sets. Fix sets X and Y with | X | = n and | Y | = k . We know there is a bijection y :{1, 2, …, k } → Y . The idea is to slice up the product set X × Y into k pairwise disjoint sets X i = X × { y ( i )} for 1 ≤ i k ; see Figure 7.3 for an example where k = 3, X = {1, 2, 3, 4}, and Y = {1, 2, 4}. To see that the sets X i are pairwise disjoint, fix i j between 1 and k . To get a contradiction, assume there is an ordered pair ( a , b ) X i X j . Then y ( i ) = b = y ( j ), forcing i = j since y is one-to-one. This contradicts i j . Next, we claim X × Y = X 1 X 2 X k . On one hand, fix ( a , b ) ∈ X × Y . Then a X and b = y ( i ) for some i , because y is onto. So ( a , b ) ∈ X i for some i , hence ( a , b ) i = 1 k X i . Thus X × Y i = 1 k X i ; we let the reader check the reverse set inclusion. By the lemma, | X i | = | X | = n for 1 ≤ i k . Since X × Y is the union of the pairwise disjoint sets X 1 , …, X k , the Sum Rule gives

现在我们证明两个集合的乘法法则。固定集合 X 和 Y,使得 |X| = n 且 |Y| = k。我们知道存在一个双射 y:{1, 2, …, k} → Y。思路是将乘积集 X × Y 分割成 k 个两两不相交的集合 X i = X × {y(i)},其中 1 ≤ i ≤ k;参见图 7.3,其中 k = 3,X = {1, 2, 3, 4},Y = {1, 2, 4}。为了证明集合 X i 两两不相交,固定 i ≠ j,其中 i ≠ j 介于 1 和 k 之间。为了得到矛盾,假设存在一个有序对 (a,b)∈Xi∩Xj。那么 y(i) = b = y(j),由于 y 是一一映射,所以 i = j。这与 i ≠ j 矛盾。接下来,我们断言 X×Y=X1∪X2∪⋯∪Xk。一方面,固定 (a, b) ∈ X × Y。则 a ∈ X 且对于某个 i,b = y(i),因为 y 是满射。所以对于某个 i,(a, b) ∈ X i ,因此 (a,b)∈⋃i=1kXi。故 X×Y⊆⋃i=1kXi;我们让读者验证反向集合包含关系。根据引理,对于 1 ≤ i ≤ k,|X i | = |X| = n。由于 X × Y 是两两不相交的集合 X 1 , …, X k 的并集,根据求和规则可得

| X X × × Y | = | X X 1 | + + | X X k k | = n n + + n n ( k k summands 加数 ) = n n k k .

We can now prove the Product Rule for s Sets by induction on s , using the case just proved ( s = 2) to complete the induction step. This proof is very similar to what we did in the proof of the Sum Rule for s Sets, so we leave this as an exercise.

现在我们可以用归纳法证明 s 个集合的乘法法则,利用刚才证明的情况(s = 2)来完成归纳步骤。这个证明与我们证明 s 个集合的加法法则时非常相似,所以我们把它留作练习。

7.4 Example. For n Z > 0 , how many strings of zeroes and ones of length n are there? For example, when n = 3, there are eight such strings: 000, 001, 010, 011, 100, 101, 110, and 111. The map sending a binary string a 1 a 2 a n to the ordered n -tuple ( a 1 , a 2 , …, a n ) is a bijection from the set of such strings to the product set {0, 1} n . You can check that {0, 1} is a two-element set. By the Product Rule, |{0, 1} n | = 2 n . Thus, there exists a bijection f : {1, 2, …, 2 n } → {0, 1} n .

7.4 示例。对于 n∈Z>0,长度为 n 的全 0 和 1 字符串有多少个?例如,当 n = 3 时,有 8 个这样的字符串:000、001、010、011、100、101、110 和 111。将二进制字符串 a 1 a 2 … a n 映射到有序 n 元组 (a 1 , a 2 , …, a n ) 的映射是从这些字符串的集合到乘积集 {0, 1} n 的双射。你可以验证 {0, 1} 是一个包含两个元素的集合。根据乘积法则,|{0, 1} n | = 2 n 。因此,存在一个双射 f : {1, 2, …, 2 n → {0, 1} n

fig7_3

Figure 7.3

图 7.3

Decomposing X × Y into disjoint sets X 1 , …, X k .

将 X × Y 分解为不相交的集合 X 1 , …, X k

Aided by the preceding example, we can now prove the Power Set Rule. Fix an n -element set X . We can choose a bijection x : {1, 2, …, n } → X so that X = { x (1), x (2), …, x ( n )}. Define a function g : { 0 , 1 } n P ( X ) as follows. Given ( a 1 , …, a n ) ∈ {0, 1} n , let S = g ( a 1 , …, a n ) be the set of all x ( i ) such that a i = 1. In other words, if a i = 0, then x ( i ) is not a member of S , and if a i = 1, then x ( i ) is a member of S . Next define g : P ( X ) { 0 , 1 } n as follows. Given T X , let g ′( T ) = ( b 1 , …, b n ), where b i = 1 if x ( i ) ∈ T , and b i = 0 if x ( i ) T . We ask the reader to check that g and g ′ are well-defined functions such that g g ′ and g ′ ○ g are identity maps. So g is a bijection. Composing g with the bijection f from the example, we get a bijection from {1, 2, …, 2 n } onto P ( X ) . Thus, | P ( X ) | = 2 n as needed.

借助前面的例子,我们现在可以证明幂集法则。固定一个包含 n 个元素的集合 X。我们可以选择一个双射 x : {1, 2, …, n} → X,使得 X = {x(1), x(2), …, x(n)}。定义函数 g:{0,1}n→P(X) 如下。给定 (a 1 , …, a n ) ∈ {0, 1} n ,令 S = g(a 1 , …, a n ) 为所有满足 a i = 1 的 x(i) 的集合。换句话说,如果 a i = 0,则 x(i) 不是 S 的成员;如果 a i = 1,则 x(i) 是 S 的成员。接下来定义 g′:P(X)→{0,1}n 如下。给定 T⊆ X,令 g′(T) = (b 1 , …, b n ),其中当 x(i) ∈ T 时,b i = 1;当 x(i)∉ T 时,b i = 0。请读者验证 g 和 g′ 是定义良好的函数,且 g ○ g′ 和 g′ ○ g 是恒等映射。因此,g 是一个双射。将 g 与示例中的双射 f 复合,我们得到从 {1, 2, …, 2 n 到 P(X) 的双射。因此,|P(X)|=2n,符合要求。

To illustrate this proof, suppose n = 3, X = { a , b , c }, and x (1) = a , x (2) = b , x (3) = c . We find that g (1, 0, 1) = { x (1), x (3)} = { a , c }, g (0, 1, 0) = { x (2)} = { b }, and g ( 0 , 0 , 0 ) = { } = 0 . Going the other way, g ′({ b , c }) = g ′({ x (2), x (3)}) = (0, 1, 1), g ′({ a }) = g ′({ x (1)}) = (1, 0, 0), and so on.

为了说明这个证明,假设 n = 3,X = {a, b, c},且 x(1) = a,x(2) = b,x(3) = c。我们发现 g(1, 0, 1) = {x(1), x(3)} = {a, c},g(0, 1, 0) = {x(2)} = {b},且 g(0, 0, 0) = {} = 0。反过来,g′({b, c}) = g′({x(2), x(3)}) = (0, 1, 1),g′({a}) = g′({x(1)}) = (1, 0, 0),依此类推。

Preservation of Finiteness

有限性的保持

Finally, we use the preceding results to prove part (i) of Theorem 7.3 . Informally, this item says that doing finitely many set operations on finite sets always produces another finite set. Fix s Z > 0 and finite sets X 1 , …, X s with | X i | = n i for each i . By the Product Rule, | X 1 × ⋯ × X s | = n 1 n 2 n s , so X 1 × ⋯ × X s is finite. By the Power Set Rule, | P ( X 1 ) | = 2 n 1 , so P ( X 1 ) is finite. Item 2 of the theorem immediately implies that any subset of any finite set is finite. Since i = 1 s X i is a subset of the finite set X 1 , this intersection is a finite set.

最后,我们利用前面的结果证明定理 7.3 的第 (i) 部分。通俗地说,这部分指出,对有限集进行有限次集合运算总是会得到另一个有限集。固定 s∈Z>0,并设有限集 X 1 , …, X s ,其中对于每个 i,|X i | = n i 。根据乘法法则,|X 1 × ⋯ × X s | = n 1 n 2 … n s ,因此 X 1 × ⋯ × X s 是有限的。根据幂集规则,|P(X1)|=2n1,因此 P(X1) 是有限的。定理第 2 条直接表明,任何有限集的子集也是有限的。由于 ⋂i=1sXi 是有限集 X 1 的子集,因此该交集也是有限集。

Proving finiteness of the union of the X i is a bit more complicated, since the Sum Rule for s Sets required that the given sets be pairwise disjoint. Here we use induction on s . If s = 1, i = 1 s X i = X 1 , and X 1 was assumed to be finite. Now fix s ≥ 1, assume that the union of any s finite sets is finite, and prove that the union of any s + 1 finite sets is finite. Fix finite sets X 1 , …, X s , X s +1 . By the induction hypothesis, X = X 1 X s is finite, say | X | = n . By assumption, X s +1 is also finite, say | X s +1 | = k . The intersection X X s + 1 is finite, being a subset of the finite set X s +1 ; say | X X s + 1 | = i . Now the General Sum Rule (part (e) in the theorem) tells us that | X X s + 1 | = n + k i , so i = 1 s + 1 X i = X X s + 1 is finite.

证明集合 X i 的并集的有限性稍微复杂一些,因为 s 个集合的求和规则要求给定的集合两两不相交。这里我们对 s 进行归纳。如果 s = 1,则 ⋃i=1sXi=X1,并且假设 X 1 是有限的。现在固定 s ≥ 1,假设任意 s 个有限集合的并集是有限的,并证明任意 s + 1 个有限集合的并集也是有限的。固定有限集合 X 1 , …, X s , X s +1 。根据归纳假设,X=X1∪⋯∪Xs 是有限的,记为 |X| = n。根据假设,X s +1 也是有限的,例如 |X s +1 | = k。交集 X∩Xs+1 也是有限的,它是有限集 X s +1 的子集;例如 |X∩Xs+1|=i。现在,一般求和规则(定理中的 (e) 部分)告诉我们 |X∪Xs+1|=n+k−i,因此 ⋃i=1s+1Xi=X∪Xs+1 是有限的。

Two Technical Proofs (Optional)

两份技术证明(可选)

We now prove part (a) of the Theorem on Finite Sets, which says that the size of a finite set is unique. We use induction on n to prove this statement:

现在我们证明有限集定理的(a)部分,即有限集的大小是唯一的。我们使用对n的归纳法来证明这个结论:

n n , m Z Z 0 , X X , ( | X X | = n n | X X | = m ) n n = m .

We prove the base case ( n = 0) by contradiction. Assume, to get a contradiction, that there is a set X and there exists m Z 0 with | X | = 0 and | X | = m and 0 ≠ m . On one hand, | X | = 0 means X = 0 by definition. On the other hand, | X | = m > 0 means there is a bijection f : {1, 2, …, m } → X . Then f (1) ∈ X contradicts X = 0 .

我们用反证法证明基本情况 (n = 0)。假设存在一个集合 X,且存在 m∈Z≥0,使得 |X| = 0 且 |X| = m 且 0 ≠ m。一方面,|X| = 0 根据定义意味着 X=0。另一方面,|X| = m > 0 意味着存在一个双射 f : {1, 2, …, m} → X。那么 f(1) ∈ X 与 X=0 矛盾。

For the induction step, fix n Z 0 , and assume m Z 0 , X , ( | X | = n | X | = m ) n = m . We must prove p Z 0 , Y , ( | Y | = n + 1 | Y | = p n + 1 = p . Fix p Z 0 , fix a set Y , assume | Y | = n + 1 and | Y | = p , and prove n + 1 = p . Since n + 1 > 0, we cannot have p = 0; otherwise, the argument in the previous paragraph leads to a contradiction. Thus, we have p = m + 1 for some m Z 0 . By assumption on Y , there are bijections f : {1, 2, …, n + 1} → Y and g : {1, 2, …, m + 1} → Y . Let y 0 = f ( n + 1) ∈ Y and z 0 = g ( m + 1) ∈ Y . Suppose first that z 0 = y 0 . You can check that the restrictions f ′ = f | {1,2,…, n } and g ′ = g | {1,2,…, m } are both bijections onto the set X = Y − { y 0 }. Then | X | = n and | X | = m , so the induction hypothesis applies to show that n = m , and hence n + 1 = m + 1 = p as needed.

对于归纳步骤,固定 n∈Z≥0,并假设 ∀m∈Z≥0,∀X,(|X|=n∧|X|=m)⇒n=m。我们必须证明对于任意 p∈Z≥0,对于任意 Y,(|Y|=n+1∧|Y|=p⇒n+1=p)。固定 p∈Z≥0,固定一个集合 Y,假设 |Y| = n + 1 且 |Y| = p,并证明 n + 1 = p。由于 n + 1 > 0,我们不能有 p = 0;否则,上一段的论证将导致矛盾。因此,对于某个 m∈Z≥0,我们有 p = m + 1。根据对 Y 的假设,存在双射 f : {1, 2, …, n + 1} → Y 和 g : {1, 2, …, m + 1} → Y。令 y 0 = f(n + 1) ∈ Y 且 z 0 = g(m + 1) ∈ Y。首先假设 z 0 = y 0 。你可以验证限制 f′ = f| {1,2,…, n } 和 g′ = g| {1,2,…, m } 都是到集合 X = Y − {y 0 的双射。那么 |X| = n 且 |X| = m,因此归纳假设适用,可以证明 n = m,进而证明 n + 1 = m + 1 = p,满足要求。

If z 0 y 0 , we reduce to the case just considered as follows. Define h : Y Y by setting h ( z 0 ) = y 0 , h ( y 0 ) = z 0 , and h ( y ) = y for all y Y − { y 0 , z 0 }. You can verify that h h = I d Y , so h is a bijection with h −1 = h . The composition h g is a bijection from {1, 2, …, m + 1} to Y such that ( h g )( m + 1) = h ( g ( m + 1)) = h ( z 0 ) = y 0 . Now we finish as before: the restriction of f to {1, 2, …, n } and the restriction of h g to {1, 2, …, m } are bijections from their respective domains onto X = Y − { y 0 }, so | X | = n and | X | = m , so n = m by induction hypothesis, so n + 1 = m + 1 = p . This completes the proof of part (a).

如果 z 0 ≠ y 0 ,则我们简化为刚才讨论的情况,如下所示。定义 h : Y → Y,令 h(z 0 ) = y 0 ,h(y 0 ) = z 0 ,并且对于所有 y ∈ Y − {y 0 , z 0 ,h(y) = y。您可以验证 h∘h=IdY,因此 h 是一个双射,且 h −1 = h。复合函数 h ○ g 是从 {1, 2, …, m + 1} 到 Y 的双射,使得 (h ○ g)(m + 1) = h(g(m + 1)) = h(z 0 ) = y 0 )。现在我们像之前一样完成证明:f 在 {1, 2, …, n} 上的限制以及 h ○ g 在 {1, 2, …, m} 上的限制都是从它们各自的定义域到 X = Y − {y 0 的双射,因此 |X| = n 且 |X| = m,根据归纳假设,n = m,所以 n + 1 = m + 1 = p。这就完成了 (a) 部分的证明。

We now use induction on n to prove part (b) of the Theorem on Finite Sets. We must show: n Z 0 , X , Y , ( | X | = n Y X ) ( m Z 0 , | Y | = m m n ) . For the base case, take n = 0, fix sets X and Y , and assume | X | = 0 and Y X . By definition, | X | = 0 means X = 0 . We assumed Y 0 , and we know 0 Y , so Y = 0 . Thus Y = 0 and | Y | = 0. So we may choose m = 0 in the statement to be proved.

现在我们用归纳法证明有限集定理的(b)部分。我们需要证明:∀n∈Z≥0,∀X,∀Y,(|X|=n∧Y⊆X)⇒(∃m∈Z≥0,|Y|=m∧m≤n)。对于基本情况,取n=0,固定集合X和Y,并假设|X|=0且Y⊆X。根据定义,|X|=0意味着X=0。我们假设Y⊆0,并且我们知道0⊆Y,所以Y=0。因此Y=0且|Y|=0。所以我们可以在待证明的命题中选择m=0。

For the induction step, fix n Z 0 , and assume

对于归纳步骤,固定 n∈Z≥0,并假设

X X , Y , ( | X X | = n n Y X X ) ( m Z Z 0 , | Y | = m m n n ) .

We must prove

我们必须证明

X X , Y , ( | X X | = n n + 1 Y X X ) ( p p Z Z 0 , | Y | = p p p p n n + 1 ) .

Fix sets X and Y , assume | X | = n + 1 and Y X , and prove p Z 0 , | Y | = p n + 1 . Choose a bijection f : {1, 2, …, n + 1} → X , define x 0 = f ( n + 1) ∈ X , and note f ′ = f | {1,2,…, n } is a bijection from {1, 2, …, n } onto X ′ = X − { x 0 }. Thus, | X ′| = n . We know x 0 Y or x 0 Y , so consider two cases. Case 1. Assume x 0 Y . Since we also assumed Y X , we see that Y X ′ in this case. We can apply the induction hypothesis taking Y ′ = Y to see that for some nonnegative integer m n , | Y | = m . Choosing p = m , we have | Y | = p n + 1 as needed. Case 2. Assume x 0 Y . In this case, we take Y ′ = Y − { x 0 } in the induction hypothesis, which is allowed since Y ′ is a subset of X ′. This leads to the existence of m n with | Y ′| = m , which means there is a bijection g : {1, 2, …, m } → Y ′. (If m = 0, g : 0 0 is the empty function.) Define a function h : {1, 2, …, m + 1} → Y by setting h ( i ) = g ( i ) for 1 ≤ i m , and h ( m + 1) = x 0 . We ask the reader to check that h is a bijection. Thus, | Y | = m + 1 ≤ n + 1, so that the required existence statement holds with p = m + 1. This completes the proof of part (b).

固定集合 X 和 Y,假设 |X| = n + 1 且 Y⊆ X,并证明存在 p∈Z≥0,|Y|=p≤n+1。选取一个双射 f : {1, 2, …, n + 1} → X,定义 x 0 = f(n + 1) ∈ X,并注意到 f′ = f| {1,2,…, n } 是从 {1, 2, …, n} 到 X′ = X − {x 0 的双射。因此,|X′| = n。我们知道 x0∉Y 或 x 0 ∈ Y,因此考虑两种情况。情况 1:假设 x0∉Y。由于我们也假设 Y⊆ X,因此在这种情况下 Y⊆ X′。我们可以应用归纳假设,令 Y′ = Y,从而证明对于某个非负整数 m ≤ n,|Y| = m。选择 p = m,则有 |Y| = p ≤ n + 1,符合要求。情况 2:假设 x 0 ∈ Y。在这种情况下,我们在归纳假设中取 Y′ = Y − {x 0 ,这是允许的,因为 Y′ 是 X′ 的子集。这导致存在 m ≤ n 使得 |Y′| = m,这意味着存在双射 g : {1, 2, …, m} → Y′。(如果 m = 0,则 g:0→0 为空函数。)定义函数 h : {1, 2, …, m + 1} → Y,令 h(i) = g(i),其中 1 ≤ i ≤ m,且 h(m + 1) = x 0 。我们请读者验证 h 是一个双射。因此,|Y| = m + 1 ≤ n + 1,所以当 p = m + 1 时,所需的存在性命题成立。这就完成了 (b) 部分的证明。

Section Summary

章节概要

  1. Finite Sets. A set X has n elements iff there is a bijection f : {1, 2, …, n } → X , in which case we write | X | = n . A set X is finite iff n Z 0 , | X | = n . The size n is uniquely determined by X . A set X is infinite iff X is not finite.
    有限集。集合 X 有 n 个元素当且仅当存在双射 f : {1, 2, …, n} → X,此时记为 |X| = n。集合 X 是有限的当且仅当存在 n∈Z≥0,|X|=n。n 的大小由 X 唯一确定。集合 X 是无限的当且仅当 X 不是有限的。
  2. Sum Rule. Given a finite list of pairwise disjoint finite sets X 1 , …, X s ,
    | X 1 X 2 X s | = | X 1 | + | X 2 | + + | X s | .

    Given any two finite sets X and Y , | X Y | = | X | + | Y | | X Y | .


    求和规则。给定一个两两不相交的有限集合列表 X 1 , …, X s ,|X1∪X2∪⋯∪Xs|=|X1|+|X2|+⋯+|Xs|。给定任意两个有限集合 X 和 Y,|X∪Y|=|X|+|Y|−|X∩Y|。
  3. Product Rule. Given a finite list of finite sets X 1 , …, X s ,
    | X 1 × X 2 × × X s | = i = 1 s | X i | .

    乘积规则。给定有限集合 X 1 , …, X s , |X1×X2×⋯×Xs|=∏i=1s|Xi|。
  4. Power Set Rule. Given an n -element set X , | P ( X ) | = 2 n .
    幂集规则。给定一个 n 元素集合 X,|P(X)|=2n。
  5. Preservation of Finiteness. The union, intersection, and product of finitely many finite sets are finite sets. Each subset of a finite set is finite. The power set of a finite set is finite.
    有限性守恒。有限多个有限集的并集、交集和积仍然是有限集。有限集的每个子集都是有限的。有限集的幂集也是有限的。

Exercises

练习

  1. (a) Prove: for all n Z > 0 , |{1, 2, …, n }| = n . (b) Prove: for all n Z > 0 , {0, 1, …, n − 1} has n elements. (c) Prove: for all n Z > 0 , { − 1, − 2, …, − n } has size n .
    (a)证明:对于所有 n∈Z>0,|{1, 2, …, n}| = n。(b)证明:对于所有 n∈Z>0,{0, 1, …, n − 1} 有 n 个元素。(c)证明:对于所有 n∈Z>0,{ − 1, − 2, …, − n} 的大小为 n。
  2. (a) Prove that X = {1, 3, 5, 7, 9} has 5 elements. (b) Prove that Y = {2, 4, 6, 8} has 4 elements. (c) Use (a), (b), and the proof of the Sum Rule to build a bijection h : { 1 , 2 , , 9 } X Y .
    (a) 证明 X = {1, 3, 5, 7, 9} 有 5 个元素。(b) 证明 Y = {2, 4, 6, 8} 有 4 个元素。(c) 利用 (a)、(b) 和求和法则的证明,构造一个双射 h:{1,2,…,9}→X∪Y。
  3. (a) Prove: for all sets X and Y , if X is finite, then X Y is finite. (b) Prove: for all sets X and Y , if X Y and Y are finite, then X is finite.
    (a)证明:对于任意集合 X 和 Y,如果 X 是有限的,则 X − Y 也是有限的。(b)证明:对于任意集合 X 和 Y,如果 X − Y 和 Y 都是有限的,则 X 也是有限的。
  4. Prove: for all sets X and Y , if Y X and Y is infinite, then X is infinite.
    证明:对于任意集合 X 和 Y,如果 Y⊆ X 且 Y 是无穷大,则 X 也是无穷大。
  5. Verify these details in the proof of the Sum Rule: (a) h is a well-defined function mapping {1, 2, …, n + k } into X Y . (b) h ′ is a well-defined function mapping X Y into {1, 2, …, n + k }. (c) h h = I d X Y . (d) h h = I d { 1 , 2 , , n + k } .
    在求和法则的证明中验证以下细节:(a) h 是一个定义良好的函数,将 {1, 2, …, n + k} 映射到 X∪Y。(b) h′ 是一个定义良好的函数,将 X∪Y 映射到 {1, 2, …, n + k}。(c) h∘h′=IdX∪Y。(d) h′∘h=Id{1,2,…,n+k}。
  6. Prove the set inclusion i = 1 k X i X × Y in the proof of the Product Rule.
    在乘积法则的证明中证明集合包含关系⋃i=1kXi⊆X×Y。
  7. Prove the general Product Rule (part (g) in the Theorem on Finite Sets) by induction on s .
    利用归纳法证明一般乘积法则(有限集定理中的 (g) 部分)。
  8. Finish the proof of the Power Set Rule by checking that g and g ′ are well-defined functions such that g g ′ and g ′ ○ g are identity maps.
    通过验证 g 和 g′ 是定义良好的函数,使得 g ○ g′ 和 g′ ○ g 是恒等映射,从而完成幂集规则的证明。
  9. (a) Draw the arrow diagram for a bijection f : {1, 2, …, 8} → {0, 1} 3 . (b) Use (a) and the proof of the Power Set Rule to draw an arrow diagram for a bijection from {1, 2, …, 8} onto P ( { a , b , c } ) .
    (a)画出双射 f : {1, 2, …, 8} → {0, 1} 3 的箭头图。(b)利用(a)和幂集规则的证明,画出从 {1, 2, …, 8} 到 P({a,b,c}) 的双射的箭头图。
  10. Prove carefully that the restriction of a bijection f : {1, 2, …, n + 1} → Y to {1, 2, …, n } is a bijection from {1, 2, …, n } onto Y − { f ( n + 1)}.
    仔细证明双射 f : {1, 2, …, n + 1} → Y 到 {1, 2, …, n} 的限制是从 {1, 2, …, n} 到 Y − {f(n + 1)} 的双射。
  11. Prove: for all finite sets X , Y , Z ,
    | X Y Z | = | X | + | Y | + | Z | | X Y | | X Z | | Y Z | + | X Y Z | .

    证明:对于所有有限集合 X、Y、Z,|X∪Y∪Z|=|X|+|Y|+|Z|−|X∩Y|−|X∩Z|−|Y∩Z|+|X∩Y∩Z|。
  12. Given c , d Z > 0 , define f : {0, 1, …, c − 1} × {0, 1, …, d − 1} → {0, 1, …, cd − 1} by f ( i , j ) = di + j . Aided by the Division Theorem, prove that f is injective and surjective.
    给定 c,d∈Z>0,定义 f : {0, 1, …, c − 1} × {0, 1, …, d − 1} → {0, 1, …, cd − 1},使得 f(i, j) = di + j。利用除法定理,证明 f 是单射且满射。
  13. Use the previous problem to give a new proof of part (f) in the Theorem on Finite Sets.
    利用前面的问题,给出有限集定理中 (f) 部分的新证明。
  14. Prove carefully that Z > 0 is infinite. ( Suggestion: One approach is to show by induction on n that for all n Z 0 , Z > 0 does not have size n .)
    仔细证明 Z>0 是无穷大。(提示:一种方法是对 n 进行归纳,证明对于所有 n∈Z≥0,Z>0 的大小不为 n。)
  15. Prove: for all sets X and Y and all n Z 0 , if | Y | = n and there exists an injection g : X Y , then | X | = m for some m n .
    证明:对于所有集合 X 和 Y 以及所有 n∈Z≥0,如果 |Y| = n 且存在单射 g : X → Y,则对于某些 m ≤ n,|X| = m。
  16. Prove: for all sets X and Y and all n Z 0 , if | X | = n and there exists a surjection f : X Y , then | Y | = m for some m n .
    证明:对于所有集合 X 和 Y 以及所有 n∈Z≥0,如果 |X| = n 且存在满射 f : X → Y,则 |Y| = m,其中 m ≤ n。
  17. Prove: for all sets X and Y , if X is finite and Y X and | X | = | Y |, then X = Y .
    证明:对于任意集合 X 和 Y,如果 X 是有限的,Y⊆ X 且 |X| = |Y|,则 X = Y。
  18. Prove: for all finite sets X and all f : X X , f is injective iff f is surjective.
    证明:对于所有有限集 X 和所有 f : X → X,f 是单射当且仅当 f 是满射。
  19. (a) Prove that every nonempty finite totally ordered set has a least element. (b) Prove that every finite totally ordered set is well-ordered.
    (a)证明每个非空有限全序集都有最小元素。(b)证明每个有限全序集都是良序的。

7.2 Countably Infinite Sets

7.2 可数无限集

Now that we understand finite sets, the next step is to examine countably infinite sets, which are the smallest infinite sets. Informally, a set X is countably infinite iff it is possible to list all the elements of X (with no repetitions) in a sequence a 1 , a 2 , …, a n , …, where the terms of the sequence are indexed by positive integers. Formally, this means that there exists a bijection from Z > 0 onto X . We will see that the following sets are all countably infinite: Z > 0 , Z 0 , Z , Q , Z k , and Q k (where k is any positive integer). More generally, the product of finitely many countably infinite sets is still countably infinite.

现在我们理解了有限集,下一步是考察可数无限集,即最小的无限集。通俗地说,集合 X 是可数无限集当且仅当可以列出 X 的所有元素(不重复)成一个序列 a 1 , a 2 , …, a n , …,其中序列的项由正整数索引。形式上,这意味着存在从 Z>0 到 X 的双射。我们将看到以下集合都是可数无限集:Z>0、Z≥0、Z、Q、Zk 和 Qk(其中 k 为任意正整数)。更一般地,有限多个可数无限集的乘积仍然是可数无限集。

Cardinality Definitions

基数定义

The following definition allows us to compare the size (cardinality) of two arbitrary sets.

以下定义允许我们比较两个任意集合的大小(基数)。

7.5. Definition: Cardinality. For two sets A and B :

7.5. 定义:基数。对于两个集合 A 和 B:

(a) | A | = | B | means there exists a bijection f : A B .

(a)|A| = |B| 表示存在双射 f : A → B。

(b) | A | ≤ | B | means there exists an injective function g : A B .

(b)|A| ≤ |B| 表示存在单射函数 g : A → B。

When A is finite with n elements, the symbol | A | denotes the nonnegative integer n . For an infinite set A , we do not attempt to assign a meaning to the symbol | A | by itself; only the longer expressions “| A | = | B |” and “| A | ≤ | B |” have been defined here. We now show that these concepts obey some expected rules of equality and order.

当集合 A 为包含 n 个元素的有限集合时,符号 |A| 表示非负整数 n。对于无限集合 A,我们不尝试单独赋予符号 |A| 特定的含义;这里仅定义了更长的表达式“|A| = |B|”和“|A| ≤ |B|”。现在,我们将证明这些概念遵循一些预期的相等性和顺序规则。

7.6 Theorem on Cardinal Equality and Order. For all sets A , B , C , A ′, and B ′ :

7.6 关于基数相等和序的定理。对于所有集合 A、B、C、A′ 和 B′:

(a) Reflexivity: | A | = | A |.

(a)自反性:|A| = |A|。

(b) Symmetry: If | A | = | B |, then | B | = | A |.

(b)对称性:如果 |A| = |B|,则 |B| = |A|。

(c) Transitivity: If | A | = | B | and | B | = | C |, then | A | = | C |.

(c)传递性:如果 |A| = |B| 且 |B| = |C|,则 |A| = |C|。

(d) Reflexivity of Order: | A | ≤ | A |.

(d)顺序的自反性:|A| ≤ |A|。

(e) Antisymmetry of Order: If | A | ≤ | B | and | B | ≤ | A |, then | A | = | B |.

(e)序反对称性:如果 |A| ≤ |B| 且 |B| ≤ |A|,则 |A| = |B|。

(f) Transitivity of Order: If | A | ≤ | B | and | B | ≤ | C |, then | A | ≤ | C |.

(f)顺序传递性:如果 |A| ≤ |B| 且 |B| ≤ |C|,则 |A| ≤ |C|。

(g) Disjoint Union Property: If | A | = | A ′| and | B | = | B ′| and A B = 0 = A B , then | A B | = | A B | .

(g)不相交并集性质:如果 |A| = |A′| 且 |B| = |B′| 且 A∩B=0=A′∩B′,则 |A∪B|=|A′∪B′|。

(h) Product Property: If | A | = | A ′| and | B | = | B ′|, then | A × B | = | A ′ × B ′|.

(h)乘积性质:如果 |A| = |A′| 且 |B| = |B′|,则 |A × B| = |A′ × B′|。

(i) Power Set Property: If | A | = | A ′|, then | P ( A ) | = | P ( A ) | .

(i)幂集性质:如果 |A| = |A′|,则 |P(A)|=|P(A′)|。

Proof. Fix sets A , B , C , A ′, and B ′. To prove | A | = | A |, we must find a bijection from A to A . The identity map I d A : A A is such a bijection. For part (b), assume | A | = | B | and prove | B | = | A |. We assumed there is a bijection f : A B . Then f −1 : B A is a bijection, so | B | = | A |. For part (c), assume | A | = | B | and | B | = | C |, so there exist bijections f : A B and g : B C . The composition g f is a bijection from A to C , so | A | = | C |. Parts (d) and (f) are proved similarly, replacing bijections with injections. But part (e) is much more difficult to prove. This is a famous result called the Schröder–Bernstein Theorem , which we prove in §7.3 .

证明。固定集合 A、B、C、A′ 和 B′。为了证明 |A| = |A|,我们必须找到一个从 A 到 A 的双射。恒等映射 IdA:A→A 就是这样一个双射。对于 (b) 部分,假设 |A| = |B| 并证明 |B| = |A|。我们假设存在一个双射 f : A → B。那么 f −1 : B → A 是一个双射,所以 |B| = |A|。对于 (c) 部分,假设 |A| = |B| 且 |B| = |C|,所以存在双射 f : A → B 和 g : B → C。复合映射 g ○ f 是一个从 A 到 C 的双射,所以 |A| = |C|。(d) 和 (f) 部分的证明类似,只需将双射替换为单射即可。但 (e) 部分的证明要困难得多。这是一个著名的结果,称为施罗德-伯恩斯坦定理,我们将在第 7.3 节中证明它。

For the remaining properties, assume | A | = | A ′| and | B | = | B ′|, which means there are bijections f : A A ′ and g : B B ′. To prove part (h), define h : A × B A ′ × B ′ by h ( a , b ) = ( f ( a ), g ( b )) for ( a , b ) ∈ A × B , and define h ′ : A ′ × B ′ → A × B by h ′( a ′, b ′) = ( f −1 ( a ′), g −1 ( b ′)) for ( a ′, b ′) ∈ A ′ × B ′. It is routine to check that h h = I d A × B and h h = I d A × B , so h and h ′ are bijections. Thus, | A × B | = | A ′ × B ′| as needed. To prove part (g), assume A B = 0 = A B . Define p : A B A B by letting p ( x ) = f ( x ) if x A , and p ( x ) = g ( x ) if x B . Then p is a bijection because it has a two-sided inverse p : A B A B defined by letting p ′( y ) = f −1 ( y ) if y A ′, and p ′( y ) = g −1 ( y ) if y B ′. You are asked to prove part (i) as an exercise.     □

对于其余性质,假设 |A| = |A′| 且 |B| = |B′|,这意味着存在双射 f : A → A′ 和 g : B → B′。为了证明 (h) 部分,定义 h : A × B → A′ × B′,使得 h(a, b) = (f(a), g(b)),其中 (a, b) ∈ A × B;定义 h′ : A′ × B′ → A × B,使得 h′(a′, b′) = (f(a′), g(b′)),其中 (a′, b′) ∈ A′ × B′。很容易验证 h′∘h=IdA×B 且 h∘h′=IdA′×B′,因此 h 和 h′ 是双射。由此,|A × B| = |A′ × B′|,符合要求。为了证明 (g) 部分,假设 A∩B=0=A′∩B′。定义 p:A∪B→A′∪B′,令 p(x) = f(x) 当 x ∈ A 时,p(x) = g(x) 当 x ∈ B 时。那么 p 是一个双射,因为它有一个双侧逆映射 p′:A′∪B′→A∪B′,定义为 p′(y) = f(y) 当 y ∈ A′ 时,p′(y) = g(y) 当 y ∈ B′ 时。请将证明 (i) 部分作为练习。□

Countably Infinite Sets

可数无限集

In Exercise (14) of §7.1 , you were asked to prove that Z > 0 is an infinite set. The next definition introduces countably infinite sets, which are sets with the same cardinality as Z > 0 .

在第 7.1 节的练习 (14) 中,你被要求证明 Z>0 是一个无限集。下一个定义引入了可数无限集,它是与 Z>0 具有相同基数的集合。

7.7. Definition: Countably Infinite Sets. For all sets X : X is countably infinite iff there exists a bijection f : Z > 0 X .

7.7. 定义:可数无限集。对于所有集合 X: X 是可数无限集当且仅当存在双射 f:Z>0→X。

Note that X is countably infinite iff | Z > 0 | = | X | . For all sets X and Y , if | X | = | Y | then X is countably infinite iff Y is countably infinite. This follows from symmetry and transitivity of cardinal equality. The next few propositions provide specific examples of countably infinite sets.

注意,X 是可数无限集当且仅当 |Z>0|=|X|。对于任意集合 X 和 Y,如果 |X| = |Y|,则 X 是可数无限集当且仅当 Y 是可数无限集。这由基数相等的对称性和传递性得出。接下来的几个命题给出了可数无限集的具体例子。

7.8 Proposition. For all a Z , Z > a and Z a and Z < a and Z a are countably infinite sets.

7.8 命题。对于所有 a∈Z,Z>a 和 Z≥a 以及 Z<a 和 Z≤a 都是可数无限集。

Proof. We prove the statement about Z a as an example. Fix a Z . Writing Z > 0 = Z 1 , we need to construct a bijection f : Z 1 Z a . Define f ( n ) = n + a − 1 for n Z 1 . Also define g : Z a Z 1 by g ( m ) = m − ( a − 1) for m Z a . You can check that f and g map into their claimed codomains, and g is a two-sided inverse of f . Hence, f is the required bijection.      □

证明。我们以 Z≥a 为例来证明这个命题。固定 a∈Z。令 Z>0=Z≥1,我们需要构造一个双射 f:Z≥1→Z≥a。定义 f(n) = n + a − 1,其中 n∈Z≥1。同时定义 g:Z≥a→Z≥1,其中 g(m) = m − (a − 1),其中 m∈Z≥a。你可以验证 f 和 g 都映射到它们各自的值域,并且 g 是 f 的双侧逆。因此,f 就是所需的双射。□

7.9 Proposition. Z is countably infinite.      □

7.9 命题。Z 是可数无限的。□

Proof . We build a bijection f : Z > 0 Z as follows. Each n Z > 0 is either even or odd, but not both. If n is even, define f ( n ) = n /2, which is in the codomain Z . If n is odd, define f ( n ) = −( n − 1)/2, which is in the codomain Z . To see that f is a bijection, we describe its inverse g : Z Z > 0 . For m Z , let g ( m ) = 2 m if m > 0, and let g ( m ) = −2 m + 1 if m ≤ 0. The arrow diagram in Figure 7.4 illustrates the action of g . You can prove that g maps into Z > 0 and is the two-sided inverse for f . We check one case as an example. Given m Z with m ≤ 0, we must show f ( g ( m )) = m . First, g ( m ) = −2 m + 1, which is odd. So f ( g ( m )) = −( − 2 m + 1 − 1)/2 = 2 m /2 = m , as needed.

证明。我们构造如下双射 f:Z>0→Z。每个 n∈Z>0 要么是偶数,要么是奇数,但不能同时是偶数和奇数。如果 n 是偶数,则定义 f(n) = n/2,它在值域 Z 中。如果 n 是奇数,则定义 f(n) = −(n − 1)/2,它也在值域 Z 中。为了证明 f 是双射,我们描述它的逆映射 g:Z→Z>0。对于 m∈Z,令 g(m) = 2m(如果 m > 0),令 g(m) = −2m + 1(如果 m ≤ 0)。图 7.4 中的箭头图展示了 g 的作用。你可以证明 g 映射到 Z>0,并且是 f 的双侧逆映射。我们以一个例子来验证。给定 m∈Z 且 m ≤ 0,我们必须证明 f(g(m)) = m。首先,g(m) = −2m + 1,它是奇数。所以 f(g(m)) = −( − 2m + 1 − 1)/2 = 2m/2 = m,符合要求。

fig7_4

Figure 7.4

图 7.4

A bijection g from Z > 0 to Z 0 2 .

从 Z>0 到 Z≥02 的双射 g。

Countability of Z × Z

Z×Z 的可数性

For any set A , recall that A 2 denotes the set A × A , which consists of all ordered pairs ( x , y ) with x A and y A . We now show that the product set Z 2 and certain related sets are countably infinite. Our proof gives surprising applications of material from Chapter 4 .

对于任意集合 A,回想一下,A 2 表示集合 A × A,它由所有满足 x ∈ A 且 y ∈ A 的有序对 (x, y) 组成。现在我们证明乘积集 Z2 和某些相关集合是可数无限的。我们的证明给出了第 4 章内容的一些令人惊讶的应用。

7.10 Proposition. Z 0 2 and Z > 0 2 and Z 2 are countably infinite sets.

7.10 命题。Z≥02 且 Z>02 和 Z2 是可数无限集。

Proof. We first define a bijection h : Z 0 2 Z > 0 . Let h ( a , b ) = 2 a (2 b + 1) for all a , b Z 0 . We prove that h is one-to-one and onto. To see that h is injective, fix a , b , c , d Z 0 with h ( a , b ) = h ( c , d ), and prove ( a , b ) = ( c , d ). We have assumed 2 a (2 b + 1) = 2 c (2 d + 1). We know a > c or a < c or a = c , so consider three cases.

证明。首先定义一个双射 h: Z≥02→Z>0。令 h(a, b) = 2 a (2b + 1),其中 a, b∈Z≥0。我们证明 h 是单射且满射。为了证明 h 是单射,固定 a, b, c, d∈Z≥0,使得 h(a, b) = h(c, d),并证明 (a, b) = (c, d)。我们假设 2 a (2b + 1) = 2 c (2d + 1)。我们知道 a > c 或 a < c 或 a = c,因此考虑三种情况。

Case 1. Assume a > c . Rewrite the assumption as 2 a c (2 b + 1) = 2 d + 1. The left side of this equation is an even integer since a c > 0, but the right side is an odd integer. This contradicts the earlier theorem that no integer is both even and odd.

情况 1. 假设 a > c。将该假设改写为 2 a c (2b + 1) = 2d + 1。由于 a − c > 0,该等式左侧为偶数,但右侧为奇数。这与之前的定理“不存在既是偶数又是奇数的整数”相矛盾。

Case 2. Assume a < c . Here we can write 2 b + 1 = 2 c a (2 d + 1) with c a > 0. Again there is a contradiction because the left side is odd and the right side is even.

情况 2. 假设 a < c。这里我们可以写成 2b + 1 = 2 c a (2d + 1),其中 c − a > 0。这里又出现了矛盾,因为左边是奇数,右边是偶数。

Case 3. Assume a = c . Dividing the assumption by 2 a = 2 c , we get 2 b + 1 = 2 d + 1, and hence b = d . Thus, ( a , b ) = ( c , d ) as needed.

情况 3. 假设 a = c。将该假设除以 2 a = 2 c ,我们得到 2b + 1 = 2d + 1,因此 b = d。因此,(a, b) = (c, d),符合要求。

To see that h is surjective, fix n Z > 0 ; we must find ( a , b ) Z 0 2 with h ( a , b ) = n . Using the formula for h , we need to solve 2 a (2 b + 1) = n for a and b . To do so, use the Fundamental Theorem of Arithmetic to write n = 2 e 1 p 2 e 2 p k e k , where k ≥ 1, e 1 , …, e k ≥ 0, and p 2 , …, p k are distinct odd primes. Comparing to the previous expression, we choose a = e 1 and b = ( p 2 e 2 p k e k 1 ) / 2 (which is in Z 0 ) to achieve h ( a , b ) = n . Note that b = 0 when k = 1.

为了证明 h 是满射,固定 n∈Z>0;我们需要找到 (a,b)∈Z≥02 使得 h(a, b) = n。利用 h 的公式,我们需要求解 2 a (2b + 1) = n 以求得 a 和 b。为此,利用算术基本定理,将 n 写成 2e1p2e2⋯pkek 的形式,其中 k ≥ 1,e 1 , …, e k ≥ 0,且 p 2 , …, p k 是互不相同的奇素数。与之前的表达式比较,我们选择 a = e 1 和 b=(p2e2⋯pkek−1)/2(它在 Z≥0 中),从而得到 h(a, b) = n。注意,当 k = 1 时,b = 0。

So far, we have proved that | Z 0 2 | = | Z > 0 | . We also know | Z | = | Z > 0 | = | Z 0 | , so it follows from part (h) of Theorem 7.6 that | Z × Z | = | Z > 0 × Z > 0 | = | Z 0 × Z 0 | = | Z > 0 | . By transitivity and symmetry, we conclude that Z 2 , Z > 0 2 , and Z 0 2 are all countably infinite sets.

到目前为止,我们已经证明了 |Z≥02|=|Z>0|。我们也知道 |Z|=|Z>0|=|Z≥0|,因此由定理 7.6 的 (h) 部分可知 |Z×Z|=|Z>0×Z>0|=|Z≥0×Z≥0|=|Z>0|。由传递性和对称性可知,Z2、Z>02 和 Z≥02 都是可数无限集。

We may visualize the map h in the previous proof by labeling each point ( a , b ) in a picture of Z 0 2 with the corresponding positive integer h ( a , b ), as shown on the left in Figure 7.5 . Pictures like this one provide powerful intuition for defining bijections on infinite sets. For example, the picture on the right in Figure 7.5 suggests another possible bijection k : Z 0 2 Z > 0 , obtained by traversing integer points in the first quadrant one diagonal at a time.

我们可以通过给 Z≥02 中的每个点 (a, b) 标记对应的正整数 h(a, b) 来可视化前面证明中的映射 h,如图 7.5 左侧所示。这样的图为定义无限集上的双射提供了强有力的直观理解。例如,图 7.5 右侧的图展示了另一个可能的双射 k:Z≥02→Z>0,它是通过每次沿对角线遍历第一象限中的整数点得到的。

fig7_5

Figure 7.5

图 7.5

Pictures of two bijections from Z > 0 to Z 0 2 .

从 Z>0 到 Z≥02 的两个双射的图像。

The new figure gives compelling visual insight into why there is a bijection from Z 0 2 to Z > 0 , but how do we translate the figure into a rigorous proof? One approach is to give a recursive definition of the inverse f : Z > 0 Z 0 2 for k . For the base case, define f (1) = (0, 0). For the recursive case, fix n Z > 0 and suppose f ( n ) = ( a , b ) has already been defined. Motivated by the figure, we set f ( n + 1) = ( a + 1, b − 1) if b > 0, and we set f ( n + 1) = (0, a + 1) if b = 0. Intuitively, the first case moves one step down and right along the current diagonal. The second case occurs when n appears at the lower-right end of one diagonal, so that n + 1 must be placed at the upper-left end of the next diagonal. It can now be proved by induction that f is one-to-one and onto. Remarkably, the map k = f −1 is given by a simple non-recursive formula: k ( a , b ) = (( a + b ) 2 + 3 a + b )/2 + 1 for ( a , b ) Z 0 2 . You can check that k ( f ( n )) = n for all n Z > 0 by induction on n , and that f ( k ( a , b )) = ( a , b ) for all a , b Z 0 2 by induction on a + b , with an inner induction on a .

新图清晰地展示了为什么存在从 Z≥0² 到 Z>0 的双射,但我们如何将图中的元素转化为严格的证明呢?一种方法是给出逆映射 f:Z>0→Z≥0² 的递归定义,其中 k 为任意正整数。对于基本情况,定义 f(1) = (0, 0)。对于递归情况,固定 n∈Z>0,并假设 f(n) = (a, b) 已定义。受图中的启发,我们定义当 b > 0 时 f(n + 1) = (a + 1, b − 1),当 b = 0 时 f(n + 1) = (0, a + 1)。直观地说,第一种情况沿着当前对角线向下向右移动一步。第二种情况发生在 n 出现在一条对角线的右下角时,因此 n + 1 必须位于下一条对角线的左上角。现在可以通过归纳法证明 f 是单射且满射。值得注意的是,映射 k = f −1 由一个简单的非递归公式给出:k(a, b) = ((a + b) 2 + 3a + b)/2 + 1,其中 (a,b)∈Z≥02。你可以通过对 n 进行归纳来验证对于所有 n∈Z>0,k(f(n)) = n;并且可以通过对 a + b 进行归纳(其中 a 为内归纳),验证对于所有 a,b∈Z≥02,f(k(a, b)) = (a, b)。

Countability of Q and Finite Products

Q 的可数性和有限积

Now that we know Z × Z is countably infinite, we can show that the set of rational numbers is countably infinite.

既然我们知道 Z×Z 是可数无限的,我们就可以证明有理数集是可数无限的。

7.11 Proposition. Q is countably infinite.

7.11 命题。Q 是可数无限的。

Proof. By the Schröder-Bernstein Theorem (part (e) of Theorem 7.6 ), it suffices to show | Z > 0 | | Q | and | Q | | Z > 0 | . [In the next section, we give another proof of the countability of Q not relying on this hard result.] On one hand, the inclusion map i : Z > 0 Q given by i ( n ) = n /1 for all n Z > 0 is one-to-one, so | Z > 0 | | Q | holds by definition. On the other hand, define f : Q Z × Z as follows. Given q Q , there exist unique integers m and n with n > 0, gcd( m , n ) = 1, and q = m / n ; we define f ( q ) = ( m , n ). Now f is injective, because if q , r Q satisfy f ( q ) = ( m , n ) = f ( r ), then q = m / n = r . Since Z × Z is known to be countably infinite, there is a bijection g : Z × Z Z > 0 . Then g f : Q Z > 0 is injective, being a composition of two injective functions. Therefore | Q | | Z > 0 | , as needed.      □

证明。根据 Schröder-Bernstein 定理(定理 7.6 的 (e) 部分),只需证明 |Z>0|≤|Q| 且 |Q|≤|Z>0|。[下一节我们将给出 Q 可数性的另一种证明,该证明不依赖于这个困难的结果。] 一方面,包含映射 i:Z>0→Q 由 i(n) = n/1 给出,对于所有 n∈Z>0,它是一一对应的,因此根据定义,|Z>0|≤|Q| 成立。另一方面,定义 f:Q→Z×Z 如下。给定 q∈Q,存在唯一的整数 m 和 n,使得 n > 0,gcd(m, n) = 1,且 q = m/n;我们定义 f(q) = (m, n)。现在 f 是单射,因为如果 q,r∈Q 满足 f(q) = (m, n) = f(r),则 q = m/n = r。由于已知 Z×Z 是可数无限集,因此存在双射 g:Z×Z→Z>0。那么 g∘f:Q→Z>0 也是单射,因为它是两个单射函数的复合。因此 |Q|≤|Z>0|,符合要求。□

Next we prove that a product of finitely many countably infinite sets is also countably infinite.

接下来,我们将证明有限多个可数无限集的乘积也是可数无限集。

7.12 Theorem on Countability of Products. For all k Z > 0 and all sets X 1 , …, X k , if every X i is countably infinite, then X 1 × X 2 × ⋯ × X k is countably infinite.

7.12 关于乘积可数性的定理。对于所有 k∈Z>0 和所有集合 X 1 , …, X k ,如果每个 X i 都是可数无限的,那么 X 1 × X 2 × ⋯ × X k 也是可数无限的。

Proof. We use induction on k . The base case k = 1 is immediate. Fix k Z > 0 , and assume the product of any k countably infinite sets is countably infinite. Next, fix k + 1 countably infinite sets X 1 , …, X k , X k +1 , and prove X = X 1 × ⋯ × X k × X k +1 is countably infinite. Let Y = X 1 × ⋯ × X k , which is countably infinite by the induction hypothesis. Define f : X Y × X k +1 by letting f send ( x 1 , …, x k , x k +1 ) to (( x 1 , …, x k ), x k +1 ) for ( x 1 , …, x k +1 ) ∈ X . You can check that f is a bijection, so | X | = | Y × X k +1 |. We know | Y | = | Z > 0 | and | X k + 1 | = | Z > 0 | since Y and X k +1 are countably infinite. By part (h) of Theorem 7.6 , we conclude that | Y × X k + 1 | = | Z > 0 × Z > 0 | . Now Z > 0 2 is countably infinite, so | Z > 0 × Z > 0 | = | Z > 0 | . Combining all these cardinal equalities using transitivity, we conclude that | X | = | Z > 0 | , as needed.     □

证明。我们对 k 进行归纳。基本情况 k = 1 显然成立。固定 k∈Z>0,并假设任意 k 个可数无限集的乘积也是可数无限集。接下来,固定 k + 1 个可数无限集 X 1 , …, X k , X k +1 ,并证明 X = X 1 × ⋯ × X k × X k +1 也是可数无限集。令 Y = X 1 × ⋯ × X k ,根据归纳假设,Y 也是可数无限集。定义 f : X → Y × X k +1 ,使得 f 将 (x 1 , …, x k , x k +1 ) 映射到 ((x 1 , …, x k ), x k +1 ),其中 (x 1 , …, x k +1 ) ∈ X。可以验证 f 是一个双射,因此 |X| = |Y × X k +1 |。我们知道 |Y|=|Z>0| 且 |Xk+1|=|Z>0|,因为 Y 和 X k +1 是可数无穷的。根据定理 7.6 的 (h) 部分,我们得出结论 |Y×Xk+1|=|Z>0×Z>0|。现在 Z>02 是可数无穷的,所以 |Z>0×Z>0|=|Z>0|。结合所有这些基数等式并利用传递性,我们得出 |X|=|Z>0|,符合要求。□

7.13 Corollary. For any countably infinite set X and any k Z > 0 , X k is countably infinite. In particular, Z > 0 k and Z k and Q k are countably infinite.     □

7.13 推论。对于任意可数无限集 X 和任意 k∈Z>0,X k 是可数无限集。特别地,Z>0k 且 Zk 和 Qk 是可数无限集。□

Section Summary

章节概要

  1. Cardinal Equality. For all sets A and B , | A | = | B | means there exists a bijection f : A B . Cardinal equality is reflexive, symmetric, and transitive. If | A | = | A ′| and | B | = | B ′|, then | A × B | = | A ′ × B ′| and | P ( A ) | = | P ( A ) | ; if also A B = 0 = A B , then | A B | = | A B | .
    基数相等。对于任意集合 A 和 B,|A| = |B| 表示存在一个双射 f : A → B。基数相等具有自反性、对称性和传递性。如果 |A| = |A′| 且 |B| = |B′|,则 |A × B| = |A′ × B′| 且 |P(A)| = |P(A′)|;如果 A∩B = 0 = A′∩B′,则 |A∪B| = |A′∪B′|。
  2. Cardinal Ordering. For all sets A and B , | A | ≤ | B | means there exists an injection g : A B . Cardinal ordering is reflexive, antisymmetric, and transitive. Antisymmetry (for all A and B , if | A | ≤ | B | and | B | ≤ | A |, then | A | = | B |) is a hard result called the Schröder–Bernstein Theorem.
    基数序。对于任意集合 A 和 B,|A| ≤ |B| 表示存在单射 g : A → B。基数序具有自反性、反对称性和传递性。反对称性(对于任意集合 A 和 B,如果 |A| ≤ |B| 且 |B| ≤ |A|,则 |A| = |B|)是一个难解结果,称为施罗德-伯恩斯坦定理。
  3. Countably Infinite Sets. A set X is countably infinite iff | Z > 0 | = | X | . The following sets are countably infinite: Z , Z a for any a Z , Z × Z , Q , products of finitely many countably infinite sets, Z k , and Q k (where k Z > 0 ).
    可数无限集。集合 X 是可数无限集当且仅当 |Z>0|=|X|。下列集合是可数无限集:Z、对于任意 a∈Z 都有 Z≥a、Z×Z、Q、有限多个可数无限集的乘积、Zk 和 Qk(其中 k∈Z>0)。

Exercises

练习

  1. Prove: for all sets X and Y , if | X | = | Y |, then | X | ≤ | Y |.
    证明:对于所有集合 X 和 Y,如果 |X| = |Y|,则 |X| ≤ |Y|。
  2. (a) Prove: for all sets X , Y , and Z , if | X | = | Y | and | Y | ≤ | Z |, then | X | ≤ | Z |. (b) Prove: for all sets X , Y , and Z , if | X | ≤ | Y | and | Y | = | Z |, then | X | ≤ | Z |.
    (a)证明:对于所有集合 X、Y 和 Z,如果 |X| = |Y| 且 |Y| ≤ |Z|,则 |X| ≤ |Z|。(b)证明:对于所有集合 X、Y 和 Z,如果 |X| ≤ |Y| 且 |Y| = |Z|,则 |X| ≤ |Z|。
  3. (a) Prove: for all sets X and Y , if X Y , then | X | ≤ | Y |. (b) Prove or disprove the converse of part (a).
    (a)证明:对于任意集合 X 和 Y,如果 X⊆ Y,则 |X| ≤ |Y|。(b)证明或反证(a)部分的逆命题。
  4. (a) Complete the proof of part (h) of Theorem 7.6 by showing h ′ ○ h and h h ′ are identity maps. (b) Finish the proof of part (h) without using h ′, by proving that h is one-to-one and onto.
    (a) 完成定理 7.6 的 (h) 部分的证明,证明 h′ ○ h 和 h ○ h′ 是恒等映射。(b) 不使用 h′,完成定理 7.6 的 (h) 部分的证明,证明 h 是单射且满射。
  5. Complete the proof of part (g) of Theorem 7.6 by showing p and p ′ are well-defined functions such that p ′ ○ p and p p ′ are identity maps. Where is the assumption A B = 0 = A B needed?
    完成定理 7.6 的 (g) 部分的证明,证明 p 和 p′ 是定义良好的函数,且 p′ ○ p 和 p ○ p′ 是恒等映射。假设 A∩B=0=A′∩B′ 在哪里需要用到?
  6. Prove part (i) of Theorem 7.6 .
    证明定理 7.6 的第 (i) 部分。
  7. Prove from the definition that each set below is countably infinite: (a) the set of odd positive integers; (b) the set of positive multiples of a fixed k Z ; (c) the set of all even integers.
    根据定义证明下列各集合是可数无限的:(a)奇数正整数的集合;(b)固定 k∈Z 的正整数倍的集合;(c)偶数的集合。
  8. Explain why every countably infinite set actually is infinite (not finite).
    解释为什么每个可数无限集实际上都是无限的(而不是有限的)。
  9. Complete the proof of Proposition 7.8 .
    完成命题 7.8 的证明。
  10. Complete the proof of Proposition 7.9 .
    完成命题 7.9 的证明。
  11. (a) Show that f : Z 0 3 Z > 0 defined by f ( a , b , c ) = 2 a 3 b 5 c for a , b , c Z 0 is injective. (b) Trace through proofs from this section to find an explicit formula for a bijection g : Z 0 3 Z > 0 .
    (a) 证明由 f(a, b, c) = 2352 定义的 f:Z≥03→Z>0(其中 a, b, c∈Z≥0)是单射。(b) 梳理本节的证明,找到双射 g:Z≥03→Z>0 的显式公式。
  12. (a) Prove that the map f defined recursively below Figure 7.5 is one-to-one and onto. (b) Prove that k ( a , b ) = (( a + b ) 2 + 3 a + b )/2 + 1 is the two-sided inverse of f .
    (a)证明图 7.5 下方递归定义的映射 f 是单射且满射。(b)证明 k(a, b) = ((a + b) 2 + 3a + b)/2 + 1 是 f 的双边逆。
  13. The diagram below depicts yet another bijection p : Z 0 2 Z . Formally define the function p (perhaps recursively), and prove p is a bijection.

    ufig7_1


    下图展示了另一个双射 p:Z≥02→Z。请正式定义函数 p(可以使用递归方法),并证明 p 是一个双射。 ufig7_1
  14. Draw a diagram (similar to the one in the previous problem) depicting a bijection q : Z × Z Z . Define q formally, and prove q is a bijection.
    画一个图(类似于上一个问题中的图),表示双射 q:Z×Z→Z。正式定义 q,并证明 q 是一个双射。
  15. Prove: for all nonempty sets A and B , | A | ≤ | B | iff there exists a surjection h : B A .
    证明:对于所有非空集合 A 和 B,|A| ≤ |B| 当且仅当存在满射 h : B → A。
  16. Suppose I is any index set, and for each i I , A i and A i ′ are sets such that | A i | = | A i ′|. Prove: if for all i j in I , A i A j = 0 = A i A j , then | i I A i | = | i I A i | .
    假设 I 是任意索引集,对于每个 i ∈ I,A i 和 A i ′ 是满足 |A i | = |A i ′| 的集合。证明:如果对于 I 中的所有 i ≠ j,Ai∩Aj=0=Ai′∩Aj′,则 |⋃i∈IAi|=|⋃i∈IAi′|。
  17. Suppose k Z > 0 , and for 1 ≤ i k , A i and A i ′ are sets such that | A i | = | A i ′|. (a) Prove | A 1 × ⋯ × A k | = | A 1 ′ × ⋯ × A k ′| by constructing a bijection. (b) Prove the result in (a) by induction on k , using part (h) of Theorem 7.6 .
    假设 k∈Z>0,且对于 1 ≤ i ≤ k,A i 和 A i ′ 是满足 |A i | = |A i ′| 的集合。(a) 通过构造双射证明 |A 1 × ⋯ × A k | = |A 1 ′ × ⋯ × A k ′|。(b) 利用定理 7.6 的 (h) 部分,对 k 进行归纳证明 (a) 中的结果。
  18. (a) Prove that the union of two countably infinite sets is countably infinite. (b) Prove: for all k Z > 0 and all sets X 1 , …, X k , if every X i is countably infinite, then i = 1 k X i is countably infinite.
    (a)证明两个可数无限集的并集是可数无限集。(b)证明:对于所有 k∈Z>0 和所有集合 X 1 , …, X k ,如果每个 X i 都是可数无限集,则 ⋃i=1kXi 也是可数无限集。
  19. Suppose that for every n Z > 0 , X n is a given countably infinite set. Prove that X = n Z > 0 X n is countably infinite. [ Hint: Use the fact that Z > 0 2 is countably infinite.]
    假设对于任意 n∈Z>0,X n 是一个给定的可数无限集。证明 X=⋃n∈Z>0Xn 也是可数无限集。[提示:利用 Z>02 是可数无限集这一事实。]
  20. Let X be any countably infinite set. (a) Prove there exists f : X X that is injective but not surjective. (b) Prove there exists g : X X that is surjective but not injective.
    设 X 为任意可数无限集。(a) 证明存在函数 f : X → X,它是单射但不是满射。(b) 证明存在函数 g : X → X,它是满射但不是单射。

7.3 Countable Sets

7.3 可数集

This section studies countable sets, which are sets that are either finite or countably infinite. Our main results are: any subset of a countable set is countable; products of finitely many countable sets are countable; and unions of countably many countable sets are countable.

本节研究可数集,即有限集或可数无限集。我们的主要结论是:可数集的任何子集都是可数的;有限个可数集的乘积是可数的;可数个可数集的并集也是可数的。

Countable Sets and Their Subsets

可数集及其子集

7.14. Definition: Countable Sets. For all sets X , X is countable iff X is finite or X is countably infinite.

7.14. 定义:可数集。对于所有集合 X,X 是可数的当且仅当 X 是有限的或 X 是可数无限的。

We intend to show that any subset of any countable set is countable. First we consider the special case of subsets of Z > 0 .

我们的目的是证明任何可数集的任意子集都是可数的。首先,我们考虑 Z>0 的子集这一特殊情况。

7.15 Lemma on Subsets of Z > 0 . Every subset of Z > 0 is countable.

7.15 关于 Z>0 的子集的引理。Z>0 的每个子集都是可数的。

Proof . Fix an arbitrary subset T of Z > 0 . To prove T is countable, assume T is not finite and prove T is countably infinite. We construct a bijective function f : Z > 0 T recursively. Intuitively, f ( n ) = t will mean that t is the n th smallest element of T . To make this precise, recall from §6.5 the Well-Ordering Property of Z > 0 , which tells us that every nonempty subset of Z > 0 has a least element. Since T is certainly nonempty, we may define f (1) to be the least element of T . Now fix n Z > 0 , and assume that f (1), f (2), …, f ( n ) have already been defined. We also assume f (1), …, f ( n ) are pairwise distinct members of T . Note that T n = T − { f (1), f (2), …, f ( n )} is a subset of Z > 0 that is nonempty. For if T n were empty, then T ⊆{ f (1), f (2), …, f ( n )}, so that T would be finite, contrary to our assumption. We define f ( n + 1) to be the least element of T n , which is a member of T different from f ( i ) for all i between 1 and n . This completes the recursive definition of f . Figure 7.6 illustrates the definition of f for the set T = {3, 6, 10, 11, 13, …}.

证明。固定 Z>0 的任意子集 T。为了证明 T 是可数的,假设 T 不是有限的,并证明 T 是可数无限的。我们递归地构造一个双射函数 f:Z>0→T。直观地说,f(n) = t 表示 t 是 T 的第 n 个最小元素。为了更精确地表达,回顾 §6.5 中 Z>0 的良序性质,该性质告诉我们 Z>0 的每个非空子集都有一个最小元素。由于 T 显然非空,我们可以定义 f(1) 为 T 的最小元素。现在固定 n∈Z>0,并假设 f(1), f(2), …, f(n) 已经定义。我们还假设 f(1), …, f(n) 是 T 中两两不同的元素。注意,T n = T − {f(1), f(2), …, f(n)} 是 Z>0 的一个非空子集。如果 T n 为空,则 T⊆{f(1), f(2), …, f(n)},因此 T 将是有限的,这与我们的假设相反。我们将 f(n + 1) 定义为 T n 中除 f(i) 外的最小元素,其中 i 介于 1 和 n 之间。这就完成了 f 的递归定义。图 7.6 展示了集合 T = {3, 6, 10, 11, 13, …} 中 f 的定义。

fig7_6

Figure 7.6

图 7.6

Example of the bijection f : Z > 0 T .

双射 f:Z>0→T 的例子。

To show that f is injective, fix m , p Z > 0 , assume m p , and prove f ( m ) ≠ f ( p ). We know m < p or p < m . In the case where m < p , f ( p ) ≠ f ( m ) since f ( m ) is in the set { f (1), …, f ( p − 1)}, but f ( p ) is not in this set by definition. In the case where p < m , f ( p ) ≠ f ( m ) since f ( p ) is in the set { f (1), …, f ( m − 1)}, but f ( m ) is not in this set.

为了证明 f 是单射,固定 m, p∈Z>0,假设 m ≠ p,并证明 f(m) ≠ f(p)。我们知道 m < p 或 p < m。当 m < p 时,f(p) ≠ f(m),因为 f(m) 属于集合 {f(1), …, f(p − 1)},而根据定义,f(p) 不属于该集合。当 p < m 时,f(p) ≠ f(m),因为 f(p) 属于集合 {f(1), …, f(m − 1)},而 f(m) 不属于该集合。

To show that f is surjective, we first prove that f ( n ) ≥ n for all n Z > 0 . We use induction on n . For the base case, f (1) is a member of T , so f ( 1 ) Z > 0 , so f (1) ≥ 1. Now fix n Z > 0 , assume f ( n ) ≥ n , and prove f ( n + 1) ≥ n + 1. Assume, to get a contradiction, that f ( n + 1) < n + 1. Since f ( n + 1) is an integer, it follows that f ( n + 1) ≤ n f ( n ). Now, f ( n + 1) is the least element of T − { f (1), f (2), …, f ( n )}, so f ( n + 1) ≠ f ( n ). We deduce that f ( n + 1) < f ( n ). But now f ( n + 1) is an element of the set T − { f (1), …, f ( n − 1)} strictly smaller than f ( n ), which contradicts the definition of f ( n ) as the least element of this set. So f ( n + 1) ≥ n + 1, completing the induction.

为了证明 f 是满射,我们首先证明对于所有 n∈Z>0,都有 f(n) ≥ n。我们对 n 使用归纳法。对于基本情况,f(1) 是 T 的元素,所以 f(1)∈Z>0,因此 f(1) ≥ 1。现在固定 n∈Z>0,假设 f(n) ≥ n,并证明 f(n + 1) ≥ n + 1。为了得出矛盾,假设 f(n + 1) < n + 1。由于 f(n + 1) 是整数,因此 f(n + 1) ≤ n ≤ f(n)。现在,f(n + 1) 是 T − {f(1), f(2), …, f(n)} 中的最小元素,所以 f(n + 1) ≠ f(n)。由此我们得出 f(n + 1) < f(n)。但现在 f(n + 1) 是集合 T − {f(1), …, f(n − 1)} 中严格小于 f(n) 的元素,这与 f(n) 作为该集合的最小元素的定义相矛盾。因此 f(n + 1) ≥ n + 1,归纳完成。

Now we prove f is surjective. Fix t T ; we must show there exists n Z > 0 with f ( n ) = t . We have shown that f ( t ) ≥ t . If f ( t ) = t , we may choose n = t . If f ( t ) > t , then the least element of the set S = T − { f (1), f (2), …, f ( t − 1)} (which is f ( t ) by definition) is larger than t . It follows that t cannot belong to S . But t does belong to T , which forces t = f ( n ) for some n ∈ {1, 2, …, t − 1}. Thus, f is surjective.     □

现在我们证明 f 是满射。固定 t ∈ T;我们必须证明存在 n ∈ Z > 0 使得 f(n) = t。我们已经证明 f(t) ≥ t。如果 f(t) = t,我们可以选择 n = t。如果 f(t) > t,那么集合 S = T − {f(1), f(2), …, f(t − 1)} 中的最小元素(根据定义是 f(t))大于 t。由此可知 t 不能属于 S。但 t 属于 T,这迫使存在某个 n ∈ {1, 2, …, t − 1} 使得 t = f(n)。因此,f 是满射。□

7.16 Theorem on Countable Sets. For all sets X and Y :

7.16 可数集定理。对于任意集合 X 和 Y:

(a) If | X | = | Y |, then X is countable iff Y is countable.

(a)如果 |X| = |Y|,则 X 可数当且仅当 Y 可数。

(b) | X | ≤ | Y | iff W , W Y and | X | = | W |.

(b)|X| ≤ |Y| 当且仅当存在 W,W⊆Y 且 |X| = |W|。

(c) X is countable iff | X | | Z > 0 | .

(c)X 是可数的当且仅当 |X|≤|Z>0|。

(d) If | X | ≤ | Y | and Y is countable, then X is countable.

(d)如果 |X| ≤ |Y| 且 Y 是可数的,那么 X 也是可数的。

(e) If X Y and Y is countable, then X is countable.

(e)如果 X⊆ Y 且 Y 是可数的,那么 X 也是可数的。

Proof. We prove parts (b) through (e), leaving part (a) as an exercise. For part (b), first assume | X | ≤ | Y |, which means there is an injection f : X Y . Choose W = f [ X ] to be the image of f , which is a subset of Y . Let f ′ : X W be obtained from f by restricting the codomain from Y to W . Then f ′ is injective since f is, and f ′ is surjective by construction. So | X | = | W |. Conversely, assume | X | = | W | for some W Y . This means there is a bijection g : X W . Composing g with the injective inclusion map i : W Y , we obtain an injection i g : X Y , which proves that | X | ≤ | Y |.

证明。我们证明 (b) 到 (e) 部分,将 (a) 部分留作练习。对于 (b) 部分,首先假设 |X| ≤ |Y|,这意味着存在单射 f : X → Y。选择 W = f[X] 为 f 的像,它是 Y 的一个子集。令 f′ : X → W 是通过将 f 的值域从 Y 限制到 W 而得到的。由于 f 是单射,因此 f′ 也是单射,并且根据构造,f′ 也是满射。所以 |X| = |W|。反之,假设对于某个 W⊆ Y,|X| = |W|。这意味着存在双射 g : X → W。将 g 与单射包含映射 i : W → Y 复合,我们得到单射 i ○ g : X → Y,这证明了 |X| ≤ |Y|。

For part (c), first assume X is countable. In the case where X is finite, we know there exists n Z 0 with | X | = |{1, 2, …, n }|. Since {1, 2, …, n } is a subset of Z > 0 , part (b) applies to show that | X | | Z > 0 | . Similarly, in the case where X is countably infinite, we know | X | = | Z > 0 | . Since Z > 0 is a subset of itself, part (b) applies to show that | X | | Z > 0 | . Conversely, assume | X | | Z > 0 | . By part (b), there exists W Z > 0 with | X | = | W |. By Lemma 7.15 , W is countable. By part (a), X is also countable.

对于 (c) 部分,首先假设 X 是可数的。当 X 是有限集时,我们知道存在 n∈Z≥0 使得 |X| = |{1, 2, …, n}|。由于 {1, 2, …, n} 是 Z>0 的子集,因此 (b) 部分成立,可以证明 |X|≤|Z>0|。类似地,当 X 是可数无限集时,我们知道 |X|=|Z>0|。由于 Z>0 是它自身的子集,因此 (b) 部分成立,可以证明 |X|≤|Z>0|。反之,假设 |X|≤|Z>0|。由 (b) 部分可知,存在 W⊆Z>0 使得 |X| = |W|。由引理 7.15 可知,W 是可数的。由 (a) 部分可知,X 也是可数的。

For part (d), assume | X | ≤ | Y | and Y is countable. Then | Y | | Z > 0 | by part (c), so | X | | Z > 0 | by transitivity, so X is countable by part (c). For part (e), assume X Y and Y is countable. The inclusion i : X Y is injective, so | X | ≤ | Y |. Thus, X is countable by part (d).     □      □

对于 (d) 部分,假设 |X| ≤ |Y| 且 Y 是可数的。那么根据 (c) 部分,|Y|≤|Z>0|,因此根据传递性,|X|≤|Z>0|,所以根据 (c) 部分,X 是可数的。对于 (e) 部分,假设 X⊆ Y 且 Y 是可数的。包含关系 i : X → Y 是单射,所以 |X| ≤ |Y|。因此,根据 (d) 部分,X 是可数的。□ □

Using this theorem, we can see that Q is countably infinite without invoking the Schröder–Bernstein Theorem. Recall that we showed | Q | | Z × Z | in the proof of Proposition 7.11 . Since | Z × Z | is known to be countably infinite and hence countable, we conclude that Q is countable by part (d) of Theorem 7.16 . Now Q is not finite (since it contains the infinite subset Z > 0 ), so Q is countably infinite.

利用该定理,我们可以无需借助施罗德-伯恩斯坦定理即可证明 Q 是可数无限的。回想一下,我们在命题 7.11 的证明中证明了 |Q|≤|Z×Z|。由于已知 |Z×Z| 是可数无限的,因此 Q 也是可数的,根据定理 7.16 的 (d) 部分,我们得出 Q 是可数的。现在 Q 不是有限的(因为它包含无限子集 Z>0),所以 Q 是可数无限的。

Products of Countable Sets

可数集的乘积

We have seen that the product of finitely many finite sets is finite, and the product of finitely many countably infinite sets is countably infinite. We now establish the corresponding result for countable sets.

我们已经看到,有限个有限集的乘积是有限的,有限个可数无限集的乘积是可数无限的。现在我们证明可数集的相应结果。

7.17 Theorem on Products of Countable Sets. For all positive integers k and all sets X 1 , …, X k , Y 1 , …, Y k :

7.17 可数集乘积定理。对于所有正整数 k 和所有集合 X 1 , …, X k , Y 1 , …, Y k :

(a) If | X i | ≤ | Y i | for 1 ≤ i k , then | X 1 × X 2 × ⋯ × X k | ≤ | Y 1 × Y 2 × ⋯ × Y k |.

(a)如果对于 1 ≤ i ≤ k,有 |X i | ≤ |Y i |,则 |X 1 × X 2 × ⋯ × X k | ≤ |Y 1 × Y 2 × ⋯ × Y k |。

(b) If every X i is countable, then X 1 × X 2 × ⋯ × X k is countable.

#

(b)如果每个 X i 都是可数的,那么 X 1 × X 2 × ⋯ × X k 也是可数的。

Proof. Fix k Z > 0 and sets X i , Y i for 1 ≤ i k . Let X = X 1 × ⋯ × X k and Y = Y 1 × ⋯ × Y k . For part (a), assume | X i | ≤ | Y i | for all i , which means there exist injections f i : X i Y i for all i . We must build an injection f : X Y . Define f ( x 1 , …, x k ) = ( f 1 ( x 1 ), …, f k ( x k )) for ( x 1 , …, x k ) ∈ X . To see that f is injective, fix x = ( x 1 , …, x k ) ∈ X , x ′ = ( x 1 ′, …, x k ′) ∈ X , assume f ( x ) = f ( x ′), and prove x = x ′. We have assumed ( f 1 ( x 1 ), …, f k ( x k )) = ( f 1 ( x 1 ′), …, f k ( x k ′)), so f i ( x i ) = f i ( x i ′) for 1 ≤ i k . Since each f i is injective, we conclude that x i = x i ′ for 1 ≤ i k , which means x = x ′.

证明。固定 k∈Z>0,并令 X i , Y i ,其中 1 ≤ i ≤ k。令 X = X 1 × ⋯ × X k ,Y = Y 1 × ⋯ × Y k 。对于 (a) 部分,假设对于所有 i,|X i | ≤ |Y i |,这意味着对于所有 i,存在单射 f i : X i → Y i 。我们必须构造一个单射 f : X → Y。定义 f(x 1 , …, x k ) = (f 1 (x 1 ), …, f k (x k )),其中 (x 1 , …, x k ) ∈ X。为了证明 f 是单射,固定 x = (x 1 , …, x k ) ∈ X,x′ = (x 1 ′, …, x k ′) ∈ X,假设 f(x) = f(x′),并证明 x = x′。我们假设 (f 1 (x 1 ), …, f k (x k )) = (f 1 (x 1 ′), …, f k (x k ′)),因此对于 1 ≤ i ≤ k,f i (x i ) = f i (x i ′)。由于每个 f i 都是单射,我们得出结论:对于 1 ≤ i ≤ k,x i = x i ′,这意味着 x = x′。

For part (b), assume every X i is countable. By part (c) of Theorem 7.16 , | X i | | Z > 0 | for 1 ≤ i k . By what we proved in the last paragraph, | X | = | X 1 × × X k | | Z > 0 × × Z > 0 | = | Z > 0 k | = | Z > 0 | , so | X | | Z > 0 | . By part (c) of Theorem 7.16 , X is countable.

对于 (b) 部分,假设每个 X i 都是可数的。根据定理 7.16 的 (c) 部分,对于 1 ≤ i ≤ k,有 |Xi|≤|Z>0|。根据我们在上一段的证明,|X|=|X1×⋯×Xk|≤|Z>0×⋯×Z>0|=|Z>0k|=|Z>0|,因此 |X|≤|Z>0|。根据定理 7.16 的 (c) 部分,X 是可数的。

Unions of Countable Sets

可数集的并集

We have seen that the union of finitely many finite sets is finite. Now we can show that the union of countably many countable sets is countable.

我们已经看到,有限个有限集的并集是有限的。现在我们可以证明,可数个可数集的并集是可数的。

7.18 Theorem on Unions of Countable Sets.

7.18 可数集的并集定理。

(a) Suppose X k is a countable set for each k Z > 0 . Then k = 1 X k is countable.

(a)假设对于每个 k∈Z>0,X k 都是可数集。那么 ⋃k=1∞Xk 是可数的。

(b) For all n Z > 0 and all sets Y 1 , …, Y n , if every Y i is countable, then k = 1 n Y k is countable.

(b)对于所有 n∈Z>0 和所有集合 Y 1 , …, Y n ,如果每个 Y i 都是可数的,那么 ⋃k=1nYk 也是可数的。

Proof. We first prove part (a) under the additional assumption that the sets X k are pairwise disjoint and countable. We know | X k | | Z > 0 | for each k , so there exist injections f k : X k Z > 0 for each k Z > 0 . Let X = k = 1 X k , and define f : X Z > 0 2 as follows. Given x X , there exists a unique k Z > 0 with x X k , because the sets X k are pairwise disjoint. Define f ( x ) = ( k , f k ( x ) ) Z > 0 2 . Figure 7.7 illustrates how f sends each set X k into the k th vertical slice of Z > 0 2 . In this figure, | X 1 | = 4, | X 3 | = 3, and X 2 and X 4 are countably infinite.

证明。我们首先在集合 X k 两两不相交且可数的附加假设下证明 (a) 部分。我们知道对于每个 k,|Xk|≤|Z>0|,因此对于每个 k∈Z>0,都存在单射 fk:Xk→Z>0。令 X=⋃k=1∞Xk,并定义 f:X→Z>02 如下。给定 x ∈ X,存在唯一的 k∈Z>0 使得 x ∈ X k ,因为集合 X k 两两不相交。定义 f(x)=(k,fk(x))∈Z>02。图 7.7 展示了 f 如何将每个集合 X k 映射到 Z>02 的第 k 个垂直切片。在该图中,|X 1 | = 4,|X 3 | = 3,并且 X 2 和 X 4 是可数无限的。

fig7_7

Figure 7.7

图 7.7

Illustration of the map f : k = 1 X k Z > 0 2 .

地图 f:⋃k=1∞Xk→Z>02 的示意图。

We now show that f is injective. Fix x , y X , assume f ( x ) = f ( y ), and prove x = y . We have f ( x ) = ( k , f k ( x )) and f ( y ) = ( j , f j ( y )) for some j , k Z > 0 . Since f ( x ) = f ( y ), we have k = j and f k ( x ) = f j ( y ) = f k ( y ). Because f k is injective, x = y follows. Because f is one-to-one and Z > 0 2 is countably infinite, we conclude that | X | | Z > 0 2 | = | Z > 0 | . So | X | | Z > 0 | , hence X is countable by part (c) of Theorem 7.16 .

现在我们证明 f 是单射。固定 x, y ∈ X,假设 f(x) = f(y),并证明 x = y。我们有 f(x) = (k, f k (x)) 且 f(y) = (j, f j (y)),其中 j, k ∈ Z>0。由于 f(x) = f(y),我们有 k = j 且 f k (x) = f j (y) = f k (y)。因为 f k 是单射,所以 x = y 成立。因为 f 是单射且 Z>02 是可数无穷的,所以我们得出 |X|≤|Z>02|=|Z>0|。因此 |X|≤|Z>0|,由定理 7.16 的 (c) 部分可知,X 是可数的。

Now we prove the general version of part (a). Assume each X k (for k Z > 0 ) is a countable set. Define X 1 ′ = X 1 and, for each k > 1, X k = X k i = 1 k 1 X i . For each k > 0, X k ′ ⊆ X k , so X k ′ is countable. You can check that k = 1 X k = k = 1 X k and that the sets X k ′ are pairwise disjoint. Since the union of the sets X k ′ is countable by the special case of (a) already proved, we conclude that the union of the sets X k is countable, as needed.

现在我们证明 (a) 部分的一般形式。假设每个 X k (对于 k∈Z>0)都是可数集。定义 X 1 ′ = X 1 ,并且对于每个 k > 1,Xk′=Xk−⋃i=1k−1Xi。对于每个 k > 0,X k ′ ⊆ X k ,因此 X k ′ 是可数的。你可以验证 ⋃k=1∞Xk=⋃k=1∞Xk′,并且集合 X k ′ 两两不相交。由于集合 X k ′ 的并集由已证明的 (a) 特殊情况可数化,因此我们得出结论,集合 X k 的并集是可数化的,这是必要的。

Part (b) follows from part (a) by taking X k = Y k for 1 ≤ k n , and X k = 0 for all k > n .     □

(b) 部分由 (a) 部分得出,只需令当 1 ≤ k ≤ n 时,X k = Y k ,且当 k > n 时,Xk=0。□

Section Summary

章节概要

  1. Countable Sets. A set X is countable iff X is finite or countably infinite. X is countable iff | X | | Z > 0 | iff there is a bijection from X onto a subset of Z > 0 . If | X | ≤ | Y | and Y is countable, then X is countable.
    可数集。集合 X 是可数的当且仅当 X 是有限集或可数无限集。X 是可数的当且仅当 |X|≤|Z>0| 当且仅当存在从 X 到 Z>0 的子集的双射。如果 |X|≤|Y| 且 Y 是可数的,则 X 是可数的。
  2. Operations on Countable Sets. Any subset of a countable set is countable. A product of finitely many countable sets is countable. A union of countably many countable sets is countable.
    可数集的运算。可数集的任何子集都是可数的。有限个可数集的乘积是可数的。可数个可数集的并集是可数的。

Exercises

练习

  1. Prove part (a) of Theorem 7.16 .
    证明定理 7.16 的 (a) 部分。
  2. Complete the proof of Theorem 7.18 by showing that the sets X k ′ are pairwise disjoint, and k = 1 X k = k = 1 X k .
    完成定理 7.18 的证明,证明集合 X k ′ 两两不相交,且 ⋃k=1∞Xk=⋃k=1∞Xk′。
  3. (a) Prove: for all sets X and Y , if X is countable, then X Y is countable. (b)Prove: for all sets X and Y , if X is countably infinite and X Y is finite, then X Y is countably infinite.
    (a)证明:对于任意集合 X 和 Y,如果 X 是可数的,则 X − Y 也是可数的。(b)证明:对于任意集合 X 和 Y,如果 X 是可数无限的且 X∩Y 是有限的,则 X − Y 也是可数无限的。
  4. Give a specific example of countably infinite sets X and Y such that Y X and X Y is countably infinite.
    给出可数无限集合 X 和 Y 的具体例子,使得 Y⊆ X 且 X − Y 是可数无限的。
  5. (a) Give a specific example of pairwise disjoint countably infinite sets X k (for each k Z > 0 ) such that Z = k = 1 X k . (b) Give a specific example of pairwise disjoint countably infinite sets Y k (for each k Z > 0 ) such that Q = k = 1 Y k .
    (a) 给出一个两两不相交的可数无限集 X k (对于每个 k∈Z>0)的具体例子,使得 Z=⋃k=1∞Xk。(b) 给出一个两两不相交的可数无限集 Y k (对于每个 k∈Z>0)的具体例子,使得 Q=⋃k=1∞Yk。
  6. A finite sequence with values in a set X is a function f : {1, 2, …, k } → X for some k Z 0 . Show: for all countable sets X , the set of all finite sequences with values in X is countably infinite.
    取值于集合 X 的有限序列是一个函数 f : {1, 2, …, k} → X,其中 k∈Z≥0。证明:对于所有可数集合 X,取值于 X 的所有有限序列构成的集合是可数无限的。
  7. Show: for all countable sets X , the set of all finite subsets of X is countable.
    证明:对于所有可数集 X,X 的所有有限子集构成的集合是可数的。
  8. Let T be an infinite subset of Z > 0 . In the proof of Lemma 7.15 , we constructed a bijection f : Z > 0 T . Prove that the function g : T Z > 0 , given by g ( t ) = | T { 1 , 2 , , t } | for t T , is the two-sided inverse of f .
    设 T 是 Z>0 的一个无限子集。在引理 7.15 的证明中,我们构造了一个双射 f:Z>0→T。证明函数 g:T→Z>0,对于 t ∈ T,g(t)=|T∩{1,2,…,t}|,是 f 的双边逆。
  9. (a) Suppose we try to prove that every subset of Z is countable by replacing Z > 0 by Z everywhere in the proof of Lemma 7.15 . Explain precisely why the proof does not work. (b) Repeat part (a), but now replace Z > 0 by R 0 .
    (a) 假设我们尝试通过在引理 7.15 的证明中将 Z>0 全部替换为 Z 来证明 Z 的每个子集都是可数的。请准确解释为什么该证明不成立。(b) 重复 (a) 部分,但这次将 Z>0 替换为 R≥0。
  10. Let W be a countably infinite well-ordered set (see §6.5 ), and let T be an infinite subset of W . (a) Show that the recursive construction of f : Z > 0 T in the proof of Lemma 7.15 extends to this situation and produces a well-defined injective function f . (b) The proof that f is surjective does not extend to this setting. Explain the exact point in the proof that fails. (c) Consider the specific example W = { 0 , 1 } × Z > 0 , ordered lexicographically (see Exercise 12 in §6.5 ). Prove that the function f in part (a) is surjective iff T ( { 1 } × Z > 0 ) = 0 .
    设 W 为可数无限良序集(见 §6.5),T 为 W 的无限子集。(a) 证明引理 7.15 证明中 f:Z>0→T 的递归构造可以推广到这种情况,并产生一个定义良好的单射函数 f。(b) f 是满射的证明不能推广到这种情况。解释证明中具体哪里失效。(c) 考虑具体例子 W={0,1}×Z>0,按字典序排列(见 §6.5 中的练习 12)。证明 (a) 部分中的函数 f 是满射当且仅当 T∩({1}×Z>0)=0。
  11. The proof that a union of countably many countable sets is countable requires the Axiom of Choice. Find the exact point in the proof where this axiom is used.
    证明可数个可数集的并集仍然是可数集需要用到选择公理。请找出证明过程中使用选择公理的具体位置。

7.4 Uncountable Sets

7.4 不可数集

One of the unintuitive aspects of cardinality theory is that there are many different “sizes” of infinite sets. We have already studied countably infinite sets, which are the “smallest” infinite sets. In this section, we produce some concrete examples of uncountably infinite sets, including: the set R of real numbers; any open interval ( a , b ) or closed interval [ a , b ] with a < b ; the set P ( Z > 0 ) of all subsets of Z > 0 ; and the set of infinite sequences of zeroes and ones. We also prove Cantor’s Theorem, which says that P ( X ) always has strictly larger cardinality than X , and the Schröder–Bernstein Theorem, which asserts the antisymmetry of the relation | X | ≤ | Y |.

基数理论中一个反直觉的方面是,无限集存在许多不同的“大小”。我们已经研究过可数无限集,它们是“最小的”无限集。在本节中,我们将给出一些不可数无限集的具体例子,包括:实数集 R;任意开区间 (a, b) 或闭区间 [a, b],其中 a < b;Z>0 的所有子集构成的集合 P(Z>0);以及由 0 和 1 构成的无限序列构成的集合。我们还将证明康托尔定理,该定理指出 P(X) 的基数总是严格大于 X,以及施罗德-伯恩斯坦定理,该定理断言关系 |X| ≤ |Y| 的反对称性。

Definition of Uncountable Sets

不可数集的定义

7.19. Definition: Uncountable Sets. For all sets X , X is uncountable iff X is not countable.

7.19. 定义:不可数集。对于所有集合 X,X 是不可数的当且仅当 X 不是可数的。

The following theorem is readily proved, using Theorem 7.16 .

利用定理 7.16,可以很容易地证明以下定理。

7.20 Theorem on Uncountable Sets. For all sets X and Y :

7.20 关于不可数集的定理。对于所有集合 X 和 Y:

(a) If | X | = | Y |, then X is uncountable iff Y is uncountable.

(a)如果 |X| = |Y|,则 X 是不可数的当且仅当 Y 是不可数的。

(b) If | X | ≤ | Y | and X is uncountable, then Y is uncountable.

(b)如果 |X| ≤ |Y| 且 X 是不可数的,那么 Y 也是不可数的。

(c) If Y has an uncountable subset, then Y is uncountable.

(c)如果 Y 有一个不可数子集,那么 Y 就是不可数的。

Uncountable Sets of Sequences

不可数序列集

To prove that an infinite set X is countable, it suffices to construct one particular bijection from Z > 0 to X . On the other hand, proving the uncountability of X is harder, since we must show that every function from Z > 0 to X fails to be bijective. To provide a specific example of an uncountable set, we need the notion of an infinite sequence.

要证明无限集 X 是可数的,只需构造一个从 Z>0 到 X 的特定双射即可。另一方面,证明 X 是不可数的则更难,因为我们必须证明从 Z>0 到 X 的每个函数都不是双射。为了给出不可数集的具体例子,我们需要用到无限序列的概念。

7.21. Definition: Infinite Sequences. For any set X , an infinite sequence with values in X is a function f : Z > 0 X . When thinking of f as a sequence, we often write f n instead of f ( n ) and call f n the n th term of the sequence .

7.21. 定义:无穷序列。对于任意集合 X,取值于 X 的无穷序列是一个函数 f:Z>0→X。当把 f 看作一个序列时,我们通常用 f n 代替 f(n),并将 f n 称为该序列的第 n 项。

7.22 Theorem on Uncountable Sets of Sequences. For any set X with more than one element, the set S of all infinite sequences with values in X is uncountable.

7.22 关于不可数序列集的定理。对于任何包含多个元素的集合 X,取值于 X 的所有无限序列构成的集合 S 是不可数的。

Proof Fix a set X with more than one element, and fix two elements a b in X . To prove S is not countable, we must show S is not finite and S is not countably infinite. You can prove that S is not finite by exhibiting an injection from the infinite set Z > 0 into S (see Exercise 5). To show that S is not countably infinite, we must prove there does not exist a bijection from Z > 0 onto S . Our strategy is to prove that every function g : Z > 0 S is not surjective (and hence not bijective).

证明:设 X 包含多个元素,且 X 中有两个元素 a ≠ b。为了证明 S 是不可数的,我们必须证明 S 不是有限的,也不是可数无限的。你可以通过证明从无限集 Z>0 到 S 的单射来证明 S 不是有限的(参见练习 5)。为了证明 S 不是可数无限的,我们必须证明不存在从 Z>0 到 S 的双射。我们的策略是证明每个函数 g: Z>0→S 都不是满射(因此也不是双射)。

Fix a function g : Z > 0 S . Surjectivity of g means f S , m Z > 0 , g ( m ) = f . Negating this, we must prove f S , m Z > 0 , g ( m ) f . Note that for each m > 0, g ( m ) is an infinite sequence, i.e., a function from Z > 0 to X . The n th term of the sequence g ( m ) is g ( m )( n ), which we abbreviate as g ( m ) n or g m , n . We can visualize g itself as an infinite table, where row m of the table contains the terms of the sequence g ( m ), as shown here:

设函数 g: Z>0→S。g 的满射性意味着对于任意 f∈S,∃m∈Z>0,g(m)=f。否定此结论,我们必须证明对于任意 m∈Z>0,∃f∈S,g(m)≠f。注意,对于每个 m > 0,g(m) 是一个无穷序列,即一个从 Z>0 到 X 的函数。序列 g(m) 的第 n 项是 g(m)(n),我们将其简写为 g(m) n 或 g m , n 。我们可以将 g 本身可视化为一个无穷表格,其中表格的第 m 行包含序列 g(m) 的项,如下所示:

n = 1

n = 1

n = 2

n = 2

n = 3

n = 3

n = 4

n = 4

n = 5

n = 5

g (1):

g(1):

g 1,1

g 1,1

g 1,2

g 1,2

g 1,3

g 1,3

g 1,4

g 1,4

g 1,5

g 1,5

g (2):

g(2):

g 2,1

g 2,1

g 2,2

g 2,2

g 2,3

g 2,3

g 2,4

g 2,4

g 2,5

g 2,5

g (3):

g(3):

g 3,1

g 3,1

g 3,2

g 3,2

g 3,3

g 3,3

g 3,4

g 3,4

g 3,5

g 3,5

g (4):

g(4):

g 4,1

g 4,1

g 4,2

g 4,2

g 4,3

g 4,3

g 4,4

g 4,4

g 4,5

g 4,5

g (5):

g(5):

g 5,1

g 5,1

g 5,2

g 5,2

g 5,3

g 5,3

g 5,4

g 5,4

g 5,5

g 5,5

Our goal is to build a sequence f that is unequal to every sequence in this table. For each n Z > 0 , define f n = b if g n , n = a , and define f n = a if g n , n a . Fix an arbitrary m Z > 0 . We must prove g ( m ) ≠ f . To prove that the sequence g ( m ) is unequal to the sequence f , it is enough to find one position where these sequences disagree (by definition of equality of functions). By construction of f , we have f m g m , m = g ( m ) m , so that the sequence f and the sequence g ( m ) disagree at position m . Thus g ( m ) ≠ f , as needed.

我们的目标是构造一个序列 f,使其与表中所有序列都不相等。对于每个 n∈Z>0,定义 f n = b 当且仅当 g n , n = a,定义 f n = a 当且仅当 g n , n ≠ a。固定任意 m∈Z>0。我们需要证明 g(m) ≠ f。为了证明序列 g(m) 与序列 f 不相等,只需找到这两个序列不一致的位置(根据函数相等性的定义)。根据 f 的构造,我们有 f m ≠ g m , m = g(m) m ,因此序列 f 和序列 g(m) 在位置 m 处不一致。所以 g(m) ≠ f,证毕。

Here is an example illustrating the preceding proof. Suppose X = {1, 2, 3}, and the table for g looks like this:

下面举一个例子来说明前面的证明。假设 X = {1, 2, 3},且 g 的表格如下所示:

n = 1

n = 1

n = 2

n = 2

n = 3

n = 3

n = 4

n = 4

n = 5

n = 5

g (1):

g(1):

1

1

1

1

1

g (2):

g(2):

2

2

2

2

2

g (3):

g(3):

3

3

3

3

3

g (4):

g(4):

1

2

1

2

1

g (5):

g(5):

3

1

2

2

1

To build a sequence f not appearing in this table, we read down the main diagonal of the table and change every entry. Taking a = 1 and b = 2, the sequence f for this g has first five terms f 1 = 2, f 2 = 1, f 3 = 1, f 4 = 1, and f 5 = 2. We see that f g (1) since these sequences disagree in position 1; f g (2) since these sequences disagree in position 2; and so on. This construction is called Cantor’s diagonal argument .

为了构造一个表中未出现的序列 f,我们沿着表格的主对角线向下读取,并更改每个元素。取 a = 1 和 b = 2,则此 g 对应的序列 f 的前五项分别为 f 1 = 2,f 2 = 1,f 3 = 1,f 4 = 1 和 f 5 = 2。我们可以看出,f ≠ g(1),因为这两个序列在位置 1 处不一致;f ≠ g(2),因为这两个序列在位置 2 处不一致;依此类推。这种构造方法称为康托尔对角线论证。

Uncountable Sets of Real Numbers

不可数实数集

Now that we have some uncountable sets available, we can use bijections and injections to show that certain subsets of R are uncountable.

现在我们有了一些不可数集,我们可以利用双射和单射来证明 R 的某些子集是不可数的。

7.23 Theorem on Uncountable Subsets of R The following sets are uncountable: the set R of real numbers; the open interval (0, 1); the open interval ( a , b ) for all a < b in R ; and the closed interval [ a , b ] for all a < b in R .

7.23 关于 R 的不可数子集的定理 下列集合是不可数的:实数集 R;开区间 (0, 1);对于 R 中的所有 a < b,开区间 (a, b);以及对于 R 中的所有 a < b,闭区间 [a, b]。

Proof. First we show (0, 1) is uncountable. We need the following fact about decimal representations of real numbers, which we state without proof: for every real number r ∈ (0, 1), there exists exactly one infinite sequence f with values in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} such that r = k = 1 f ( k ) 10 k , and ¬ k , m k , f ( m ) = 9 . Informally, f ( k ) is the k th digit after the decimal point in the decimal expansion of r . To ensure uniqueness, we forbid expansions that end in an infinite string of 9s. For example, taking r = 1/3, the sequence f has f ( k ) = 3 for all k Z > 0 . Taking r = 1/2 = 0.5000 … =0.4999 … , the sequence f has f (1) = 5 and f ( k ) = 0 for all k > 1.

证明。首先,我们证明 (0, 1) 是不可数的。我们需要以下关于实数十进制表示的事实(此处不作证明):对于每个实数 r ∈ (0, 1),存在唯一一个取值于 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 的无限序列 f,使得 r = ∑k=1∞ f(k)10−k,且对于所有 m≥k,都有 f(m) = 9。通俗地说,f(k) 是 r 的十进制展开式中小数点后第 k 位数字。为了确保唯一性,我们禁止以无限个 9 结尾的展开式。例如,取 r = 1/3,则对于所有 k∈Z>0,序列 f 都有 f(k) = 3。取 r = 1/2 = 0.5000 … =0.4999 …,则序列 f 有 f(1) = 5 且对于所有 k > 1,f(k) = 0。

Now let S be the set of all infinite sequences with values in X = {4, 7}. We already know S is uncountable. Define G : S → (0, 1) by setting G ( f ) = k = 1 f ( k ) 10 k for f S . Using calculus, you can check that this infinite series does converge to some number in (0, 1). The fact cited above ensures that G is injective, so | S | ≤ |(0, 1)|. By part (b) of Theorem 7.20 , (0, 1) is uncountable.

现在令 S 为所有取值于 X = {4, 7} 的无穷序列的集合。我们已知 S 是不可数的。定义 G : S → (0, 1),令 G(f) = ∑k=1∞f(k)10−k,其中 f ∈ S。利用微积分,可以验证该无穷级数收敛于 (0, 1) 中的某个数。上述事实保证了 G 是单射,因此 |S| ≤ |(0, 1)|。根据定理 7.20 的 (b) 部分,(0, 1) 是不可数的。

Next, fix a , b R with a < b . The map h : (0, 1) → ( a , b ), defined by h ( t ) = a + ( b a ) t for 0 < t < 1, is readily seen to be a bijection. Thus |( a , b )| = |(0, 1)|, so the open interval ( a , b ) is uncountable. It follows that [ a , b ] and R itself are uncountable, since the uncountable set ( a , b ) is a subset of each of these sets.

接下来,固定 a, b∈R,且 a < b。映射 h : (0, 1) → (a, b),定义为 h(t) = a + (b − a)t,其中 0 < t < 1,显然是双射。因此,|(a, b)| = |(0, 1)|,所以开区间 (a, b) 是不可数的。由此可知,[a, b] 和 R 本身也是不可数的,因为不可数集 (a, b) 分别是它们的子集。

We now present another proof that closed intervals in R are uncountable. This proof does not rely on the existence and uniqueness of decimal expansions of real numbers. Instead, we need a much more elementary fact called the Nested Interval Property. This property says that for any sequence of nonempty closed intervals [ a n , b n ] such that [ a n +1 , b n +1 ]⊆ [ a n , b n ] for all n Z > 0 , the intersection n = 1 [ a n , b n ] is nonempty. See §8.5 for a proof of this property.

现在我们给出实数集 R 中闭区间不可数的另一种证明。此证明不依赖于实数十进制展开式的存在性和唯一性。相反,我们需要一个更基本的性质,称为嵌套区间性质。该性质指出,对于任意非空闭区间序列 [a n , b n ],如果对于所有 n∈Z>0,都有 [a n +1 , b n +1 ]⊆ [a n , b n ],则它们的交集 ⋂n=1∞[an,bn] 非空。有关此性质的证明,请参见 §8.5。

7.24 Theorem on Uncountability of Closed Intervals. For all real a < b , [ a , b ] is uncountable.

7.24 闭区间不可数性定理。对于所有实数 a < b,[a, b] 是不可数的。

Proof. Fix real a < b . We fix an arbitrary function (or sequence) f : Z > 0 [ a , b ] and prove that f cannot be surjective. We recursively construct a sequence of nested intervals [ a n , b n ] for n Z 0 with the property that [ a n , b n ]⊆ [ a , b ], a n < b n , and [ a n , b n ] { f ( 1 ) , f ( 2 ) , , f ( n ) } = 0 for all n ≥ 0. To begin, let a 0 = a and b 0 = b . Now, fix n Z 0 and assume a 0 , a 1 , …, a n , b 0 , b 1 , …, b n have already been chosen in a way that satisfies the property above. Let d = b n a n > 0 be the length of the interval [ a n , b n ], and consider the three subintervals [ a n , a n + d /3], [ a n + d /3, a n + 2 d /3], and [ a n + 2 d /3, b n ]. These three intervals are subsets of [ a n , b n ] and hence do not contain any of the real numbers f (1), …, f ( n ). Furthermore, at most two of the three subintervals can contain f ( n + 1). [It is possible for f ( n + 1) to belong to two of the intervals; this happens when f ( n + 1) = a n + d /3 or f ( n + 1) = a n + 2 d /3.] Choose [ a n +1 , b n +1 ] to be the first of the three intervals that does not contain f ( n + 1). Then [ a n +1 , b n +1 ]⊆ [ a n , b n ], a n +1 < b n +1 , and [ a n + 1 , b n + 1 ] { f ( 1 ) , , f ( n ) , f ( n + 1 ) } = 0 . By recursion, this defines [ a n , b n ] for all n Z 0 . To finish, invoke the Nested Interval Property to see that there must exist a real number r in the intersection of all the intervals [ a n , b n ]. For fixed n Z > 0 , we know r ∈ [ a n , b n ] and hence r f ( n ) by construction. Since n was arbitrary, we see that r is a number in [ a , b ] that is outside the image of f . Thus, f is not surjective and so not bijective. So [ a , b ] must be uncountable.     □

证明。固定实数 a < b。我们固定任意函数(或序列)f:Z>0→[a,b],并证明 f 不可能是满射。我们递归构造一个嵌套区间序列 [a n , b n ],其中 n∈Z≥0,且满足 [a n , b n ]⊆ [a, b],a n < b n ,以及对于所有 n ≥ 0,[an,bn]∩{f(1),f(2),…,f(n)}=0。首先,令 a 0 = a 且 b 0 = b。现在,固定 n∈Z≥0,并假设 a 0 , a 1 , …, a n , b 0 , b 1 , …, b n 已经以满足上述性质的方式选择。令 d = b n − a n > 0 为区间 [a n , b n ] 的长度,并考虑三个子区间 [a n , a n + d/3]、[a n + d/3, a n + 2d/3] 和 [a n + 2d/3, b n ]。这三个区间是 [a n , b n ] 的子集,因此不包含任何实数 f(1), …, f(n)。此外,这三个子区间中至多有两个包含 f(n + 1)。 [f(n + 1) 有可能属于两个区间;这种情况发生在 f(n + 1) = a n + d/3 或 f(n + 1) = a n + 2d/3 时。] 选择 [a n +1 , b n +1 ] 作为三个区间中第一个不包含 f(n + 1) 的区间。那么 [a n +1 , b n +1 ]⊆ [a n , b n ],a n +1 < b n +1 ,并且 [an+1,bn+1]∩{f(1),…,f(n),f(n+1)}=0。通过递归,对于所有 n∈Z≥0,都有 [a n , b n ]。最后,利用嵌套区间性质可知,所有区间 [a n , b n ] 的交集中必然存在一个实数 r。对于固定的 n∈Z>0,我们知道 r ∈ [a n , b n ],因此根据构造可知 r ≠ f(n)。由于 n 是任意的,我们看到 r 是 [a, b] 中一个不在 f 的像集上的数。因此,f 不是满射,所以也不是双射。所以 [a, b] 一定是不可数的。□

Cantor’s Theorem

康托尔定理

We now prove the remarkable fact that for every set X , the power set P ( X ) (the set of all subsets of X ) has strictly larger cardinality than X .

现在我们证明了一个惊人的事实:对于每个集合 X,幂集 P(X)(X 的所有子集的集合)的基数严格大于 X。

7.25 Cantor’s Theorem. For any set X , | X | < | P ( X ) | .

7.25 康托尔定理。对于任意集合 X,|X|<|P(X)|。

Proof. Fix a set X . We must show | X | | P ( X ) | and | X | | P ( X ) | . On one hand, the map k : X P ( X ) given by k ( x ) = { x } for x X is readily seen to be an injection. So | X | | P ( X ) | . On the other hand, to show | X | | P ( X ) | , we must prove that every function f : X P ( X ) is non-bijective. Fix such a function f ; it suffices to prove f is non-surjective. So, we must show S P ( X ) , x X , S f ( x ) .

证明。固定一个集合 X。我们必须证明 |X|≤|P(X)| 且 |X|≠|P(X)|。一方面,映射 k:X→P(X),对于 x ∈ X,由 k(x) = {x} 给出,显然是单射。因此 |X|≤|P(X)|。另一方面,为了证明 |X|≠|P(X)|,我们必须证明每个函数 f:X→P(X) 都是非双射的。固定这样一个函数 f;只需证明 f 是非满射即可。因此,我们必须证明存在 S∈P(X),∀x∈X,S≠f(x)。

The rest of the proof is a variation of Cantor’s diagonal argument for sequences. We choose S = { z X : z f ( z ) } , which is a subset of X and hence a member of P ( X ) . We prove x X , S f ( x ) by contradiction. Assume, to get a contradiction, that x 0 X , S = f ( x 0 ) . Let us ask whether the object z = x 0 is in S . By definition of S and the fact that S = f ( x 0 ), x 0 S iff x 0 f ( x 0 ) iff x 0 S . We now have reached the contradiction “ x 0 S x 0 S .”

证明的其余部分是康托尔对数列的对角线论证的变体。我们选择 S={z∈X:z∉f(z)},它是 X 的一个子集,因此也是 P(X) 的成员。我们用反证法证明 ∀x∈X,S≠f(x)。为了得到矛盾,假设 ∃x0∈X,S=f(x0)。我们来问对象 z = x 0 是否在 S 中。根据 S 的定义以及 S = f(x 0 ),x 0 ∈ S 当且仅当 x0∉f(x0) 当且仅当 x0∉S。现在我们得到了矛盾“x0∈S⇔x0∉S”。

Taking X = Z > 0 in this theorem, we see that P ( Z > 0 ) is uncountable. As another example, just as the uncountable set R is a larger infinite set than Z or Q , we see from Cantor’s Theorem that P ( R ) is an even larger set than R . Iterating the theorem, we get a chain of uncountable sets, each strictly larger in cardinality than the previous one:

在该定理中取 X=Z>0,我们可以看出 P(Z>0) 是不可数的。再举一个例子,正如不可数集 R 比 Z 或 Q 更大一样,根据康托尔定理,P(R) 比 R 更大。重复该定理,我们得到一个不可数集链,每个集的基数都严格大于前一个集:

| R | < | P P ( R ) | < | P P ( P P ( R ) ) | < | P P ( P P ( P P ( R ) ) ) | < .

The mind-boggling tower of infinities never ends; for if we tentatively assumed there were a largest set X , Cantor’s Theorem would provide an even larger set P ( X ) .

无穷大之塔令人难以置信,永无止境;因为如果我们暂时假设存在一个最大的集合 X,康托尔定理将提供一个更大的集合 P(X)。

Proof of the Schröder–Bernstein Theorem (Optional)

施罗德-伯恩斯坦定理的证明(可选)

We conclude our introduction to cardinality by proving the Schröder–Bernstein Theorem. We begin with a special case of the main result.

我们以证明施罗德-伯恩斯坦定理来结束对基数的介绍。我们首先从主要结果的一个特例开始。

7.26 Lemma. For all sets X and W , if W X and there exists an injective function f : X W , then there exists a bijective function h : X W .

7.26 引理。对于所有集合 X 和 W,如果 W⊆ X 且存在单射函数 f : X → W,则存在双射函数 h : X → W。

Proof . Fix arbitrary sets X and W . Assume W X and f : X W is a one-to-one function. When X is infinite, the function f may not be onto. The following ingenious construction uses f to build a bijection h : X W . Recursively define a sequence of sets

证明。固定任意集合 X 和 W。假设 W⊆ X 且 f : X → W 是一个单射函数。当 X 为无穷集时,函数 f 可能不是满射。以下巧妙的构造利用 f 构造了一个双射 h : X → W。递归地定义一个集合序列

A 一个 0 A 一个 1 A 一个 2 A 一个 n n A 一个 n n + 1

by letting A 0 = X , A 1 = W and A n +2 = f [ A n ] for all n Z 0 . We check by induction on n that A n +1 A n for all n Z 0 . This is true when n = 0 because W X by assumption. This is true when n = 1 because A 2 = f [ X ], A 1 = W , and we assumed f maps X into the codomain W . Now fix n Z 2 , and assume that A n A n −1 A n −2 ⊆ … ⊆ A 0 is already known. We must prove A n +1 A n . Since A n −1 A n −2 , we conclude that f [ A n −1 ]⊆ f [ A n −2 ], so A n +1 A n as needed.

令 A 0 = X,A 1 = W,A n +2 = f[A n ],其中 n∈Z≥0。我们对 n 进行归纳检验,可知对于所有 n∈Z≥0,A n +1 ⊆ A n 。当 n = 0 时,此结论成立,因为根据假设,W⊆ X。当 n = 1 时,此结论成立,因为 A 2 = f[X],A 1 = W,并且我们假设 f 将 X 映射到值域 W。现在固定 n∈Z≥2,并假设 A n ⊆ A n −1 ⊆ A n −2 ⊆ … ⊆ A 0 已知。我们必须证明 A n +1 ⊆ A n 。由于 A n −1 ⊆ A n −2 ,我们得出结论 f[A n −1 ]⊆ f[A n −2 ],因此 A n +1 ⊆ A n ,符合要求。

The next step is to define B n = A n A n +1 for all n Z 0 and B = n = 0 A n . It is routine to check that the sets B 0 , B 1 , …, B n , …, B are pairwise disjoint and have union X , whereas the sets B 1 , B 2 , …, B n , …, B are pairwise disjoint and have union W . See Figure 7.8 , where X is the entire box, and W is the part of the box below the thick line.

下一步是定义 B n = A n − A n +1 ,其中 n∈Z≥0,且 B=⋂n=0∞An。可以简单地验证集合 B 0 , B 1 , …, B n , …, B 两两不相交且其并集为 X,而集合 B 1 , B 2 , …, B n , …, B 两两不相交且其并集为 W。参见图 7.8,其中 X 为整个方框,W 为粗线以下的方框部分。

fig7_8

Figure 7.8

图 7.8

Visualizing the proof of the Schröder–Bernstein Theorem.

将施罗德-伯恩斯坦定理的证明可视化。

A crucial observation is that f [ B n ] = B n +2 for all n Z 0 . To verify this, note that injectivity of f implies f [ A C ] = f [ A ] − f [ C ] for all A , C X (see Exercise 14 in §5.8 ). So for each n Z 0 ,

一个关键的观察结果是,对于所有 n∈Z≥0,都有 f[B n ] = B n +2 。为了验证这一点,注意到 f 的单射性意味着对于所有 A, C⊆ X,f[A − C] = f[A] − f[C](参见 §5.8 中的练习 14)。因此,对于每个 n∈Z≥0,

f f [ B B n n ] = f f [ A 一个 n n A 一个 n n + 1 ] = f f [ A 一个 n n ] f f [ A 一个 n n + 1 ] = A 一个 n n + 2 A 一个 n n + 3 = B B n n + 2 .

Next, using injectivity of f , f [ B ] = B follows from the computation

接下来,利用 f 的单射性,可得 f[B] = B。

f f [ B B ] = f f [ n n = 0 A 一个 n n ] = n n = 0 f f [ A 一个 n n ] = n n = 0 A 一个 n n + 2 = j j = 2 A 一个 j j = B B .

(The last equality uses the fact that A 2 A 1 A 0 .)

(最后一个等式利用了 A 2 ⊆ A 1 ⊆ A 0 这一事实。)

Since f is injective, all the restrictions f | B n and f | B are also injective. Since f [ B ] = B and f [ B n ] = B n +2 for all n Z 0 , we can restrict the codomains of f | B n and f | B to get bijections f n : B n B n +2 and f : B B . Now comes the final miraculous idea: define a function h by

由于 f 是单射,所有限制 f|Bn 和 f| B 也都是单射。由于对于所有 n∈Z≥0,都有 f[B] = B 且 f[B n ] = B n +2 ,我们可以限制 f|Bn 和 f| B 的值域,得到双射 f n : B n → B n +2 和 f∗:B→B。现在,最后一个奇妙的想法出现了:定义一个函数 h,

h h ( x x ) = { f f n n ( x x ) if 如果 x x B B n n where 在哪里 n n Z Z 0 is even; 是偶数; x x if 如果 x x B B m where 在哪里 m Z Z > 0 is odd; 很奇怪; f f * ( x x ) if 如果 x x B B .

As shown on the left edge of the box in Figure 7.8 , h is defined by gluing together the bijections f n : B n B n +2 for n even, I d B m : B m B m for m odd, and f : B B . The domains of these bijections are the pairwise disjoint sets B 0 , B 1 , B 2 , B 3 , B 4 , …, B whose union is X ; and the codomains of these bijections are the pairwise disjoint sets B 2 , B 1 , B 4 , B 3 , B 6 , …, B whose union is W . We see at once that h is a well-defined function mapping the domain X onto the codomain W , and h is one-to-one. So we have constructed the required bijection h : X W .      □

如图 7.8 中方框左侧所示,h 定义为将以下双射粘合在一起:当 n 为偶数时,f n : B n → B n +2 ;当 m 为奇数时,IdBm:Bm→Bm;以及 f∗:B→B。这些双射的定义域是两两不相交的集合 B 0 , B 1 , B 2 , B 3 , B 4 , …, B,它们的并集为 X;这些双射的陪域是两两不相交的集合 B 2 , B 1 , B 4 , B 3 , B 6 , …, B,它们的并集为 W。我们立即看出 h 是一个定义良好的函数,它将定义域 X 映射到陪域 W,并且 h 是一一对应的。因此,我们构造了所需的双射 h : X → W。□

7.27 Schröder–Bernstein Theorem. For all sets X and Y , if | X | ≤ | Y | and | Y | ≤ | X | then | X | = | Y |.

7.27 施罗德-伯恩斯坦定理。对于所有集合 X 和 Y,如果 |X| ≤ |Y| 且 |Y| ≤ |X|,则 |X| = |Y|。

Proof. Fix sets X and Y . Assume | X | ≤ | Y | and | Y | ≤ | X |, which means there are injections F : X Y and G : Y X . We must prove | X | = | Y |, which means there is a bijection H : X Y . Define W = G [ Y ], which is a subset of X . Define f : X W by f ( x ) = G ( F ( x )) for all x X . Note that f is injective, since it is obtained from the injective function g F : X X by shrinking the codomain to W . Similarly, if we shrink the codomain of the injective function G to W = G [ Y ], we obtain a bijective function G 1 : Y W . Since W X and f : X W is injective, Lemma 7.26 shows that there is a bijection h : X W . Then H = G 1 1 h : X Y is a bijection, since it is the composition of the bijections h and G 1 1 .

证明。固定集合 X 和 Y。假设 |X| ≤ |Y| 且 |Y| ≤ |X|,这意味着存在单射 F : X → Y 和 G : Y → X。我们必须证明 |X| = |Y|,这意味着存在双射 H : X → Y。定义 W = G[Y],它是 X 的一个子集。定义 f : X → W,使得对于所有 x ∈ X,f(x) = G(F(x))。注意 f 是单射,因为它是由单射函数 g ○ F : X → X 通过将值域缩小到 W 而得到的。类似地,如果我们将单射函数 G 的值域缩小到 W = G[Y],则得到一个双射函数 G 1 : Y → W。由于 W⊆ X 且 f : X → W 是单射,引理 7.26 表明存在双射 h : X → W。因此,H=G1−1∘h:X→Y 是一个双射,因为它是双射 h 和 G1−1 的复合。

7.28 Corollary: No-Universe Theorem. There is no set of all sets.

7.28 推论:无宇宙定理。不存在包含所有集合的集合。

Proof. Assume, to get a contradiction, that X is a set such that for all sets Y , Y X . Consider the power set P ( X ) consisting of all subsets of X . Every Y P ( X ) is a set, so every such Y is in X by hypothesis, so P ( X ) X . This implies | P ( X ) | | X | , and we saw earlier that | X | | P ( X ) | . Now | X | = | P ( X ) | follows from the Schröder–Bernstein Theorem; but this contradicts Cantor’s Theorem.

证明。假设为了得出矛盾,X 是一个集合,使得对于所有集合 Y,都有 Y ∈ X。考虑由 X 的所有子集构成的幂集 P(X)。每个 Y∈P(X) 都是一个集合,因此根据假设,每个这样的 Y 都属于 X,所以 P(X)⊆X。这意味着 |P(X)|≤|X|,而我们之前已经看到 |X|≤|P(X)|。现在,根据 Schröder-Bernstein 定理,|X|=|P(X)| 成立;但这与康托尔定理相矛盾。

Section Summary

章节概要

  1. Uncountable Sets. A set X is uncountable iff X is not countable. To prove that an infinite set is uncountable, it suffices to prove that every f : Z > 0 X is not surjective. If | X | = | Y |, then X is uncountable iff Y is uncountable. If | X | ≤ | Y | and X is uncountable, then Y is uncountable. If Y has an uncountable subset, then Y is uncountable.
    不可数集。集合 X 是不可数的当且仅当 X 不可数。要证明一个无限集是不可数的,只需证明每个 f:Z>0→X 都不是满射。如果 |X| = |Y|,则 X 是不可数的当且仅当 Y 不可数。如果 |X| ≤ |Y| 且 X 不可数,则 Y 不可数。如果 Y 有一个不可数子集,则 Y 不可数。
  2. Examples of Uncountable Sets. R is uncountable. For all a < b in R , the intervals ( a , b ) and [ a , b ] are uncountable. The set P ( Z > 0 ) is uncountable. If X has more than one element, the set of infinite sequences with values in X is uncountable.
    不可数集的例子。实数集 R 是不可数的。对于 R 中所有 a < b,区间 (a, b) 和 [a, b] 都是不可数的。集合 P(Z>0) 是不可数的。如果 X 包含多个元素,则取值于 X 的所有无限序列构成的集合也是不可数的。
  3. Cantor’s Theorem. For every set X , | X | < | P ( X ) | . The proof uses the diagonal method, starting with any given function f : X P ( X ) and constructing a subset of X not in the image of f . One consequence of Cantor’s Theorem is that there is no set of all sets.
    康托尔定理。对于任意集合 X,|X|<|P(X)|。证明采用对角线方法,从任意给定的函数 f:X→P(X) 出发,构造 X 的一个子集,该子集不在 f 的像中。康托尔定理的一个推论是:不存在包含所有集合的集合。
  4. Schröder–Bernstein Theorem. For all sets X and Y , if | X | ≤ | Y | and | Y | ≤ | X |, then | X | = | Y |. The proof combines injections F : X Y and G : Y X to build a bijection H : X Y . When Y X and G : Y X is the inclusion map, the construction of H is visualized in Figure 7.8 .
    施罗德-伯恩斯坦定理。对于任意集合 X 和 Y,若 |X| ≤ |Y| 且 |Y| ≤ |X|,则 |X| = |Y|。证明结合了单射 F : X → Y 和 G : Y → X 来构造双射 H : X → Y。当 Y⊆ X 且 G : Y → X 为包含映射时,H 的构造过程如图 7.8 所示。

Exercises

练习

  1. For fixed a < b in R , show that h : (0, 1) → ( a , b ), given by h ( t ) = a + ( b a ) t for t ∈ (0, 1), is a bijection.
    对于 R 中固定的 a < b,证明由 h(t) = a + (b − a)t 给出的映射 h : (0, 1) → (a, b),其中 t ∈ (0, 1),是一个双射。
  2. Given that the interval [0, 1] is uncountable, prove that every open interval ( a , b ) is uncountable without using Theorem 7.23 .
    已知区间 [0, 1] 是不可数的,证明每个开区间 (a, b) 都是不可数的,不要使用定理 7.23。
  3. Show that f : ( 1 , 1 ) R , given by f ( x ) = x /(1 − x 2 ) for x ∈ ( − 1, 1), is a bijection.
    证明 f:(−1,1)→R,其中 f(x) = x/(1 − x 2 ),x ∈ ( − 1, 1),是一个双射。
  4. Prove Theorem 7.20 .
    证明定理 7.20。
  5. (a) Complete the proof of Theorem 7.22 by constructing an injection g : Z > 0 S . (b) The proof shows that the map g in part (a) cannot be surjective. Find the specific sequence f constructed by the proof that is outside the image of g .
    (a) 通过构造单射 g:Z>0→S 完成定理 7.22 的证明。(b) 证明表明 (a) 部分中的映射 g 不可能是满射。找出证明中构造的、位于 g 的像之外的特定序列 f。
  6. Use calculus to show that the series k = 1 f ( k ) 10 k in the proof of Theorem 7.23 converges to a real number in (0, 1). [ Hint: Compare to the geometric series k 1 ( 4 / 10 ) k and k 1 ( 7 / 10 ) k .]
    利用微积分证明定理 7.23 证明中的级数 ∑k=1∞f(k)10−k 收敛于 (0, 1) 中的一个实数。[提示:与几何级数 ∑k≥1(4/10)k 和 ∑k≥1(7/10)k 进行比较。]
  7. (a) What goes wrong with the proof of Theorem 7.23 if we use X = {0, 4} instead of X = {4, 7}? (b) What goes wrong with the proof if we use X = {7, 9} instead of X = {4, 7}?
    (a) 如果用 X = {0, 4} 代替 X = {4, 7},定理 7.23 的证明会出什么问题?(b) 如果用 X = {7, 9} 代替 X = {4, 7},证明会出什么问题?
  8. Let X be a fixed countably infinite set, and let Y be a fixed uncountable set. Decide (with proof) whether each set of functions is countable or uncountable. (a) the set of all f : {1, 2, 3} → X (b) the set of all g : {1, 2, 3} → Y (c) the set of all h : X → {1, 2, 3} (d) the set of all k : Y → {1, 2, 3}
    设 X 为一个固定的可数无限集,Y 为一个固定的不可数集。判断(并证明)下列函数集是可数集还是不可数集。 (a) 所有 f : {1, 2, 3} → X 的集合 (b) 所有 g : {1, 2, 3} → Y 的集合 (c) 所有 h : X → {1, 2, 3 的集合} (d) 所有 k : Y → {1, 2, 3 的集合}
  9. In the proof of Lemma 7.26 , check that the sets B 0 , B 1 , …, B n , …, B are pairwise disjoint, the union of these sets is X , and the union of these sets excluding B 0 is W .
    在引理 7.26 的证明中,验证集合 B 0 , B 1 , …, B n , …, B 两两不相交,这些集合的并集为 X,这些集合除 B 0 外的并集为 W。
  10. (a) Prove that for any injective function f : X Y and all sets A n X , f [ n = 0 A n ] = n = 0 f [ A n ] . (b) In the proof of Lemma 7.26 , explain in detail why j = 2 A j = B .
    (a)证明对于任意单射函数 f : X → Y 和所有集合 A n ⊆ X,f[⋂n=0∞An]=⋂n=0∞f[An]。(b)在引理 7.26 的证明中,详细解释为什么 ⋂j=2∞Aj=B。
  11. Let W = { 2 k : k Z } , and define an injection f : Z W by f ( n ) = 4 n for all n Z . In the proof of Lemma 7.26 , describe the sets A n , B n , B , and the bijection h : Z W .
    令 W={2k:k∈Z},并定义一个单射 f:Z→W,使得对于所有 n∈Z,f(n) = 4n。在引理 7.26 的证明中,描述集合 A n 、B n 、B 和双射 h:Z→W。
  12. Define an injection f : Z 0 Z 1 by f ( n ) = n + 3 for all n Z 0 . In the proof of Lemma 7.26 , describe the sets A n , B n , B , and the bijection h : Z 0 Z 1 .
    定义一射 f:Z≥0→Z≥1,使得对于所有 n∈Z≥0,f(n) = n + 3。在引理 7.26 的证明中,描述集合 A n 、B n 、B 和双射 h:Z≥0→Z≥1。
  13. In Lemma 7.26 , suppose f : X W happens to be a bijection. Describe the sets A n , B n , B and the bijection h : X W in this case.
    在引理 7.26 中,假设 f : X → W 恰好是一个双射。描述在这种情况下集合 A n 、B n 、B 和双射 h : X → W。
  14. In Lemma 7.26 , suppose W = X . Describe the sets A n , B n , B , and the bijection h : X W in this case.
    在引理 7.26 中,假设 W = X。描述此情况下的集合 A n 、B n 、B 和双射 h : X → W。
  15. Define F , G : Z > 0 Z > 0 by F ( n ) = 2 n and G ( n ) = 2 n − 1 for n Z > 0 . Trace through the proof of the Schröder–Bernstein Theorem to construct a bijection H : Z > 0 Z > 0 from F and G .
    定义 F,G:Z>0→Z>0,其中 F(n) = 2n,G(n) = 2n − 1,n∈Z>0。通过 Schröder–Bernstein 定理的证明,构造从 F 和 G 到 Z>0 的双射 H:Z>0→Z>0。
  16. For any sets X and Y , let Y X be the set of all functions f : X Y . Prove: if | X | = k and | Y | = n , then | Y X | = n k .
    对于任意集合 X 和 Y,令 Y X 为所有函数 f : X → Y 的集合。证明:如果 |X| = k 且 |Y| = n,则 |Y X | = n k
  17. Use a bijection to prove: for all sets X , Y , Z , if X Y = 0 then | Z X Y | = | Z X × Z Y | .
    利用双射证明:对于所有集合 X、Y、Z,如果 X∩Y=0,则 |ZX∪Y|=|ZX×ZY|。
  18. Use a bijection to prove: for all sets X , Y , Z , |( Z Y ) X | = | Z X × Y |.
    利用双射证明:对于所有集合 X、Y、Z,|(Z Y ) X | = |Z X × Y |。
  19. Given sets X and Y , a function f : X Y , and an injection g : Y X , show there exist sets A , B , C , D such that { A , B } is a set partition of X , { C , D } is a set partition of Y , C = f [ A ], and B = g [ D ]. ( Hint: Let Z = X g [ Y ] and h = g f ; then define A to be the intersection of all sets U X with Z h [ U ] U .)
    给定集合 X 和 Y,函数 f : X → Y,以及单射 g : Y → X,证明存在集合 A、B、C、D,使得 {A, B} 是 X 的一个集合划分,{C, D} 是 Y 的一个集合划分,C = f[A],B = g[D]。(提示:令 Z = X − g[Y],h = g ○ f;然后定义 A 为所有满足 Z∪h[U]⊆U 的集合 U⊆ X 的交集。)
  20. Use the previous exercise to give another proof of the Schröder–Bernstein Theorem.
    利用前面的练习,给出施罗德-伯恩斯坦定理的另一个证明。
  21. A set S Z > 0 is closed under multiplication iff a , b S , a b S . Let X be the set of all subsets of Z > 0 that are closed under multiplication. Is X countable or uncountable? Prove your answer.
    集合 S⊆Z>0 对乘法封闭当且仅当对于任意 a,b∈S,a⋅b∈S。设 X 为 Z>0 的所有对乘法封闭的子集构成的集合。X 是可数集还是不可数集?证明你的答案。
  22. A set S Z > 0 is closed under addition iff a , b S , a + b S . Let Y be the set of all subsets of Z > 0 that are closed under addition. Is Y countable or uncountable? Prove your answer.
    设集合 S⊆Z>0 对加法封闭当且仅当对于任意 a,b∈S,a+b∈S。令 Y 为 Z>0 的所有对加法封闭的子集构成的集合。Y 是可数集还是不可数集?证明你的答案。
#

Review of Functions, Relations, and Cardinality

函数、关系和基数回顾

Tables 7.1–7.6 on the following pages review the main definitions and theorems covered in the preceding chapters on functions, relations, and cardinality. All unquantified variables in the tables represent arbitrary objects, sets, functions, or relations (as required by context) .

以下各页的表 7.1 至 7.6 回顾了前几章中关于函数、关系和基数的主要定义和定理。表中所有未量化的变量均代表任意对象、集合、函数或关系(视上下文而定)。

TABLE 7.1 Function Definitions. Here, f is a function with domain A and codomain B .

表 7.1 函数定义。其中,f 是一个定义域为 A,值域为 B 的函数。

Concept

概念

Definition

定义

function [formal version]

函数[正式版本]

ordered triple f = ( A , B , G ) where A and B are sets,

有序三元组 f = (A, B, G) 其中 A 和 B 是集合,

G A × B , and x A , ! y , ( x , y ) G .

G ⊆ A × B,且 ∀x∈A,∃!y,(x,y)∈G。

A is the domain of f , B is the codomain of f , and

A 是函数 f 的定义域,B 是函数 f 的值域,

G is the graph of f ; y = f ( x ) means ( x , y ) ∈ G .

G 是函数 f 的图像;y = f(x) 表示 (x, y) ∈ G。

f : A B

f : A → B

f is a function with domain A and codomain B , meaning:

f 是一个定义域为 A,值域为 B 的函数,这意味着:

x , y , y = f ( x ) ( x A y B ) , and

∀x,∀y,y=f(x)⇒(x∈A∧y∈B),且

x A , ! y , y = f ( x ) .

∀x∈A,∃!y,y=f(x)。

graph of f : A B

函数 f : A → B 的图像

G = {( x , f ( x )) : x A }.

G = {(x, f(x)) : x ∈ A}。

f is injective (one-to-one)

f 是单射(一对一)

a A , c A , f ( a ) = f ( c ) a = c .

∀a∈A,∀c∈A,f(a)=f(c)⇒a=c。

f is surjective (onto)

f 是满射(到)

y B , x A , y = f ( x ) .

∀y∈B,∃x∈A,y=f(x)。

f is bijective

f 是双射

f is one-to-one and f is onto; i.e., y B , ! x A , y = f ( x ) .

f 是单射且 f 是满射;即,∀y∈B,∃!x∈A,y=f(x)。

direct image f [ C ]

直接图像 f[C]

y f [ C ] iff x C , y = f ( x ) .

y ∈ f[C] 当且仅当 ∃x∈C,y=f(x)。

preimage f −1 [ D ]

原像 f −1 [D]

x f −1 [ D ] iff x A and f ( x ) ∈ D .

x ∈ f −1 [D] 当且仅当 x ∈ A 且 f(x) ∈ D。

function equality

函数相等性

For f : A B and g : A ′ → B ′,

对于 f : A → B 和 g : A′ → B′,

f = g iff A = A ′ and B = B ′ and x A , f ( x ) = g ( x ) .

f = g 当且仅当 A = A′ 且 B = B′ 且 ∀x∈A,f(x)=g(x)。

identity function Id A

恒等函数 Id A

Id A : A A where Id A ( x ) = x for all x A .

IdA:A→A 其中对于所有 x ∈ A,IdA(x)=x。

constant function c

常数函数 c

f : A B where f ( x ) = c for all x A ( c B is fixed).

f : A → B 其中 f(x) = c 对所有 x ∈ A 成立(c ∈ B 是固定的)。

inclusion map j A , B

包含映射 j A , B

j : A B where j ( x ) = x for all x A .

j : A → B 其中 j(x) = x 对于所有 x ∈ A。

pointwise sum f + h

逐点求和 f + h

( f + h )( x ) = f ( x ) + h ( x ) for x A (codomain is R ).

对于 x ∈ A(值域为 R),(f + h)(x) = f(x) + h(x)。

pointwise product f · h

逐点乘积 f · h

( f · h )( x ) = f ( x ) h ( x ) for x A (codomain is R ).

(f · h)(x) = f(x)h(x),其中 x ∈ A(值域为 R)。

composition g f

组成 g ∘ f

For f : A B and g : B C ,

对于 f : A → B 和 g : B → C,

g f : A C satisfies ( g f )( x ) = g ( f ( x )) for x A .

g ∘ f : A → C 满足 (g ∘ f)(x) = g(f(x)),其中 x ∈ A。

inverse of f : A B

f 的逆函数:A → B

f −1 : B A where x A , y B , y = f ( x ) x = f 1 ( y ) .

f −1 : B → A 其中 ∀x∈A,∀y∈B,y=f(x)⇔x=f−1(y)。

right inverse of f

f 的右倒数

g : B A with f g = Id B .

g : B → A,其中 f∘g=IdB。

left inverse of f

f 的左逆函数

h : B A with h f = Id A .

h : B → A,其中 h∘f=IdA。

invertible function

可逆函数

function that has a (two-sided) inverse.

具有(双边)逆函数的函数。

restriction f | S (for S A )

限制 f| S (对于 S ⊆ A)

f | S : S B where f | S ( x ) = f ( x ) for x S .

f| S : S → B 其中 f| S (x) = f(x) 对于 x ∈ S。

characteristic function χ S

特征函数 χ S

χ S ( x ) = 1 for x S , χ S ( x ) = 0 for x S .

对于 x ∈ S,χ S (x) = 1;对于 x∉S,χ S (x) = 0。

h = glue ( h i : i I )

h=glue(hi:i∈I)

For h i : A i B i with graphs H i agreeing on overlaps,

对于 h i : A i → B i ,其中图 H i 在重叠方面达成一致,

h : i I A i i I B i has graph i I H i .

h:⋃i∈IAi→⋃i∈IBi 有图 ⋃i∈IHi。

So h ( x ) = h i ( x ) for all x A i .

因此,对于所有 x ∈ A i ,h(x) = h i (x)。

fiber of f over y

y 上的 f 纤维

f −1 [{ y }] = { x A : f ( x ) = y }.

f −1 [{y}] = {x ∈ A : f(x) = y}。

TABLE 7.2
表 7.2 Relation Definitions.
关系定义。

Concept

概念

Definition

定义

R is a relation from X to Y

R 是从 X 到 Y 的关系。

R X × Y .

R ⊆ X × Y。

R is a relation on X

R 是 X 上的一个关系。

R X × X .

R ⊆ X × X。

infix notation aRb

中缀表示法 aRb

aRb iff ( a , b ) ∈ R .

aRb 当且仅当 (a, b) ∈ R。

image R [ C ]

图像 R[C]

y R [ C ] iff x C , ( x , y ) R .

y ∈ R[C] 当且仅当存在 x∈C,(x,y)∈R。

identity relation on X

X 上的恒等关系

( a , b ) ∈ I X iff a X and a = b .

(a, b) ∈ I X 当且仅当 a ∈ X 且 a = b。

inverse of relation R

关系 R 的逆

( y , x ) ∈ R −1 iff ( x , y ) ∈ R .

(y, x) ∈ R −1 当且仅当 (x, y) ∈ R。

composition of R and S

R 和 S 的组成

( x , z ) ∈ S R iff y , ( x , y ) R ( y , z ) S .

(x, z) ∈ S ∘ R 当且仅当 ∃y,(x,y)∈R∧(y,z)∈S。

R is reflexive on X

R 对 X 具有自反性

a X , ( a , a ) R .

∀a∈X,(a,a)∈R。

R is symmetric

R 是对称的。

a , b , ( a , b ) R ( b , a ) R .

∀a,∀b,(a,b)∈R⇒(b,a)∈R。

R is transitive

R 是传递的

a , b , c , ( ( a , b ) R ( b , c ) R ) ( a , c ) R .

∀a,∀b,∀c,((a,b)∈R∧(b,c)∈R)⇒(a,c)∈R。

R is antisymmetric

R 是反对称的

a , b , ( ( a , b ) R ( b , a ) R ) a = b .

∀a,∀b,((a,b)∈R∧(b,a)∈R)⇒a=b。

R is irreflexive on X

R 对 X 是无自反的

a X , ( a , a ) R .

∀a∈X,(a,a)∉R。

equivalence relation on X

X 上的等价关系

symmetric, transitive, reflexive relation on X .

X 上的对称、传递、自反关系。

equivalence class of a

的等价类

x ∈ [ a ] R iff xRa (iff aRx ).

x ∈ [a] R 当且仅当 xRa(当且仅当 aRx)。

quotient set X / R

商集 X/R

S X / R iff a X , S = [ a ] R .

S ∈ X/R 当且仅当 ∃a∈X,S=[a]R。

equivalence relation ∼ f

等价关系∼ f

for f : X Y and u , v X , u f v iff f ( u ) = f ( v ).

对于 f : X → Y 和 u, v ∈ X,u ∼ f v 当且仅当 f(u) = f(v)。

congruence modulo n

模 n 的同余

For a , b Z , a n b iff n divides a b .

对于 a,b∈Z,a ≡ n b 当且仅当 n 整除 a − b。

integers modulo n

整数模 n

Z / n = { [ a ] n : a Z } = { [ 0 ] n , [ 1 ] n , , [ n 1 ] n } ,

Z/n={[a]≡n:a∈Z}={[0]n,[1]n,…,[n−1]n},

where [ r ] n = { r + k n : k Z } .

其中 [r]n={r+kn:k∈Z}。

addition modulo n

加法模n

[ a ] n n [ b ] n = [ a + b ] n for a , b Z .

[a]n⊕n[b]n=[a+b]n,其中 a,b∈Z。

multiplication modulo n

模 n 乘法

[ a ] n n [ b ] n = [ ab ] n for a , b Z .

[a] n n [b] n = [ab] n 对于 a,b∈Z。

set partition of X

X 的划分

a set P of nonempty subsets of X where

集合 P 是 X 的非空子集,其中

x X , S P , x S and S , T P , S T S T = .

∀x∈X,∃S∈P,x∈S 且 ∀S,T∈P,S≠T⇒S∩T=∅。

eqv. rel. ∼ P of set ptn. P

eqv. rel. ∼ P of set ptn. P

For a , b X , a P b iff S P , a S b S .

对于 a, b ∈ X,a ∼ P b 当且仅当 ∃S∈P,a∈S∧b∈S。

partial order on X

X 上的偏序

antisymmetric, transitive, reflexive relation on X .

X 上的反对称、传递、自反关系。

partially ordered set (poset)

偏序集(poset)

pair ( X , ≤ ) where ≤ is a partial order on X .

对 (X, ≤),其中 ≤ 是 X 上的偏序关系。

total order on X

X 的总订单

partial order where x , y X , x y y x .

偏序关系,其中 ∀x,y∈X,x≤y∨y≤x。

z is greatest element of S

z 是 S 的最大元素。

z S y S , y z .

z∈S∧∀y∈S,y≤z。

z is least element of S

z 是 S 的最小元素。

z S y S , z y .

z∈S∧∀y∈S,z≤y。

well-ordering on X

X 上的良好秩序

partial order on X where each nonempty subset

X 上的偏序关系,其中每个非空子集

of X has a least element.

X 的最小元素是多少?

TABLE 7.3
表 7.3 Cardinality Definitions.
基数定义。

Concept

概念

Definition

定义

| X | = | Y |

|X| = |Y|

There exists a bijection f : X Y .

存在一个双射 f : X → Y。

| X | ≤ | Y |

|X| ≤ |Y|

There exists an injection g : X Y .

存在单射 g : X → Y。

X has size n ( n Z 0 ).

X 的大小为 n (n∈Z≥0)。

| X | = n iff there exists a bijection f :{1, 2, …, n } → X .

|X| = n 当且仅当存在双射 f:{1, 2, …, n} → X。

X is finite.

X 是有限的。

n Z 0 , | X | = n .

∃n∈Z≥0,|X|=n。

X is infinite.

X 为无穷大。

X is not finite.

X 不是有限的。

X is countably infinite.

X 是可数无穷大。

There exists a bijection f : Z > 0 X .

存在一一射 f:Z>0→X。

X is countable.

X 是可数的。

X is finite or X is countably infinite.

X 是有限的,或者 X 是可数无限的。

X is uncountable.

X 是不可数的。

X is not countable.

X 是不可数的。

TABLE 7.4
表 7.4 Theorems on Relations.
关于关系的定理。

Direct Images

直接图像

Setup: R is a relation (or function) from X to Y .

设置:R 是从 X 到 Y 的关系(或函数)。

Empty Image

空白图像

R [ ] = .

R[∅]=∅。

Codomain Property

协域属性

R [ A ]⊆ Y .

R[A]⊆ Y。

Monotonicity

单调性

If A C , then R [ A ]⊆ R [ C ].

如果 A ⊆ C,则 R[A]⊆ R[C]。

Image of Union

联盟的形象

R [ i I A i ] = i I R [ A i ] .

R[⋃i∈IAi]=⋃i∈IR[Ai].

Image of Intersection

交叉路口图像

R [ i I A i ] i I R [ A i ] .

R[⋂iÎIAi]⊆⋂iÎIR[Ai].

Image of Difference

差异图像

R [ A ] − R [ C ]⊆ R [ A C ].

R[A] − R[C]⊆ R[A − C].

Inverses and Compositions

逆运算和复合运算

Setup: R is a relation from X to Y .

设置:R 是从 X 到 Y 的关系。

Inverse of Identity

单位元的逆元

I X 1 = I X .

IX−1=IX。

Double Inverse

双逆

( R −1 ) −1 = R .

(R −1 −1 = R。

Inverse of Union

并集的逆

( S T ) 1 = S 1 T 1 = T 1 S 1 .

(S∪T)−1=S−1∪T−1=T−1∪S−1。

Inverse of Intersection

交集的逆运算

( S T ) 1 = S 1 T 1 = T 1 S 1 .

(S∩T)−1=S−1∩T−1=T−1∩S−1。

Composition with Identity

具有身份认同的构成

R I X = R and I Y R = R .

R ∘ I X = R 且 I Y ∘ R = R。

Associativity

结合律

T ∘ ( S R ) = ( T S R .

T ∘ (S ∘ R) = (T ∘ S)° R。

Inverse of Composition

合成的逆

( R S ) −1 = S −1 R −1 .

(R ∘ S) −1 = S −1 ∘ R −1

Monotonicity Properties

单调性

X Y I X I Y ; if R S , then

X ⊆ Y ↠ I X ⊆ I Y ; 如果 R ⊆ S,则

R −1 S −1 , T R T S , and R T S T .

R −1 ⊆ S −1 ,T ∘ R ⊆ T ∘ S,且 R ∘ T ⊆ S ∘ T。

Empty Set Properties

空集属性

I = and 1 = and R = = R .

I∅=∅ 且 ∅−1=∅ 且 R∘∅=∅=∅∘R。

Distributive Laws

分配法

( S T ) R = ( S R ) ( T R ) .

(S∪T)∘R=(S∘R)∪(T∘R)。

T ( S R ) = ( T S ) ( T R ) .

T∘(S∪R)=(T∘S)∪(T∘R)。

Partial Distributive Laws

部分分配律

( S T ) R ( S R ) ( T R ) .

(S∩T)∘R⊆(S∘R)∩(T∘R)。

T ( S R ) ( T S ) ( T R ) .

T∘(S∩R)⊆(T∘S)∩(T∘R)。

TABLE 7.5 Theorems on Functions.

表 7.5 函数定理。

Preimages

原像

Setup: f : X Y .

设置:f:X → Y。

Empty Preimage

空原像

f 1 [ ] = .

f−1[∅]=∅。

Domain Property

域属性

f −1 [ B ]⊆ X .

f −1 [B]⊆ X。

Preimage of Codomain

协域的原像

f −1 [ Y ] = X .

f −1 [Y] = X。

Monotonicity

单调性

If B D , then f −1 [ B ]⊆ f −1 [ D ].

如果 B ⊆ D,则 f −1 [B]⊆ f −1 [D]。

Preimage of Union

联合的先兆

f 1 [ i I B i ] = i I f 1 [ B i ] .

f−1[⋃i∈IBi]=⋃i∈If−1[Bi].

Preimage of Intersection

交集的原像

f 1 [ i I B i ] = i I f 1 [ B i ] .

f−1[⋂i∈IBi]=⋂i∈If−1[Bi].

Preimage of Difference

差异的原像

f −1 [ B D ] = f −1 [ B ] − f −1 [ D ].

f −1 [B − D] = f −1 [B] − f −1 [D].

Image of Preimage

原像的图像

B Y , f [ f −1 [ B ]]⊆ B ; equality holds if f is surjective.

∀B⊆Y,f[f −1 [B]]⊆ B;等式成立当且仅当 f 是满射。

Preimage of Image

图像的原像

A X , A f −1 [ f [ A ]]; equality holds if f is injective.

∀A⊆X, A ⊆ f −1 [f[A]]; 等式成立当且仅当 f 是单射。

Function Composition

功能组合

Setup: f : X Y , g : Y Z , h : Z W .

设置:f:X → Y,g:Y → Z,h:Z → W。

Associativity

结合律

( h g f = h ∘ ( g f ).

(h ∘ g)° f = h ∘ (g ∘ f)。

Identity

身份

f Id X = f = Id Y f .

f∘IdX=f=IdY∘f。

Image under Composition

构图下的图像

( g f )[ A ] = g [ f [ A ]].

(g ∘ f)[A] = g[f[A]].

Preimage under Composition

构图下的原像

( g f ) −1 [ C ] = f −1 [ g −1 [ C ]].

(g ∘ f) −1 [C] = f −1 [g −1 [C]].

Injections/Surjections/Bijections

注射/注射/双注射

Setup: f : X Y and g : Y Z .

设置:f:X → Y 和 g:Y → Z。

Composing Injections

注射剂组成

If f and g are injective, then g f is injective.

如果 f 和 g 是单射,那么 g ∘ f 也是单射。

Composing Surjections

组成喷射

If f and g are surjective, then g f is surjective.

如果 f 和 g 是满射,那么 g ∘ f 也是满射。

Composing Bijections

组合双射

If f and g are bijective, then g f is bijective.

如果 f 和 g 是双射,那么 g ∘ f 也是双射。

Backwards Rule for Injections

注射的逆向规则

If g f is injective, then f is injective.

如果 g ∘ f 是单射,那么 f 也是单射。

Backwards Rule for Surjections

向后定理

If g f is surjective, then g is surjective.

如果 g ∘ f 是满射,那么 g 也是满射。

Inverse Functions

反函数

Setup: f : X Y , g : Y X , h : Y Z .

设置:f:X → Y,g:Y → X,h:Y → Z。

Two-Sided Inverse

双边逆

g = f −1 iff f g = Id Y and g f = Id X .

g = f −1 当且仅当 f∘g=IdY 且 g∘f=IdX。

Bijectivity and Invertibility

双射性和可逆性

f −1 exists iff f is bijective;

f −1 存在当且仅当 f 是双射;

in this case, f −1 is unique and bijective.

在这种情况下,f −1 是唯一的且双射的。

Inverse of Identity

单位元的逆元

Id X is invertible, and Id X 1 = Id X .

IdX 是可逆的,且 IdX−1=IdX。

Double Inverse

双逆

If f −1 exists, then ( f −1 ) −1 = f .

如果 f −1 存在,则 (f −1 ) −1 = f。

Inverse of Composition

合成的逆

If f and h are invertible, then ( h f ) −1 = f −1 h −1 .

如果 f 和 h 可逆,则 (h ∘ f) −1 = f −1 ∘ h −1

Left Inverse Criterion

左逆准则

For X , f has a left inverse iff f is injective.

对于 X≠∅,f 有左逆当且仅当 f 是单射。

Right Inverse Criterion

右逆准则

f has a right inverse iff f is surjective.

f 有右逆函数当且仅当 f 是满射。

Pointwise Operations

逐点操作

Setup: f , g , h : X R .

设置:f、g、h:X→R。

Commutativity

交换律

f + g = g + f and f · g = g · f .

f + g = g + f 且 f · g = g · f。

Associativity

结合律

( f + g ) + h = f + ( g + h ) and ( f · g ) · h = f · ( g · h ).

(f + g)+ h = f + (g + h)且(f · g)· h = f · (g · h)。

Identity

身份

f + 0 = f = 0 + f and f · 1 = f = 1 · f .

f + 0 = f = 0 + f 且 f · 1 = f = 1 · f。

Inverses

f + ( − f ) = 0 = ( − f ) + f and

f + ( − f) = 0 = ( − f) + f 且

(when g ( x ) is never zero) g · (1/ g ) = 1 = (1/ g ) · g .

(当 g(x) 永远不为零时)g · (1/g) = 1 = (1/g) · g。

Distributive Law

分配法

f · ( g + h ) = ( f · g ) + ( f · h ).

f · (g + h) = (f · g) + (f · h)。

TABLE 7.6
表 7.6 Theorems on Cardinality.
关于基数的定理。

Cardinal Equality:

基数相等:

Reflexivity

反身性

| A | = | A |.

|A| = |A|。

Symmetry

对称

| A | = | B | ↠ | B | = | A |.

|A| = |B| ↠ |B| = |A|。

Transitivity

传递性

( | A | = | B | | B | = | C | ) | A | = | C | .

(|A|=|B|∧|B|=|C|)⇒|A|=|C|。

Disjoint Unions

不相交的工会

If | A i | = | A i ′| for all i I and A i A j = = A i A j

如果对于所有 i ∈ I,|A i | = |A i ′|,并且 Ai∩Aj=∅=Ai′∩Aj′

for all i j in I , then | i I A i | = | i I A i | .

对于 I 中的所有 i ≠ j,则 |⋃i∈IAi|=|⋃i∈IAi′|。

Products

产品

If | A i | = | A i ′| for 1 ≤ i n ,

如果对于 1 ≤ i ≤ n,|A i | = |A i ′|,

then | A 1 × · · · × A n | = | A 1 ′ × · · · × A n ′|.

那么 |A 1 × · · · × A n | = |A 1 ′ × · · · × A n ′|。

Power Sets

力量集

If | A | = | B |, then | P ( A ) | = | P ( B ) | .

如果 |A| = |B|,则 |P(A)|=|P(B)|。

Cardinal Ordering:

基本顺序:

Reflexivity

反身性

| A | ≤ | A |.

|A| ≤ |A|。

Antisymmetry

反对称性

( | A | | B | | B | | A | ) | A | = | B | (Schröder–Bernstein Theorem).

(|A|≤|B|∧|B|≤|A|)⇒|A|=|B|(施罗德-伯恩斯坦定理)。

Transitivity

传递性

( | A | | B | | B | | C | ) | A | | C | .

(|A|≤|B|∧|B|≤|C|)⇒|A|≤|C|。

Products

产品

If | A i | ≤ | A i ′| for 1 ≤ i n ,

如果对于 1 ≤ i ≤ n,|A i | ≤ |A i ′|,

then | A 1 × · · · × A n | ≤ | A 1 ′ × · · · × A n ′|.

那么 |A 1 × · · · × A n | ≤ |A 1 ′ × · · · × A n ′|。

Finite Sets:

有限集:

Uniqueness of Size

尺寸的独特性

For m , n Z 0 , | X | = m and | X | = n imply m = n .

对于 m,n∈Z≥0,|X| = m 和 |X| = n 蕴含 m = n。

Sum Rule

求和规则

For pairwise disjoint finite X 1 , …, X s , | k = 1 s X k | = k = 1 s | X k | .

对于两两不相交的有限集合 X 1 , …, X s , |⋃k=1sXk|=∑k=1s|Xk|。

Product Rule

乘积法则

For finite X 1 , …, X s , | X 1 × × X s | = k = 1 s | X k | .

对于有限的 X 1 , …, X s , |X1×⋯×Xs|=∏k=1s|Xk|。

Power Set Rule

权力集规则

For finite X , | P ( X ) | = 2 | X | .

对于有限的 X,|P(X)|=2|X|。

Preserving Finiteness

保持有限性

For finite X , X 1 , …, X s , k = 1 s X k and k = 1 s X k

对于有限的 X,X 1 , …, X s ,⋃k=1sXk 和 ⋂k=1sXk

and X 1 × · · · × X k and P ( X ) and all subsets of X are finite.

且 X 1 × · · · × X k 和 P(X) 以及 X 的所有子集都是有限的。

Countable Sets:

可数集合:

Countability Criteria

可计数性标准

X is countably infinite iff | X | = | Z > 0 | .

X 是可数无穷的当且仅当 |X|=|Z>0|。

X is countable iff | X | | Z > 0 | iff bijection f : X Y Z > 0 .

X 是可数的当且仅当 |X|≤|Z>0| 当且仅当存在双射 f:X→Y⊆Z>0。

Countably Infinite Sets

可数无限集

Z a , Z a , Z , Q , Z k , Q k (where k Z > 0 ) are countably infinite.

Z≥a、Z≤a、Z、Q、Zk、Qk(其中k∈Z>0)是可数无穷的。

Ordering Property

排序属性

If | X | ≤ | Y | and Y is countable, then X is countable.

如果 |X| ≤ |Y| 且 Y 是可数的,那么 X 也是可数的。

Subset Property

子集属性

Any subset of a countable set is countable.

可数集的任何子集都是可数的。

Product Property

产品属性

A product of finitely many countable sets is countable.

有限个可数集的乘积是可数的。

Union Property

联合财产

A union of countably many countable sets is countable.

可数个可数集合的并集是可数的。

Uncountable Sets:

不可数集:

Ordering Property

排序属性

If | X | ≤ | Y | and X is uncountable, then Y is uncountable.

如果 |X| ≤ |Y| 且 X 是不可数的,那么 Y 也是不可数的。

Uncountable Sets

不可数集

R , ( a , b ), [ a , b ], and P ( Z > 0 ) are uncountable (for real a < b ).

R、(a, b)、[a, b] 和 P(Z>0) 是不可数的(对于实数 a < b)。

For | X | > 1, the set of sequences f : Z > 0 X is uncountable.

对于 |X| > 1,序列集 f:Z>0→X 是不可数的。

Cantor’s Theorem

康托尔定理

For all X , | X | < | P ( X ) | .

对于所有 X,|X|<|P(X)|。

Functions and Relations

函数与关系

  1. Arrow Diagrams. We can visualize a relation R from X to Y by drawing an arrow from a dot labeled x to a dot labeled y for each ( x , y ) ∈ R . The relation is the graph of a function f : X Y iff each x X has exactly one arrow leaving it. Such a function f is injective iff each y Y has at most one arrow entering it; f is surjective iff each y Y has at least one arrow entering it; and f is bijective iff each y Y has exactly one arrow entering it. The relation R −1 is found by reversing all arrows. We get the arrow diagram of a composition S R by concatenating R -arrows followed by S -arrows.
    箭头图。我们可以将关系 R 从 X 到 Y 可视化,方法是对于每个 (x, y) ∈ R,从标记为 x 的点到标记为 y 的点绘制一条箭头。该关系是函数 f : X → Y 的图像,当且仅当每个 x ∈ X 都恰好有一条箭头从它发出。这样的函数 f 是单射,当且仅当每个 y ∈ Y 至多有一条箭头进入它;f 是满射,当且仅当每个 y ∈ Y 至少有一条箭头进入它;f 是双射,当且仅当每个 y ∈ Y 恰好有一条箭头进入它。关系 R −1 可以通过反转所有箭头得到。复合 S ∘ R 的箭头图可以通过连接 R 箭头和 S 箭头得到。
  2. Cartesian Graphs. We can visualize a relation R on R by the graph consisting of all points ( x , y ) in the plane with ( x , y ) ∈ R . The relation is the graph of a function f : X Y iff the points drawn all lie within the rectangle X × Y , and for each x 0 X the vertical line x = x 0 intersects the graph in exactly one point. Such a function f is injective (resp. surjective, bijective) iff for all y 0 Y , the horizontal line y = y 0 intersects the graph in at most one (resp. at least one, exactly one) point. The relation R −1 is found by reflecting the graph in the line y = x .
    笛卡尔坐标图。我们可以用平面上所有满足 (x, y) ∈ R 的点 (x, y) 构成的图来可视化 R 上的关系 R。该关系是函数 f : X → Y 的图像,当且仅当所绘制的点都位于矩形 X × Y 内,并且对于每个 x ∈ X,垂直线 x = x 与该图像恰好相交于一点。这样的函数 f 是单射(或满射,或双射),当且仅当对于所有 y ∈ Y,水平线 y = y 与该图像至多相交于一点(或至少相交于一点,或恰好相交于一点)。关系 R 可以通过将该图像关于直线 y = x 反射得到。
  3. Digraphs. We can visualize a relation R on X by drawing a single copy of X with arrows from x to y for each ( x , y ) ∈ R . R is reflexive on X iff there is a loop at each vertex x X ; R is symmetric iff every arrow from x to y is accompanied by the reverse arrow from y to x ; and R is transitive iff whenever we can go from x to y to z by following two arrows, the arrow directly from x to z is also present.
    有向图。我们可以通过在 X 上绘制一个包含从 x 到 y 的箭头的副本来可视化关系 R,其中 (x, y) ∈ R。R 在 X 上是自反的,当且仅当每个顶点 x ∈ X 处都存在一个环;R 是对称的,当且仅当每个从 x 到 y 的箭头都伴随着一个从 y 到 x 的反向箭头;R 是传递的,当且仅当当我们可以通过两条箭头从 x 到 y 再到 z 时,也存在一条直接从 x 到 z 的箭头。
  4. Proving Functions are Well-Defined. We often define functions g : A B by giving a formula specifying g ( x ) for each x A . To be sure we do have a well-defined function, we must check three things: g ( x ) is in the codomain B for all x A ; for each x A there exists at least one associated output g ( x ); and for all x 1 , x 2 A , if x 1 = x 2 , then g ( x 1 ) = g ( x 2 ). The last condition can be restated: for all x , y , z , if ( x , y ) and ( x , z ) belong to the graph of g , then y = z . It is especially crucial to prove this single-valuedness condition when members of the domain A have multiple names (e.g., if A is a set of equivalence classes). If f ( x ) is computed using one of the names of x , we must verify that changing the name of the input x does not change the value of the output f ( x ) (although the name of the output could change).
    证明函数定义良好。我们通常通过给出一个公式来定义函数 g : A → B,该公式指定了对于每个 x ∈ A,g(x) 的值。为了确保我们确实有一个定义良好的函数,我们必须检查三件事:对于所有 x ∈ A,g(x) 都在值域 B 中;对于每个 x ∈ A,至少存在一个对应的输出 g(x);并且对于所有 x 1 , x 2 ∈ A,如果 x 1 = x 2 ,则 g(x 1 ) = g(x 2 )。最后一个条件可以重新表述为:对于所有 x, y, z,如果 (x, y) 和 (x, z) 属于函数 g 的图像,则 y = z。当域 A 的成员有多个名称时(例如,如果 A 是等价类的集合),证明单值性条件尤为重要。如果 f(x) 是使用 x 的某个名称计算的,我们必须验证更改输入 x 的名称不会改变输出 f(x) 的值(尽管输出的名称可能会改变)。
  5. Proving Function Equality. To prove that functions f and g are equal, first check that f and g share the same domain (say X ) and the same codomain (say Y ). Then fix an arbitrary x 0 X , and use the definitions of f and g to prove f ( x 0 ) = g ( x 0 ).
    证明函数相等。要证明函数 f 和 g 相等,首先检查 f 和 g 是否具有相同的定义域(例如 X)和相同的值域(例如 Y)。然后固定任意 x ∈ X,并利用 f 和 g 的定义来证明 f(x) = g(x)。
  6. Proving Invertibility. We can prove that a function f : X Y is invertible by proving f is one-to-one and onto. Alternatively, we can prove invertibility by defining a proposed inverse g : Y X , then proving f g = Id Y and g f = Id X .
    证明可逆性。我们可以通过证明函数 f : X → Y 是单射且满射来证明 f 是可逆的。或者,我们可以通过定义一个假定的逆函数 g : Y → X 来证明可逆性,然后证明 f∘g=IdY 和 g∘f=IdX。
  7. Common Pitfalls. Remember that composition of relations (or functions) is usually not commutative: R S S R for most choices of the relations (or functions) R and S . Do not confuse the preimage notation f −1 [ B ] with the notation f −1 for an inverse function; preimages are defined for any function f , even when the inverse function f −1 does not exist. The two-sided inverse of f is unique when it exists, but left inverses and right inverses are not unique in general. Two functions with the same domain and the same values are not equal unless their codomains also agree.
    常见陷阱。请记住,关系(或函数)的复合通常不满足交换律:对于大多数关系(或函数)R 和 S,R ∘ S ≠ S ∘ R。不要将原像符号 f −1 [B] 与反函数符号 f −1 混淆;原像对于任何函数 f 都有定义,即使反函数 f −1 不存在。f 的双侧逆函数存在时是唯一的,但左逆函数和右逆函数通常不唯一。定义域相同且值域相同的两个函数不相等,除非它们的值域也相同。

Equivalence Relations, Set Partitions, and Partial Orders

等价关系、集合划分和偏序关系

  1. Equivalence Relations. A relation R on a set X is an equivalence relation iff R is reflexive on X , symmetric, and transitive. This means that for all a , b , c X , aRa ; if aRb then bRa ; and if aRb and bRc then aRc . The equivalence class [ a ] R is the set of all elements of X related to a under R : [ a ] R = { x X : xRa }.
    等价关系。集合 X 上的关系 R 是等价关系当且仅当 R 在 X 上具有自反性、对称性和传递性。这意味着对于所有 a, b, c ∈ X,a ∈ R<sup>a</sup>;如果 a ∈ R<sup>b</sup> 则 b ∈ R<sup>b</sup>;并且如果 a ∈ R<sup>b</sup> 且 b ∈ R<sup>c</sup> 则 a ∈ R<sup>c</sup>。等价类 [a] R 是 X 中所有与 R 下的 a 相关的元素的集合:[a] R = {x ∈ X: x ∈ R<sup>a</sup>}。
  2. Theorem on Equivalence Classes. Given an equivalence relation R on a set X and a , b X , [ a ] R = [ b ] R iff aRb iff bRa iff b ∈ [ a ] R iff a ∈ [ b ] R . Two equivalence classes of R are either disjoint or equal.
    等价类定理。给定集合 X 上的等价关系 R 和 a, b ∈ X,[a] R = [b] R 当且仅当 aRb 当且仅当 bRa 当且仅当 b ∈ [a] R 当且仅当 a ∈ [b] R 。R 的两个等价类要么互不相交,要么相等。
  3. Set Partitions. A set partition of X is a set of nonempty subsets of X such that every a X belongs to exactly one set (block) in the set partition.
    集合划分。集合 X 的一个集合划分是 X 的非空子集的集合,使得每个 a ∈ X 都恰好属于集合划分中的一个集合(块)。
  4. Correspondence Between Equivalence Relations and Set Partitions. Given an equivalence relation R on a set X , the quotient set X / R = {[ a ] R : a X } is a set partition of X . Given a set partition P of X , the relation ∼ P , defined for a , b X by a P b iff a and b belong to the same block of P , is an equivalence relation on X . The function sending R to X / R is a bijection from the set of equivalence relations on X to the set of set partitions of X . The inverse function sends a set partition P of X to ∼ P .
    等价关系与集合划分的对应关系。给定集合 X 上的等价关系 R,商集 X/R = {[a] R : a ∈ X} 是 X 的一个集合划分。给定 X 的一个集合划分 P,关系 ∼ P (对于 a, b ∈ X,定义为 a ∼ P b 当且仅当 a 和 b 属于 P 的同一个块)是 X 上的一个等价关系。将 R 映射到 X/R 的函数是从 X 上的等价关系集合到 X 的集合划分集合的双射。其逆函数将 X 的一个集合划分 P 映射到 ∼ P
  5. Examples of Posets. Number systems such as R are totally ordered by ≤. A collection of sets is partially ordered by ⊆. A set of positive integers is partially ordered by divisibility. Other examples include subposets, product posets, and concatenation posets.
    偏序集的例子。像 R 这样的数系是按 ≤ 关系全序的。集合的集合是按 ⊆ 关系偏序的。正整数的集合是按整除关系偏序的。其他例子包括子偏序集、乘积偏序集和连接偏序集。
  6. Well-Ordering Property. The poset ( Z > 0 , ) is well-ordered, meaning that every nonempty subset of positive integers has a least element. This property is equivalent to the Induction Axiom.
    良序性质。偏序集 (Z>0,≤) 是良序的,这意味着每个非空正整数子集都存在最小元素。此性质等价于归纳公理。

Review Problems

复习题

  1. Complete the following definitions and proof templates:

    请完成以下定义和证明模板:

    (a) “ R is a relation from X to Z ” means...

    (a)“R 是从 X 到 Z 的关系”意味着……

    (b) For a relation S and set A and object w , “ w S [ A ]” means...

    (b) 对于关系 S、集合 A 和对象 w,“w ∈ S[A]”表示……

    (c) For relations R and S and objects x , y , “( x , y ) ∈ S R ” means...

    (c) 对于关系 R 和 S 以及对象 x、y,“(x, y) ∈ S ∘ R”表示……

    (d) For a relation T and ordered pair ( a , b ), “( a , b ) ∈ T −1 ” means...

    (d) 对于关系 T 和有序对 (a, b),“(a, b) ∈ T −1 ” 表示……

    (e) “ F is a function from X to Y ” means...

    (e)“F 是从 X 到 Y 的函数”意味着……

    (f) Given a function g with graph G and objects u , w , “ w = g ( u )” means...

    (f) 给定一个函数 g,其图像为 G,对象为 u 和 w,“w = g(u)”表示……

    (g) Given a set Z and objects k , m , “( k , m ) ∈ I Z ” means...

    (g) 给定集合 Z 和对象 k、m,“(k, m) ∈ I Z ” 表示……

    (h) Given f : X Y and A X and an object y , “ y f [ A ]” means...

    (h) 给定函数 f : X → Y 和 A ⊆ X 以及对象 y,“y ∈ f[A]” 表示……

    (i) Given f : X Y and B Y and an object x X , “ x f −1 [ B ]” means...

    (i) 给定函数 f : X → Y 和 B ⊆ Y 以及对象 x ∈ X,“x ∈ f −1 [B]” 表示……

    (j) “ h : A C is one-to-one” means...

    (j)“h : A → C 是一对一的”意味着……

    (k) “ p : D E is onto” means...

    (k)“p : D → E 是满射”意味着……

    (l) Given F : Z W and G : Z W , outline a proof that F = G .

    (l)已知 F : Z → W 和 G : Z → W,简述证明 F = G。

    (m) Given relations R and S from X to Y , describe two ways to prove that R = S .

    (m)给定从 X 到 Y 的关系 R 和 S,描述两种证明 R = S 的方法。

    (n) Given f , g , h : R R and t R , define ( f · g )( t + 1), ( f g )( t + 1), and ( g + f )( t + 1).

    (n)给定 f,g,h:R→R 和 t∈R,定义 (f · g)(t + 1)、(f ∘ g)(t + 1) 和 (g + f)(t + 1)。

    (o) “ R is reflexive on X ” means...

    (o)“R 对 X 是自反的”意味着……

    (p) “ S is symmetric” means...

    (p)“S 是对称的”意味着……

    (q) “ T is transitive” means...

    (q)“T 是及物动词”意味着……

    (r) “ X has n elements” means...

    (r)“X 有 n 个元素”意味着……

    (s) “| U | ≤ | V |” means...

    (s)“|U| ≤ |V|”表示……

  2. Fix a , b R with a ≠ 0. Define f : R R by f ( x ) = ax + b for all x R .

    固定 a,b∈R,其中 a≠0。定义 f:R→R,使得对于所有 x∈R,f(x) = ax + b。

    (a) Is f one-to-one? Is f onto? Prove your answers.

    (a)f 是单射吗?f 是满射吗?证明你的答案。

    (b) Repeat (a) replacing R by Q everywhere in the setup.

    (b)重复(a),将设置中的所有 R 替换为 Q。

    (c) Repeat (a) replacing R by Z everywhere in the setup. Does the answer to (c) depend on the particular value of a ?

    (c) 重复 (a) 的步骤,将设置中的所有 R 替换为 Z。(c) 的答案是否取决于 a 的具体值?

  3. For each set X and equivalence relation R on X , explicitly describe the requested equivalence classes. (a) X = { a , b , c , d }, R = {( a , a ), ( b , b ), ( c , c ), ( d , d ), ( b , c ), ( c , b ), ( a , d ), ( d , a )}; find [ a ] R , [ b ] R , [ c ] R , and [ d ] R . (b) X = Z , R is congruence mod 5; find [7] R . (c) X = {1, 2, 3, 4, 5}, R = I X ; find [3] R . (d) X = R 0 × R 0 , (( x , y ), ( u , v )) ∈ R means x + y = u + v ; draw [(1, 2)] R .
    对于每个集合 X 和 X 上的等价关系 R,明确描述所要求的等价类。 (a) X = {a, b, c, d}, R = {(a, a), (b, b), (c, c), (d, d), (b, c), (c, b), (a, d), (d, a)};求 [a] R , [b] R , [c] R , 和 [d] R (b) X=Z, R 模 5 同余;找到 [7] R . (c) X = {1, 2, 3, 4, 5}, R = I X ; 找到 [3] R . (d) X=R≥0×R≥0, ((x, y), (u, v)) ∈ R 表示 x + y = u + v; 绘制 [(1, 2)] R
  4. Give a specific example of a function f : Z > 0 Z > 0 with the stated properties. Prove that your examples work. (a) f is one-to-one and not onto. (b) f is onto and not one-to-one. (c) f is not one-to-one and not onto and not constant. (d) f is one-to-one and onto and f Id Z > 0 .
    给出一个满足所述性质的函数 f:Z>0→Z>0 的具体例子。证明你的例子成立。(a) f 是单射且不是满射。(b) f 是满射且不是单射。(c) f 不是单射、不是满射且不是常数。(d) f 是单射、满射且 f≠IdZ>0。
  5. Suppose S , T , U are finite sets with | S | = a and | T | = b and U S . Which formulas are always true? Explain. (a) | S T | = a + b . (b) | S × T | = ab . (c) | U | = c for some integer c < a . (d) | P ( S ) | = 2 a . (e) | S T | = a b .
    假设 S、T、U 是有限集合,且 |S| = a,|T| = b,U ⊆ S。哪些公式总是成立的?请解释。(a) |S∪T|=a+b。(b) |S × T| = ab。(c) |U| = c,其中 c < a 为某个整数。(d) |P(S)|=2a。(e) |S − T| = a − b。
  6. Which of the following sets are countably infinite? (a) (b) {1, 2, 3} (c) Z 0 (d) Z (e) Q (f) Z × Z (g) R (h) the set of all infinite sequences of 0s and 1s (i) P ( Z ) .
    下列哪些集合是可数无限的?(a)∅(b){1, 2, 3}(c)Z≥0(d)Z(e)Q(f)Z×Z(g)R(h)所有由0和1组成的无限序列的集合(i)P(Z)。
  7. (i) Draw an arrow diagram and a digraph for each relation. (ii) Is each relation reflexive on X = {1, 2, 3, 4}? symmetric? transitive? Explain. (iii) Compute R [{1, 3}] and R R −1 for each R .

    (i) 为每个关系绘制箭头图和有向图。(ii) 每个关系在 X = {1, 2, 3, 4} 上是否具有自反性?对称性?传递性?请解释。(iii) 计算每个关系 R 的 R[{1, 3}] 和 R ∘ R −1

    (a) R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 4), (1, 4), (2, 4)}.

    (a)R = {(1,1),(2,2),(3,3),(4,4),(1,3),(3,4),(1,4),(2,4)}。

    (b) R = {(1, 2), (2, 1), (3, 4), (4, 3)}.

    (b)R = {(1,2),(2,1),(3,4),(4,3)}。

    (c) R = ( X × X ) − I X .

    (c)R = (X × X) − I X

    (d) R = {(1, 1), (3, 3), (1, 3), (3, 1), (2, 2), (4, 4), (2, 4), (4, 2)}.

    (d)R = {(1,1),(3,3),(1,3),(3,1),(2,2),(4,4),(2,4),(4,2)}。

  8. (a) Prove: for all f , g , h : R R , ( f · g h = ( f h ) · ( g h ).

    (a)证明:对于所有 f,g,h:R→R,(f · g)° h =(f ∘ h)·(g ∘ h)。

    (b) Disprove: for all f , g , h : R R , f ∘ ( g · h ) = ( f g ) · ( f h ).

    (b)反证:对于所有 f,g,h:R→R,f ∘ (g · h) = (f ∘ g) · (f ∘ h)。

    (c) Give four examples of functions f : R R such that for all g , h : R R , f ∘ ( g · h ) = ( f g ) · ( f h ).

    (c)给出四个函数 f:R→R 的例子,使得对于所有 g,h:R→R,f ∘ (g · h) = (f ∘ g) · (f ∘ h)。

  9. Suppose R is an equivalence relation on X = {1, 2, 3, 4, 5, 6} containing the ordered pairs (1, 4), (6, 3), and (3, 2). (a) Draw the digraph of R . Give the answer with the fewest possible arrows. (b) Find all distinct equivalence classes of the relation R in (a). (c) What is the set partition X / R ?
    假设 R 是集合 X = {1, 2, 3, 4, 5, 6} 上的一个等价关系,包含有序对 (1, 4)、(6, 3) 和 (3, 2)。(a) 画出 R 的有向图。用最少的箭头给出答案。(b) 找出 (a) 中关系 R 的所有不同的等价类。(c) 求 X/R 的集合划分。
  10. (a) Define f : R 0 R by f ( x ) = 3 x 1 x for all real x ≠ 0. Prove f is one-to-one but not onto. What is f [ R 0 ] ? (b) Define g : R × R R × R by g (( x , y )) = ( x + y , 2 − y ) for all ( x , y ) R × R . Is g one-to-one? Is g onto? Prove your answers.
    (a) 定义函数 f:R≠0→R,使得对于所有实数 x ≠ 0,f(x)=3x−1x。证明 f 是单射但不是满射。求 f[R≠0]。(b) 定义函数 g:R×R→R×R,使得对于所有 (x,y)∈R×R,g((x, y)) = (x + y, 2 − y)。g 是单射吗?g 是满射吗?证明你的答案。
  11. Suppose X and I are fixed sets, R i is a fixed relation on X for i I , and R = i I R i . Prove or disprove each statement. (a) R is a relation on X . (b) If every R i is reflexive on X , then R must be reflexive on X . (c) If every R i is symmetric, then R must be symmetric. (d) If every R i is transitive, then R must be transitive.
    假设 X 和 I 是固定集,R i 是 X 上的一个固定关系,其中 i ∈ I,且 R=⋃i∈IRi。证明或反驳下列陈述。(a) R 是 X 上的一个关系。(b) 如果每个 R i 在 X 上都是自反的,那么 R 也一定是自反的。(c) 如果每个 R i 都是对称的,那么 R 也一定是对称的。(d) 如果每个 R i 都是传递的,那么 R 也一定是传递的。
  12. Given functions f : X U and g : Y V , define h : X × Y U × V by h (( x , y )) = ( f ( x ), g ( y )) for all ( x , y ) ∈ X × Y . (a) Prove: h is a function. (b) Prove: if f is one-to-one and g is one-to-one, then h is one-to-one. (c) Prove: if f is onto and g is onto, then h is onto. (d) Prove: if f is a bijection and g is a bijection, then h is a bijection.
    给定函数 f : X → U 和 g : Y → V,定义 h : X × Y → U × V,使得对于所有 (x, y) ∈ X × Y,都有 h((x, y)) = (f(x), g(y))。 (a) 证明:h 是一个函数。 (b) 证明:如果 f 是单射且 g 是单射,则 h 也是单射。 (c) 证明:如果 f 是满射且 g 是满射,则 h 也是满射。 (d) 证明:如果 f 是双射且 g 是双射,则 h 也是双射。
  13. (a) Prove: for all relations R and all sets A and B , R [ A ] − R [ B ]⊆ R [ A B ]. (b) Give an example to show that equality does not always hold in (a). (c) Suppose R is a relation from X to Y and A , B X . Which of the following conditions on R will force equality to hold in (a)? Select all that apply: R is the graph of a function from X to Y ; R is the inverse of the graph of a function from X to Y ; R is the graph of an injection from X to Y ; R is the graph of a surjection from X to Y .
    (a) 证明:对于所有关系 R 和所有集合 A 和 B,R[A] − R[B] ⊆ R[A − B]。 (b) 举例说明 (a) 中的等号并非总是成立。 (c) 假设 R 是从 X 到 Y 的关系,且 A, B ⊆ X。下列关于 R 的哪些条件能使 (a) 中的等号成立?选择所有适用项:R 是从 X 到 Y 的函数的图像;R 是从 X 到 Y 的函数的反函数的图像;R 是从 X 到 Y 的单射的图像;R 是从 X 到 Y 的满射的图像。
  14. Prove: for all f : X Y and g : Y Z and all C Z , ( g f ) −1 [ C ] = f −1 [ g −1 [ C ]].
    证明:对于所有 f : X → Y 和 g : Y → Z 以及所有 C ⊆ Z,(g ∘ f) −1 [C] = f −1 [g −1 [C]].
  15. Prove: for all f : X Y , f is injective iff (for all A , C X , f [ A C ] = f [ A ] f [ C ] ).
    证明:对于所有 f : X → Y,f 是单射当且仅当(对于所有 A, C ⊆ X,f[A∩C]=f[A]∩f[C])。
  16. For each set X and relation R , prove that R is an equivalence relation on X . (a) X = Z 3 , ( a , b , c ) R ( d , e , f ) means a + b + c = d + e + f for a , b , c , d , e , f Z . (b) X = R , ( x , y ) ∈ R means y x Q for x , y R . (c) X = R > 0 × R > 0 , (( x , y ), ( a , b )) ∈ R means there exists r R > 0 with a = rx and b = ry , for x , y , a , b R .
    对于每个集合 X 和关系 R,证明 R 是 X 上的等价关系。 (a) X=Z3, (a, b, c)R(d, e, f) 表示对于 a,b,c,d,e,f∈Z,有 a + b + c = d + e + f。 (b) X=R, (x, y) ∈ R 表示对于 x,y∈R,有 y−x∈Q。 (c) X=R>0×R>0, ((x, y), (a, b)) ∈ R 表示对于 x,y,a,b∈R,存在 r∈R>0,使得 a = rx 且 b = ry。
  17. Let X = [0, 3] × [0, 3]. For each equivalence relation R below, draw a picture of X and the equivalence classes [(1, 3)] R and [(2, 2)] R . (a) R is defined by: (( a , b ), ( c , d )) ∈ R iff a = c , for ( a , b ), ( c , d ) ∈ X . (b) R is defined by: (( a , b ), ( c , d )) ∈ R iff b = d , for ( a , b ), ( c , d ) ∈ X . (c) R is defined by: (( a , b ), ( c , d )) ∈ R iff a + b = c + d , for ( a , b ), ( c , d ) ∈ X . (d) R is defined by: (( a , b ), ( c , d )) ∈ R iff ab = cd , for ( a , b ), ( c , d ) ∈ X .
    令 X = [0, 3] × [0, 3].对于以下每个等价关系 R,绘制 X 以及等价类 [(1, 3)] R 和 [(2, 2)] R . 的图像。(a) R 定义为:((a, b), (c, d)) ∈ R 当且仅当 a = c,对于 (a, b), (c, d) ∈ X。 (b) R 定义为:((a, b), (c, d)) ∈ R 当且仅当 b = d,对于 (a, b), (c, d) ∈ X。 (c) R 定义为:((a, b), (c, d)) ∈ R 当且仅当 a + b = c + d,对于 (a, b), (c, d) ∈ X。 (d)R 定义为:((a, b), (c, d)) ∈ R 当且仅当 ab = cd,其中 (a, b), (c, d) ∈ X。
  18. Let X and I be fixed, nonempty sets. For each i I , let R i be a fixed equivalence relation on X . (a) Prove: R = i I R i is an equivalence relation on X . (b) Prove: for all b X , [ b ] R = i I [ b ] R i .
    设 X 和 I 为固定的非空集合。对于每个 i ∈ I,设 R i 为 X 上的一个固定的等价关系。(a) 证明:R=⋂i∈IRi 是 X 上的一个等价关系。(b) 证明:对于所有 b ∈ X,[b]R=⋂i∈I[b]Ri。
  19. Prove or disprove: for all sets X , Y , P , Q , if P is a set partition of X and Q is a set partition of Y , then P Q is a set partition of X Y .
    证明或反证:对于所有集合 X、Y、P、Q,如果 P 是 X 的一个集合划分,Q 是 Y 的一个集合划分,那么 P∪Q 是 X∪Y 的一个集合划分。
  20. Let R be a relation from X to Y . For each identity below, find a condition on R that will guarantee the truth of the identity. (Possible conditions are: R is any relation whatsoever; R is the graph of a function; R is the graph of an injection; R is the graph of a surjection; R is the graph of a bijection.) Prove that your answer works, but give examples showing that more general relations do not work. (a) For all B Y , R [ R −1 [ B ]] = B . (b) For all A X , R −1 [ R [ A ]] = A . (c) For all A , C X , R [ A C ] = R [ A ] R [ C ] .
    设 R 是从 X 到 Y 的关系。对于以下每个等式,找出 R 的一个条件,以保证该等式为真。(可能的条件包括:R 是任意关系;R 是函数的图像;R 是单射的图像;R 是满射的图像;R 是双射的图像。)证明你的答案有效,并举例说明更一般的关系不成立。 (a) 对于所有 B ⊆ Y,R[R −1 [B]] = B。 (b) 对于所有 A ⊆ X,R −1 [R[A]] = A。 (c) 对于所有 A, C ⊆ X,R[A∪C]=R[A]∪R[C]。
  21. (a) Define an explicit bijection f : Z Z > 0 . (b) Prove: for all sets X and Y , if X and Y are disjoint countably infinite sets, then X Y is countably infinite. (c) Does (b) remain true if X and Y are not disjoint?
    (a) 定义一个显式双射 f:Z→Z>0。(b) 证明:对于任意集合 X 和 Y,如果 X 和 Y 是不相交的可数无限集,则 X∪Y 也是可数无限集。(c) 如果 X 和 Y 不相交,(b) 是否仍然成立?
  22. (a) Define f : Z 0 4 Z > 0 by f ( a , b , c ) = 2 a 3 b 5 c 7 d for a , b , c , d Z 0 . Prove that f is injective but not surjective. (b) Use (a) and a theorem to prove that Z 0 4 is countably infinite.
    (a) 定义函数 f:Z≥0⁴→Z>0,使得 f(a, b, c) = 23573,其中 a, b, c, d∈Z≥0。证明 f 是单射但不是满射。(b) 利用 (a) 的结论和一个定理证明 Z≥0⁴ 是可数无限集。
  23. Let X be an infinite set, and let Z be the set of all f : X X such that f ( x ) ≠ x for all x X . Prove that Z is uncountable.
    设 X 为无限集,Z 为所有 f : X → X 的集合,使得对于所有 x ∈ X,f(x) ≠ x。证明 Z 是不可数的。
  24. Let Y be the set of all f : Z > 0 Z > 0 such that f ( n ) = n for all but finitely many n Z > 0 . Is Y countable or uncountable? Prove your answer.
    设 Y 为所有 f:Z>0→Z>0 的集合,使得对于除有限个 n∈Z>0 外的所有 n,都有 f(n) = n。Y 是可数集还是不可数集?证明你的答案。
#

8

Real Numbers (Optional)

实数(可选)

8.1  Axioms for ℝ and Properties of Addition

8.1 ℝ 的公理和加法的性质

The goal of this optional chapter is to present an axiomatic development of the real number system. Such a development is needed to put the subject of calculus (also called analysis ) analysis on a firm logical footing. We assume no prior knowledge of real numbers or any other number system (even the integers), instead deducing all needed facts from the initial axioms. We do assume that facts about logic, proofs, sets, relations, and functions are already known. Thus we may invoke any results from previous sections excluding theorems about integers and rational numbers from Chapter 4 . These theorems will become available later after we formally define ℤ and ℚ.

本选读章节的目标是提出实数系统的公理化推导。这种推导对于将微积分(也称分析)置于坚实的逻辑基础之上至关重要。我们假设读者对实数或任何其他数系(甚至整数)没有任何先验知识,而是从初始公理推导出所有必要的结论。我们假设读者已经掌握了关于逻辑、证明、集合、关系和函数的知识。因此,我们可以引用前面章节中的任何结果,但第四章中关于整数和有理数的定理除外。这些定理将在我们正式定义 ℤ 和 ℚ 之后给出。

Undefined Terms and Initial Definitions

未定义术语和初始定义

Mathematical theories based on axioms were described in §2.1. Like all such theories, the axiomatic development of the real numbers begins with certain undefined terms . Our theory uses the following eight undefined terms:

第 2.1 节描述了基于公理的数学理论。与所有此类理论一样,实数的公理化发展也始于某些未定义项。我们的理论使用了以下八个未定义项:

(a) a set ℝ called the set of real numbers

(a)集合ℝ称为实数集

(b) a function ADD : R × R R called addition

(b) 一个名为 ADD:R×R→R 的函数,称为加法

(c) a real number 0 ∈ ℝ called zero

(c) 实数 0 ∈ ℝ 称为零

(d) a function AINV : R R called additive inverse

(d) 一个称为加性逆的函数 AINV:R→R

(e) a function MULT : R × R R called multiplication

(e)一个名为 MULT:R×R→R 的函数,称为乘法

(f) a real number 1 ∈ ℝ called one

(f)实数 1 ∈ ℝ 称为一

(g) a function MINV : R 0 R called multiplicative inverse

(g) 一个称为乘法逆元的函数 MINV:R≠0→R

(h) a relation LEQ on ℝ called the ordering .

(h)ℝ 上的 LEQ 关系称为排序。

The next definition introduces the standard arithmetical symbols for the functions and relations listed above.

下一个定义介绍了上面列出的函数和关系的标准算术符号。

8.1. Definition: Arithmetical Operators.

8.1. 定义:算术运算符。

(a) For all x , y ∈ ℝ, define the sum of x and y to be x + y = ADD ( x , y ) .

(a)对于所有 x, y ∈ ℝ,定义 x 和 y 的和为 x+y=ADD(x,y)。

(b) For all x ∈ ℝ, define the negative of x to be x = AINV ( x ) .

(b)对于所有 x ∈ ℝ,定义 x 的负数为 −x=AINV(x)。

(c) For all x , y ∈ ℝ, define the difference of x and y to be x y = x + ( − y ). The function sending ( x , y ) to x y is called subtraction .

(c) 对于任意 x, y ∈ ℝ,定义 x 和 y 的差为 x − y = x + ( − y)。将 (x, y) 映射到 x − y 的函数称为减法。

(d) For all x , y ∈ ℝ, define the product of x and y to be x y = MULT ( x , y ) . We often abbreviate x · y as xy .

(d) 对于所有 x, y ∈ ℝ,定义 x 和 y 的乘积为 x⋅y=MULT(x,y)。我们通常将 x · y 简写为 xy。

(e) For all x ≠ 0 in ℝ, define the inverse of x to be x 1 = MINV ( x ) .

(e)对于 ℝ 中所有 x ≠ 0,定义 x 的逆为 x−1=MINV(x)。

(f) For all x , y ∈ ℝ with y ≠ 0, define x divided by y to be x / y = x · ( y −1 ). We also write x y for x / y . The function sending ( x , y ) to x / y is called division .

(f) 对于所有 x, y ∈ ℝ 且 y ≠ 0,定义 x 除以 y 为 x/y = x · (y −1 )。我们也用 xy 表示 x/y。将 (x, y) 赋值给 x/y 的函数称为除法。

(g) For all x , y ∈ ℝ, define x y iff ( x , y ) LEQ .

(g)对于所有 x, y ∈ ℝ,定义 x ≤ y 当且仅当 (x,y)∈LEQ。

(h) For all x , y ∈ ℝ, define x y iff y x .

(h)对于所有 x, y ∈ ℝ,定义 x ≥ y 当且仅当 y ≤ x。

(i) For all x , y ∈ ℝ, define x < y iff ( x y and x y ).

(i)对于所有 x, y ∈ ℝ,定义 x < y 当且仅当 (x ≤ y 且 x ≠ y)。

(j) For all x , y ∈ ℝ, define x > y iff y < x .

(j)对于所有 x, y ∈ ℝ,定义 x > y 当且仅当 y < x。

(k) A chain of inequalities such as x < y z < w abbreviates “ x < y and y z and z < w .”

(k)诸如 x < y ≤ z < w 之类的不等式链可简写为“x < y 且 y ≤ z 且 z < w”。

Note that addition and the other arithmetical operations are assumed to be functions with codomain ℝ. This means, for example, that given any two real numbers x and y , there exists a unique real number x + y . The fact that the output x + y must belong to the set ℝ is called closure of ℝ under addition. Thus, as part of the axiomatic setup, we are assuming that ℝ is closed under addition, additive inverses, multiplication, and multiplicative inverses of nonzero real numbers. It follows that ℝ is closed under subtraction and division by nonzero real numbers. Also, the special constants 0 and 1 belong to ℝ. Starting from 1, we can define some new real numbers as follows.

注意,加法和其他算术运算被假定为值域为 ℝ 的函数。这意味着,例如,对于任意两个实数 x 和 y,存在唯一的实数 x + y。输出 x + y 必须属于集合 ℝ 的这一事实称为 ℝ 对加法的封闭性。因此,作为公理化设置的一部分,我们假设 ℝ 对非零实数的加法、加法逆元、乘法和乘法逆元封闭。由此可知,ℝ 对非零实数的减法和除法也封闭。此外,特殊常数 0 和 1 属于 ℝ。从 1 开始,我们可以定义一些新的实数,如下所示。

8.2. Definition of Some Specific Numbers. Define 2 = 1 + 1, 3 = 2 + 1, 4 = 3 + 1, 5 = 4 + 1, 6 = 5 + 1, 7 = 6 + 1, 8 = 7 + 1, 9 = 8 + 1, and 10 = 9 + 1. By closure, the symbols 2 through 10 are real numbers, called two , three , four , five , six , seven , eight , nine , and ten (respectively).

8.2. 一些特定数字的定义。定义 2 = 1 + 1,3 = 2 + 1,4 = 3 + 1,5 = 4 + 1,6 = 5 + 1,7 = 6 + 1,8 = 7 + 1,9 = 8 + 1,10 = 9 + 1。根据封闭性,符号 2 到 10 都是实数,分别称为二、三、四、五、六、七、八、九和十。

Axioms

公理

We list all the axioms for ℝ in this subsection. We divide the axioms into four groups: axioms for addition, axioms for multiplication, basic ordering axioms, and the completeness axiom. The axioms in each group, and properties derived from them, are discussed in detail later. For emphasis, we list certain closure properties as axioms, even though these properties are already implicit in the initial descriptions of the undefined terms. We begin with the five axioms for addition.

本小节列出了 ℝ 的所有公理。我们将这些公理分为四组:加法公理、乘法公理、基本排序公理和完备性公理。每组公理及其导出的性质将在后面详细讨论。为了强调,我们将某些封闭性质列为公理,即使这些性质在未定义项的初始描述中已经隐含。我们首先来看加法的五条公理。

8.3. Axioms for Addition A1. Closure: For all x , y ∈ ℝ, x + y ∈ ℝ and 0 ∈ ℝ and − x ∈ ℝ.

8.3. 加法公理 A1. 封闭性:对于所有 x, y ∈ ℝ,x + y ∈ ℝ 和 0 ∈ ℝ 和 −x ∈ ℝ。

A2. Associativity: For all x , y , z ∈ ℝ, ( x + y ) + z = x + ( y + z ).

A2.结合律:对于所有x, y, z ∈ ℝ,(x + y) + z = x + (y + z)。

A3. Additive Identity: For all x ∈ ℝ, x + 0 = x = 0 + x .

A3. 加法单位元:对于所有 x ∈ ℝ,x + 0 = x = 0 + x。

A4. Additive Inverses: For all x ∈ ℝ, x + ( − x ) = 0 = ( − x ) + x .

A4. 加法逆元:对于所有 x ∈ ℝ,x + ( − x) = 0 = ( − x) + x。

A5. Commutativity: For all x , y ∈ ℝ, x + y = y + x .

A5. 交换律:对于所有 x, y ∈ ℝ,x + y = y + x。

There are seven axioms for multiplication. The first five are exact analogues of the five addition axioms, and the last two stipulate relationships between the operations of addition and multiplication.

乘法有七条公理。前五条公理与加法的五条公理完全对应,后两条公理规定了加法和乘法运算之间的关系。

8.4. Axioms for Multiplication

8.4 乘法公理

M1. Closure: For all x , y ∈ ℝ, x · y ∈ ℝ and 1 ∈ ℝ and if x ≠ 0, then x −1 ∈ ℝ.

M1. 闭包:对于所有 x, y ∈ ℝ,x · y ∈ ℝ 和 1 ∈ ℝ,并且如果 x ≠ 0,则 x −1 ∈ ℝ。

M2. Associativity: For all x , y , z ∈ ℝ, ( x · y ) · z = x · ( y · z ).

M2.结合律:对于所有x, y, z ∈ ℝ,(x · y) · z = x · (y · z)。

M3. Multiplicative Identity: For all x ∈ ℝ, x · 1 = x = 1 · x .

M3. 乘法单位元:对于所有 x ∈ ℝ,x · 1 = x = 1 · x。

M4. Multiplicative Inverses: For all x ∈ ℝ, if x ≠ 0, then x · ( x −1 ) = 1 = ( x −1 ) · x .

M4. 乘法逆元:对于所有 x ∈ ℝ,如果 x ≠ 0,则 x · (x −1 ) = 1 = (x −1 ) · x。

M5. Commutativity: For all x , y ∈ ℝ, x · y = y · x .

M5. 交换律:对于所有 x, y ∈ ℝ,x · y = y · x。

D. Distributive Laws: For all x , y , z ∈ ℝ, x · ( y + z ) = ( x · y ) + ( x · z ) and ( y + z ) · x = ( y · x ) + ( z · x ).

D. 分配律:对于所有 x, y, z ∈ ℝ,x · (y + z) = (x · y) + (x · z) 且 (y + z) · x = (y · x) + (z · x)。

N. Nontriviality: 0 ≠ 1.

N. 非平凡性:0 ≠ 1。

There are seven ordering axioms. The first four state that (ℝ, ≤ ) is a totally ordered set; the next two relate this ordering to the algebraic operations of addition and multiplication.

共有七条排序公理。前四条公理表明 (ℝ, ≤ ) 是一个全序集;接下来的两条公理将这种排序与加法和乘法的代数运算联系起来。

8.5. Basic Ordering Axioms

8.5 基本排序公理

O1. Reflexivity: For all x ∈ ℝ, x x .

O1. 自反性:对于所有 x ∈ ℝ,x ≤ x。

O2. Antisymmetry: For all x , y ∈ ℝ, if x y and y x , then x = y .

O2. 反对称性:对于所有 x, y ∈ ℝ,如果 x ≤ y 且 y ≤ x,则 x = y。

O3. Transitivity: For all x , y , z ∈ ℝ, if x y and y z , then x z .

O3. 传递性:对于所有 x, y, z ∈ ℝ,如果 x ≤ y 且 y ≤ z,则 x ≤ z。

O4. Total Ordering: For all x , y ∈ ℝ, x y or y x .

O4. 全序:对于所有 x, y ∈ ℝ,x ≤ y 或 y ≤ x。

O5. Additive Property: For all x , y , z ∈ ℝ, if x y , then x + z y + z .

O5. 加法性质:对于所有 x, y, z ∈ ℝ,如果 x ≤ y,则 x + z ≤ y + z。

O6. Multiplicative Property: For all x , y , z ∈ ℝ, if x y and 0 ≤ z , then x · z y · z .

O6. 乘法性质:对于所有 x, y, z ∈ ℝ,如果 x ≤ y 且 0 ≤ z,则 x · z ≤ y · z。

The final ordering axiom states a crucial technical property of ℝ called completeness. To even formulate the axiom, we need the following concepts, which make sense for any partially ordered set.

最终排序公理阐述了ℝ的一个关键技术性质,称为完备性。为了表述该公理,我们需要以下概念,这些概念对任何偏序集都适用。

8.6. Definition: Upper Bounds. For all S ⊆ℝ and z 0 ∈ ℝ, z 0 is an upper bound of S iff y S , y z 0 . For all S ⊆ℝ, S is bounded above iff z , y S , y z .

8.6. 定义:上界。对于所有 S⊆ℝ 和 z 0 ∈ ℝ,z0 是 S 的上界当且仅当 ∀y∈S,y≤z0。对于所有 S⊆ℝ,S 有上界当且仅当存在 z∈ℝ,∀y∈S,y≤z。

8.7. Definition: Least Upper Bounds. For all S ⊆ℝ and z 0 ∈ ℝ, z 0 is a least upper bound of S (written z 0 = lub S ) iff

8.7. 定义:最小上界。对于所有 S⊆ℝ 和 z 0 ∈ ℝ,z0 是 S 的最小上界(记作 z0=lubS)当且仅当

( y S S , y z z 0 ) ( x x R , if 如果 [ y S S , y x x ] then 然后 z z 0 x x ) .

8.8. Completeness Axiom O7. For all S ⊆ℝ, if S is nonempty and S is bounded above then S has a least upper bound in ℝ.

8.8. 完备性公理 O7. 对于所有 S⊆ℝ,如果 S 非空且 S 有上界,则 S 在 ℝ 中具有最小上界。

Although not explicitly listed here, we are also assuming various logical axioms governing the logical symbols such as ↠, , , and =. Regarding equality, we are assuming that = is reflexive, symmetric, and transitive, and satisfies various substitution properties such as: for all x , y , z ∈ ℝ, if y = z , then x + y = x + z . These properties will be used frequently without specifically invoking the underlying logical axioms.

虽然此处未明确列出,但我们也假设了各种逻辑公理支配着诸如 ↠、∀、∃ 和 = 等逻辑符号。关于等号,我们假设 = 具有自反性、对称性和传递性,并且满足各种替换性质,例如:对于所有 x, y, z ∈ ℝ,如果 y = z,则 x + y = x + z。这些性质将在不显式引用底层逻辑公理的情况下频繁使用。

8.9. Remark: Algebraic Structures. In abstract algebra, we study many algebraic systems that satisfy various subsets of the axioms listed above (with the set ℝ replaced throughout by an appropriate set of objects). In particular, a system satisfying axioms A1 through A5 is called a commutative group ; omitting A5 gives the axioms defining a group . A system satisfying all the additive and multiplicative axioms is called a field ; omitting M4, N, and multiplicative inverses gives the axioms for a commutative ring ; and further omitting axiom M5 gives the axioms for a ring (with identity). If we include all the axioms for ℝ except completeness (O7), we get the definition of an ordered field ; an ordered commutative ring is defined similarly. Here are a few examples using number systems not officially defined yet: ℤ (the set of integers) is an ordered commutative ring that is not a field; ℚ (the set of rational numbers) is an ordered field that does not satisfy the completeness axiom; ℂ (the set of complex numbers) is a field that cannot be made into an ordered field; and the set of 2 × 2 matrices with real entries is a non-commutative ring.

8.9. 备注:代数结构。在抽象代数中,我们研究许多满足上述公理不同子集的代数系统(其中集合ℝ被替换为适当的对象集合)。特别地,满足公理A1至A5的系统称为交换群;省略A5即可得到群的定义公理。满足所有加法和乘法公理的系统称为域;省略M4、N和乘法逆元即可得到交换环的公理;进一步省略公理M5即可得到带恒等元的环的公理。如果我们包含除完备性(O7)之外的所有ℝ的公理,即可得到有序域的定义;有序交换环的定义类似。以下是一些使用尚未正式定义的数系的例子:ℤ(整数集)是一个有序交换环,但它不是域;ℚ(有理数集)是一个有序域,但它不满足完备性公理;ℂ(复数集)是一个域,它不能被转化为有序域;而元素为实数的 2 × 2 矩阵的集合是一个非交换环。

Properties of Addition

加法性质

We begin our development of the properties of ℝ by deriving some consequences of the addition axioms A1 through A5. It is helpful to keep track of which axioms are needed to derive various results, since the same results also hold in other more general algebraic systems for which these particular axioms are true. In particular, we give proofs avoiding the commutativity axiom A5 when possible. Our first theorem is a basic but very useful uniqueness result.

我们首先推导加法公理 A1 至 A5 的一些推论,以此展开对 ℝ 性质的研究。记录推导各个结果所需的公理很有帮助,因为同样的结果也适用于其他更一般的代数系统,只要这些公理在这些系统中成立。特别地,我们尽可能避免使用交换律公理 A5 来进行证明。我们的第一个定理是一个基础但非常有用的唯一性结果。

8.10. Theorem: Left Cancellation Law for Addition.

8.10. 定理:加法的左约分法则。

For all x , y , z ∈ ℝ, if x + y = x + z , then y = z .

对于所有 x, y, z ∈ ℝ,如果 x + y = x + z,则 y = z。

Proof. Fix arbitrary x , y , z ∈ ℝ. Assume x + y = x + z ; prove y = z . By the substitution property of equality, we can add − x to both sides of the assumption x + y = x + z , obtaining ( − x ) + ( x + y ) = ( − x ) + ( x + z ). (This step also uses closure (A1), since we need to know − x ∈ ℝ, x + y ∈ ℝ, and x + z ∈ ℝ in order for the sums ( − x ) + ( x + y ) and ( − x ) + ( x + z ) to be defined. We make constant use of substitution axioms and closure axioms when manipulating complicated expressions, but we seldom mention these axioms in the sequel.) By associativity (A2), the previous equation becomes (( − x ) + x ) + y = (( − x ) + x ) + z . By the inverse axiom (A4), we get 0 + y = 0 + z . Finally, the identity axiom (A3) turns this equation into y = z , which is the required conclusion. This proof used axioms A1, A2, A3, and A4.      □

证明。固定任意 x、y、z ∈ ℝ。假设 x + y = x + z;证明 y = z。根据等式的代换性质,我们可以将 -x 加到假设 x + y = x + z 的两边,得到 ( - x) + (x + y) = ( - x) + (x + z)。(这一步也用到了封闭性 (A1),因为我们需要知道 -x ∈ ℝ、x + y ∈ ℝ 和 x + z ∈ ℝ,才能定义和式 ( - x) + (x + y) 和 ( - x) + (x + z)。我们在处理复杂表达式时经常用到代换公理和封闭性公理,但在后续内容中很少提及这些公理。)根据结合律 (A2),前面的等式变为 (( - x) + x) + y = (( - x) + x) + z。根据逆公理 (A4),我们得到 0 + y = 0 + z。最后,恒等公理 (A3) 将该等式转化为 y = z,这正是所需的结论。本证明使用了公理 A1、A2、A3 和 A4。□

There is an analogous right cancellation law : for all x , y , z ∈ ℝ, if y + x = z + x , then y = z . This law follows from the left cancellation law via commutativity (A5), but it is better to adapt the proof given above (add − x to both sides on the right ) so that the proof only uses axioms A1 through A4. The next theorem lists some fundamental facts that can be deduced as special cases of the left cancellation law. Similar facts hold based on the right cancellation law, but we do not state these explicitly.

存在一个类似的右消去律:对于任意 x, y, z ∈ ℝ,若 y + x = z + x,则 y = z。该律可由左消去律通过交换律 (A5) 推导得出,但最好对上述证明进行调整(在等式右侧两边同时加上 −x),使其仅使用公理 A1 至 A4。下一个定理列出了一些可作为左消去律特例推导出的基本事实。基于右消去律也有类似的事实,但我们在此不再赘述。

8.11. Theorem on Addition. For all x , y , a , b ∈ ℝ:

8.11. 加法定理。对于所有 x, y, a, b ∈ ℝ:

(a) Uniqueness of Additive Identity: If x + y = x , then y = 0.

(a)加法单位元的唯一性:如果 x + y = x,则 y = 0。

(b) Uniqueness of Additive Inverses: If x + y = 0, then y = − x .

(b)加法逆元的唯一性:如果 x + y = 0,则 y = −x。

(c) Negative of Zero: −0 = 0.

(c)零的负数:−0 = 0。

(d) Double Negative Rule: For all a ∈ ℝ, −( − a ) = a .

(d)双重否定规则:对于所有 a ∈ ℝ,−( − a) = a。

(e) Negative of a Sum: For all a , b ∈ ℝ, −( a + b ) = ( − b ) + ( − a ).

(e)和的负数:对于所有 a, b ∈ ℝ,−(a + b) = ( − b) + ( − a)。

Proof. For (a), fix x , y ∈ ℝ; assume x + y = x ; prove y = 0. Using the identity axiom (A3), our assumption becomes x + y = x + 0. Now use left cancellation, taking z = 0, to deduce y = 0. [This result says that zero is the only real number having the additive identity property stated in A3.]

证明。对于 (a),固定 x, y ∈ ℝ;假设 x + y = x;证明 y = 0。利用恒等式公理 (A3),我们的假设变为 x + y = x + 0。现在利用左约分,令 z = 0,即可推导出 y = 0。[此结果表明,零是唯一满足 A3 中所述的加法恒等性质的实数。]

For (b), fix x , y ∈ ℝ; assume x + y = 0; prove y = − x . Using the inverse axiom (A4), our assumption becomes x + y = x + ( − x ). Now use left cancellation, taking z = − x , to deduce y = − x . [This result says that for each fixed real x , − x is the only real number having the additive inverse property stated in A4.]

对于 (b),固定 x, y ∈ ℝ;假设 x + y = 0;证明 y = −x。利用逆公理 (A4),我们的假设变为 x + y = x + ( − x)。现在利用左约分,令 z = −x,即可推导出 y = −x。[此结果表明,对于每个固定的实数 x,−x 是唯一满足 A4 中所述的加法逆性质的实数。]

Now we prove (c). On one hand, taking x = 0 in the inverse axiom (A4) gives 0 + ( − 0) = 0. On the other hand, taking x = −0 in the identity axiom (A3) gives 0 + ( − 0) = −0. Thus, −0 = 0 + ( − 0) = 0.

现在我们证明 (c)。一方面,在逆公理 (A4) 中令 x = 0,得到 0 + ( − 0) = 0。另一方面,在恒等公理 (A3) 中令 x = −0,得到 0 + ( − 0) = −0。因此,−0 = 0 + ( − 0) = 0。

For (d), fix a ∈ ℝ. By the inverse axiom (A4), ( − a ) + a = 0. Using part (b) with x = − a and y = a , we deduce y = − x , i.e., a = −( − a ).

对于 (d),固定 a ∈ ℝ。根据逆公理 (A4),( − a) + a = 0。利用 (b) 部分,令 x = −a 和 y = a,我们推导出 y = −x,即 a = −( − a)。

For (e), fix a , b ∈ ℝ. We compute

对于 (e),固定 a, b ∈ ℝ。我们计算

( a 一个 + b b ) + ( [ b b ] + [ a 一个 ] ) = a 一个 + ( b b + ( [ b b ] + [ a 一个 ] ) ) = a 一个 + ( ( b b + [ b b ] ) + [ a 一个 ] ) = a 一个 + ( 0 + [ a 一个 ] ) = a 一个 + [ a 一个 ] = 0.

These equalities follow by associativity (A2), associativity again, the inverse axiom (A4), the identity axiom (A3), and the inverse axiom (A4). Now use part (b) with x = a + b and y = [ − b ] + [ − a ] to deduce y = − x , i.e., [ − b ] + [ − a ] = −( a + b ).

这些等式可由结合律(A2)、结合律再次成立、逆公理(A4)、恒等公理(A3)和逆公理(A4)得出。现在利用(b)部分,令 x = a + b 和 y = [ − b] + [ − a],推导出 y = −x,即 [ − b] + [ − a] = −(a + b)。

The above proofs use axioms A1 through A4 only.    □

以上证明仅使用了公理 A1 至 A4。□

8.12. Remark. If we allow ourselves to use commutativity (A5), we can rewrite part (e) of the last theorem as −( a + b ) = ( − a ) + ( − b ). We might attempt to prove this identity using the distributive axiom (D), as follows:

8.12. 备注。如果我们允许使用交换律(A5),我们可以将最后一个定理的(e)部分改写为 −(a + b) = ( − a) + ( − b)。我们可以尝试使用分配律(D)来证明这个恒等式,如下所示:

( a 一个 + b b ) = ( 1 ) ( a 一个 + b b ) = ( 1 ) a 一个 + ( 1 ) b b = ( a 一个 ) + ( b b ) .

However, this calculation requires the property − a = ( − 1) · a , which has not yet been proved. Even worse, this proof refers to multiplication to prove a property involving addition alone. Thus the proof would not work in an algebraic structure that had an addition operation but no multiplication operation.

然而,这个计算需要用到性质 −a = ( − 1) · a,而这个性质尚未被证明。更糟糕的是,这个证明竟然用乘法来证明一个只涉及加法的性质。因此,这个证明在只有加法运算而没有乘法运算的代数结构中是行不通的。

8.13. Theorem on Subtraction. For all x , y , z ∈ ℝ:

8.13. 减法定理。对于所有 x, y, z ∈ ℝ:

(a) x − 0 = x .

(a)x − 0 = x。

(b) 0 − x = − x .

(b)0 − x = −x。

(c) x x = 0.

(c)x − x = 0。

(d) −( x y ) = y x = ( − x ) + y .

(d)−(x−y)=y−x=(−x)+y。

(e) ( x y ) + ( y z ) = x z .

(e)(x − y) + (y − z) = x − z。

(f) ( x + z ) − ( y + z ) = x y .

(f)(x + z) − (y + z) = x − y。

Proof. We prove one identity and leave the others as exercises. Fix x , y ∈ ℝ. We compute

证明。我们证明一个恒等式,其余的留作练习。固定 x, y ∈ ℝ。我们计算

( x x y ) = ( x x + [ y ] ) = ( y ) + ( x x ) = y + ( x x ) = y x x ,

using the definition of subtraction, then 8.11 (e), then 8.11 (d), then the definition of subtraction. Using commutativity (A5), we can also write y x = y + ( − x ) = ( − x ) + y .

利用减法的定义,然后是 8.11(e),然后是 8.11(d),然后是减法的定义。利用交换律 (A5),我们也可以写成 y − x = y + ( − x) = ( − x) + y。

The next result illustrates how specific arithmetical identities can be proved from the axioms.

下一个结果说明了如何从公理中证明特定的算术恒等式。

8.14. Proposition. 2 + 4 = 6 and 3 − 5 = −2.

8.14.命题。2 + 4 = 6 且 3 − 5 = −2。

Proof. We gradually build up to the identity 2 + 4 = 6 by proving simpler related facts. First, 2 + 2 = 2 + (1 + 1) = (2 + 1) + 1 = 3 + 1 = 4 using the definition of 2, then associativity (A2), then the definition of 3, then the definition of 4. Second, 2 + 3 = 2 + (2 + 1) = (2 + 2) + 1 = 4 + 1 = 5 using the definition of 3, then associativity (A2), then the first result, then the definition of 5. Third, 2 + 4 = 2 + (3 + 1) = (2 + 3) + 1 = 5 + 1 = 6 using the definition of 4, then associativity (A2), then the second result, then the definition of 6.

证明。我们通过证明一些更简单的相关事实,逐步推导出恒等式 2 + 4 = 6。首先,利用 2 的定义,然后利用结合律(A2),再利用 3 的定义,最后利用 4 的定义,得出 2 + 2 = 2 + (1 + 1) = (2 + 1) + 1 = 3 + 1 = 4。其次,利用 3 的定义,然后利用结合律(A2),再利用第一个结论,最后利用 5 的定义,得出 2 + 3 = 2 + (2 + 1) = (2 + 2) + 1 = 4 + 1 = 5。最后,利用 4 的定义,再利用结合律(A2),再利用第二个结论,最后利用 6 的定义,得出 2 + 4 = 2 + (3 + 1) = (2 + 3) + 1 = 5 + 1 = 6。

Similarly, to prove 3 − 5 = −2, observe that 2 + (3 − 5) = 2 + (3 + ( − 5)) = (2 + 3) + ( − 5) = 5 + ( − 5) = 0 by definition of subtraction, then associativity (A2), then the second result above, then inverses (A4). Now use 8.11 (b) with x = 2 and y = 3 − 5 to see that y = − x , i.e., 3 − 5 = −2.      □

类似地,为了证明 3 − 5 = −2,观察到 2 + (3 − 5) = 2 + (3 + ( − 5)) = (2 + 3) + ( − 5) = 5 + ( − 5) = 0(根据减法的定义),然后是结合律 (A2),再是上面的第二个结果,最后是逆运算 (A4)。现在,使用 8.11(b) 并令 x = 2 和 y = 3 − 5,即可看出 y = −x,即 3 − 5 = −2。□

Any other arithmetical fact involving addition or subtraction of specific integers can be proved in a similar manner.

任何其他涉及特定整数加减运算的算术事实都可以用类似的方法证明。

8.15. Remark: Generalized Associativity. Given three real numbers a , b , c , the expression a + b + c is ambiguous: it could mean ( a + b ) + c or a + ( b + c ). However, associativity (A2) assures us that the latter two expressions must be equal, so it is safe to write a + b + c with no parentheses. Similarly, given four real numbers a , b , c , d , the ambiguous expression a + b + c + d could represent any of the following five quantities:

8.15. 备注:广义结合律。给定三个实数 a、b、c,表达式 a + b + c 是不确定的:它可以表示 (a + b) + c 或 a + (b + c)。然而,结合律 (A2) 保证后两个表达式必然相等,因此可以安全地写成 a + b + c,无需括号。类似地,给定四个实数 a、b、c、d,不确定的表达式 a + b + c + d 可以表示以下五个量中的任何一个:

( a 一个 + ( b b + c c ) ) + d d , ( ( a 一个 + b b ) + c c ) + d d , ( a 一个 + b b ) + ( c c + d d ) , a 一个 + ( b b + ( c c + d d ) ) , a 一个 + ( ( b b + c c ) + d d ) .

Each expression on this list can be seen to be equal to the next expression by an appropriate special case of associativity (A2). For example, letting x = a + b , y = c , and z = d in A2 shows that the second expression equals the third. Because of this fact, we often write a + b + c + d with no parentheses, and we can group adjacent terms together at will if needed in a particular calculation (as in the proof of Theorem 8.11 (e)). A similar result holds for a sum of n terms: given a 1 , a 2 , …, a n ∈ ℝ, any possible parenthesization of a 1 + a 2 + · · · + a n produces the same final output. This fact is called generalized associativity , and its proof is far from obvious, as the n = 4 case already suggests. Since we have not yet defined what a positive integer n is, we cannot even state this general fact formally, much less prove it. Luckily, in our initial development, we only need this fact for a few specific small values of n (like n = 4), where it can be justified by explicit repeated uses of A2.

通过结合律的一个特例(A2),列表中的每个表达式都与下一个表达式相等。例如,令 A2 中的 x = a + b,y = c,z = d,则第二个表达式等于第三个表达式。因此,我们通常省略括号来表示 a + b + c + d,并且在特定计算中可以根据需要将相邻项组合在一起(如定理 8.11(e) 的证明)。对于 n 项之和,也有类似的结论:给定 a 1 , a 2 , …, a n ∈ ℝ,a 1 + a 2 + … + a n 的任何可能的加括号方式都会产生相同的最终结果。这个事实被称为广义结合律,其证明远非显而易见,n = 4 的例子已经表明了这一点。由于我们尚未定义正整数 n 的含义,因此我们甚至无法正式地陈述这个一般事实,更不用说证明它了。幸运的是,在我们最初的推导过程中,我们只需要对几个特定的较小 n 值(例如 n = 4)使用这个事实,而这些值可以通过显式地重复使用 A2 来证明。

There is a related fact called generalized commutativity based on A5. It states that we can rearrange the order of terms in a 1 + a 2 + · · · + a n in any fashion without changing the final result. For example, a + b + c + d = d + b + a + c = b + c + d + a , and so on. Henceforth, we use generalized associativity and commutativity (for small values of n ) without specific comment.

基于 A5,还有一个相关的结论,称为广义交换律。它指出,我们可以以任何方式重新排列 a 1 + a 2 + · · · + a n 中的项的顺序,而不会改变最终结果。例如,a + b + c + d = d + b + a + c = b + c + d + a,依此类推。因此,我们不再赘述广义结合律和交换律(对于较小的 n 值)。

Section Summary

章节概要

  1. Undefined Terms. Our axiomatic development of the real number system uses these undefined concepts: the set ℝ, the operations x + y , − x , x · y , and z −1 , the numbers 0 and 1, and the relation x y . (Here, x , y , z ∈ ℝ and z ≠ 0.)
    未定义术语。我们对实数系统的公理化发展使用了以下未定义概念:集合ℝ,运算x + y、−x、x · y和z −1 ,数字0和1,以及关系x ≤ y。(这里,x、y、z∈ℝ且z≠0。)
  2. Axioms. Addition and multiplication satisfy the axioms of closure, associativity, identity, inverses, and commutativity. The distributive laws hold, and 0 ≠ 1. The relation ≤ is reflexive, antisymmetric, transitive, a total order, and is preserved by addition and multiplication by nonnegative numbers. Finally, the completeness axiom tells us that every nonempty subset of ℝ that is bounded above has a least upper bound in ℝ.
    公理。加法和乘法满足封闭性、结合律、恒等律、逆元律和交换律。分配律成立,且 0 ≠ 1。关系 ≤ 是自反的、反对称的、传递的、全序的,并且在非负数的加法和乘法运算下保持不变。最后,完备性公理告诉我们,ℝ 的每个非空且有上界的子集在 ℝ 中都存在最小上界。
  3. Basic Definitions. We define x y = x + ( − y ), x / z = x · z −1 , x < y iff x y and x y , x y iff y x , and x > y iff y < x (for x , y , z ∈ ℝ with z ≠ 0). Define 2 = 1 + 1, 3 = 2 + 1, and so on.
    基本定义。我们定义 x − y = x + ( − y),x/z = x · z −1 ,x < y 当且仅当 x ≤ y 且 x ≠ y,x ≥ y 当且仅当 y ≤ x,x > y 当且仅当 y < x(对于 x, y, z ∈ ℝ 且 z ≠ 0)。定义 2 = 1 + 1,3 = 2 + 1,依此类推。
  4. Properties of Addition. The additive cancellation law holds: for all x , y , z ∈ ℝ, x + y = x + z y = z . Zero is the unique additive identity; − x is the unique additive inverse of x ∈ ℝ; −0 = 0; −( − x ) = x ; and −( x + y ) = ( − y ) + ( − x ) for all x , y ∈ ℝ. Addition satisfies generalized associativity and generalized commutativity.
    加法的性质。加法消去律成立:对于任意 x, y, z ∈ ℝ,x + y = x + z ↠ y = z。零是唯一的加法单位元;−x 是 x ∈ ℝ 的唯一加法逆元;−0 = 0;−( − x) = x;且对于任意 x, y ∈ ℝ,−(x + y) = ( − y) + ( − x)。加法满足广义结合律和广义交换律。

Exercises

练习

  1. (a) Give a detailed proof of the right cancellation law without using axiom A5. (b) State and prove analogues of 8.11 (a) and 8.11 (b) based on the right cancellation law.
    (a)在不使用公理 A5 的情况下,给出右消去定律的详细证明。(b)根据右消去定律,陈述并证明 8.11(a) 和 8.11(b) 的类似物。
  2. Complete the proof of Theorem 8.13 .
    完成定理 8.13 的证明。
  3. Deduce 8.11 (c) from 8.11 (b).
    从 8.11(b) 推导出 8.11(c)。
  4. Prove that 5 + 5 = 10 and 2 − 6 = −4 and 6 − 3 = 3 (you may need to prove some other related identities first).
    证明 5 + 5 = 10 且 2 − 6 = −4 且 6 − 3 = 3 (你可能需要先证明一些其他相关的恒等式)。
  5. (a) Using only the addition axioms A1 through A5, prove: a , b , c , d R , ( a + c ) + ( b + d ) = ( a + b ) + ( c + d ) . (b) Using only axioms and theorems proved above, prove: a , b , c , d R , ( a + c ) ( b + d ) = ( a b ) + ( c d ) .
    (a) 仅使用加法公理 A1 至 A5,证明:∀a,b,c,d∈R,(a+c)+(b+d)=(a+b)+(c+d)。(b) 仅使用上述已证明的公理和定理,证明:∀a,b,c,d∈R,(a+c)−(b+d)=(a−b)+(c−d)。
  6. Suppose we give meanings to the undefined symbols of our theory by letting ℝ = { a } (where a is some fixed object), a + a = a , − a = a , a · a = a , 0 = a , 1 = a , and LEQ = { ( a , a ) } . (Multiplicative inverses are not defined since nonzero elements of ℝ do not exist.) Prove that this system satisfies all the axioms for the real numbers except the nontriviality axiom N.
    假设我们通过令 ℝ = {a}(其中 a 是某个固定对象)、a + a = a、−a = a、a · a = a、0 = a、1 = a 和 LEQ={(a,a)} 来赋予理论中未定义符号意义。(由于 ℝ 中不存在非零元素,因此乘法逆元未定义。)证明该系统满足实数域的所有公理,除了非平凡性公理 N。
  7. Let ℝ 3 = ℝ × ℝ × ℝ. Define addition, zero, and negatives on this set as follows: for all a , b , c , x , y , z ∈ ℝ,
    ( a , b , c ) + ( x , y , z ) = ( a + x , b + y , c + z ) , 0 = ( 0 , 0 , 0 ) , ( a , b , c ) = ( a , b , c ) .
    Use known properties of ℝ to show that ℝ 3 satisfies the additive axioms A1 through A5.
    令 ℝ 3 = ℝ × ℝ × ℝ。定义该集合上的加法、零和负数如下:对于所有 a, b, c, x, y, z ∈ ℝ,(a,b,c)+(x,y,z)=(a+x,b+y,c+z),0=(0,0,0),−(a,b,c)=(−a,−b,−c)。利用 ℝ 的已知性质证明 ℝ 3 满足加法公理 A1 至 A5。
  8. Suppose we redefine the multiplicative operations in ℝ by setting x · y = 0, 1 = 0, and z −1 = 0 for all x , y , z ∈ ℝ with z ≠ 0. Which of axioms M1 through M5, D, and N are now true?
    假设我们重新定义 ℝ 中的乘法运算,令 x · y = 0,1 = 0,并且对于所有 x, y, z ∈ ℝ 且 z ≠ 0,令 z −1 = 0。那么,公理 M1 到 M5、D 和 N 中哪些现在是正确的?
  9. Suppose we redefine addition on ℝ by setting x + 0 = x = 0 + x for all x ∈ ℝ, and y + z = 0 for all nonzero y , z ∈ ℝ. (a) Show that axioms A3, A4, and A5 are true, but the associativity axiom A2 is false. (b) Show that additive inverses are not unique in this system.
    假设我们重新定义 ℝ 上的加法,令对于所有 x ∈ ℝ,x + 0 = x = 0 + x,并且对于所有非零的 y, z ∈ ℝ,y + z = 0。(a)证明公理 A3、A4 和 A5 为真,但结合律公理 A2 为假。(b)证明在该系统中,加法逆元不唯一。
  10. Suppose we redefine addition on ℝ by setting x + y = x for all x , y ∈ ℝ. Which of the addition axioms A1 through A5 are true?
    假设我们重新定义 ℝ 上的加法,令对于所有 x, y ∈ ℝ,x + y = x。那么,加法公理 A1 到 A5 中哪些是正确的?
  11. For x , y ∈ ℝ, define x y = x + y + 2 . Show that the addition axioms A1 through A5 hold if we replace + by , 0 by −2, and − x by − x − 4.
    对于 x, y ∈ ℝ,定义 x⋆y=x+y+2。证明如果我们用 ⋆ 代替 +,用 −2 代替 0,用 −x − 4 代替 −x,则加法公理 A1 到 A5 成立。
  12. Let X be a fixed set. Use theorems about bijections to prove that the set G consisting of all bijections f : X X is a group if we replace addition (in axioms A1 through A4) by composition of functions, zero by Id X , and additive inverses by inverse functions.
    设 X 为一个固定的集合。利用双射定理证明,如果将公理 A1 到 A4 中的加法替换为函数复合,将零替换为 Id X ,将加法逆元替换为逆函数,则由所有双射 f:X → X 构成的集合 G 是一个群。
  13. If possible, give an example of a field in which 0 = 3.
    如果可能,请举例说明一个 0 = 3 的域。
  14. Prove: for all a , b , c ∈ ℝ, ( a + b ) + c = ( b + a ) + c = ( c + b ) + a = ( b + c ) + a = ( a + c ) + b = ( c + a ) + b . Justify every step using axioms A2 and A5 only.
    证明:对于所有 a, b, c ∈ ℝ,(a + b) + c = (b + a) + c = (c + b) + a = (b + c) + a = (a + c) + b = (c + a) + b。仅使用公理 A2 和 A5 证明每一步。
  15. (a) Define real numbers 11, 12, 13, 14, 15, 16, 17, and 18. (b) Make an addition table with rows and columns labeled 0, 1, 2, …, 9, such that the entry in row a and column b is a + b . (c) Suppose you wrote a complete proof of your answer to (b). In such a proof, how many times would a definition, axiom, or previous result be invoked? Explain.
    (a) 定义实数 11、12、13、14、15、16、17 和 18。(b) 创建一个加法表,行和列分别标记为 0、1、2、…、9,使得第 a 行第 b 列的元素为 a + b。(c) 假设你对 (b) 的答案写出了一个完整的证明。在这个证明中,定义、公理或先前的结论会被调用多少次?请解释。
  16. List all complete parenthesizations of the expression a + b + c + d + e , where a , b , c , d , e ∈ ℝ. Use the associativity axiom A2 to prove that all these expressions are equal.
    列出表达式 a + b + c + d + e 的所有完整括号形式,其中 a, b, c, d, e ∈ ℝ。利用结合律 A2 证明所有这些表达式都相等。

8.2  Algebraic Properties of Real Numbers

8.2 实数的代数性质

We continue to develop properties of the arithmetical operations — addition, subtraction, multiplication, and division — based on the algebraic axioms A1 through A5, M1 through M5, D, and N. The theorems proved using these axioms are valid in any field (such as the rational numbers, the complex numbers, or the integers modulo a prime).

我们继续基于代数公理 A1 至 A5、M1 至 M5、D 和 N 来发展算术运算(加法、减法、乘法和除法)的性质。使用这些公理证明的定理在任何领域(例如有理数、复数或模素数的整数)都有效。

Multiplicative Properties of Zero

零的乘法性质

Before developing general properties of multiplication, we need to examine the relationship between zero (the additive identity) and the multiplication operation. Our first result is very familiar, but the proof is surprisingly tricky.

在推导乘法的一般性质之前,我们需要考察零(加法单位元)与乘法运算之间的关系。我们的第一个结果非常熟悉,但证明过程却出乎意料地复杂。

8.16. Theorem on Multiplication by Zero.

8.16. 乘以零的定理。

Z1. For all a ∈ ℝ, 0 · a = 0 and a · 0 = 0.

Z1. 对于所有 a ∈ ℝ,0 · a = 0 且 a · 0 = 0。

Proof. Fix a ∈ ℝ. [How do we even begin? This is the hard part — the trick is to think of examining the quantity 0 · a + 0 · a .] We compute (0 · a ) + (0 · a ) = (0 + 0) · a = 0 · a using the distributive axiom (D) and the additive identity axiom (A3). Now recall Theorem 8.11 (a): x , y R , x + y = x y = 0 . Taking x = y = 0 · a here, we have verified that x + y = x and so we may conclude that y = 0, i.e., 0 · a = 0. We prove a · 0 = 0 similarly, starting from the quantity a · 0 + a · 0. This proof used addition axioms A1 through A4, M1 (where?), and D. [Since D and N are the only axioms connecting multiplication to addition, we can retroactively motivate the introduction of 0 · a + 0 · a by noting that we needed some expression that would allow us to use the distributive law.]      □

证明。固定 a ∈ ℝ。[我们该如何开始呢?这是难点——诀窍在于考虑考察量 0 · a + 0 · a。] 我们利用分配律公理 (D) 和加法恒等式公理 (A3) 计算 (0 · a) + (0 · a) = (0 + 0) · a = 0 · a。现在回顾定理 8.11(a):∀x,y∈R,x+y=x⇒y=0。令 x = y = 0 · a,我们验证了 x + y = x,因此可以得出 y = 0,即 0 · a = 0。类似地,我们从 a · 0 + a · 0 出发,证明 a · 0 = 0。此证明使用了加法公理 A1 至 A4、M1(具体位置?)和 D。[由于 D 和 N 是连接乘法和加法的唯一公理,我们可以追溯性地解释引入 0 · a + 0 · a 的原因,因为我们需要某种表达式来使用分配律。] □

8.17. Theorem on Closure Properties of Nonzero Real Numbers.

8.17.非零实数的封闭性质定理。

Z2. For all x , y ∈ ℝ, if x ≠ 0 and y ≠ 0, then x · y ≠ 0.

Z2. 对于所有 x, y ∈ ℝ,如果 x ≠ 0 且 y ≠ 0,则 x · y ≠ 0。

Z3. No Zero Divisors: For all x , y ∈ ℝ, if x · y = 0, then x = 0 or y = 0.

Z3. 无零因子:对于所有 x, y ∈ ℝ,如果 x · y = 0,则 x = 0 或 y = 0。

Z4. For all x , y ∈ ℝ, if x · y ≠ 0, then x ≠ 0 and y ≠ 0.

Z4. 对于所有 x, y ∈ ℝ,如果 x · y ≠ 0,则 x ≠ 0 且 y ≠ 0。

Z5. For all x ∈ ℝ, if x ≠ 0, then x −1 ≠ 0.

Z5. 对于所有 x ∈ ℝ,如果 x ≠ 0,则 x −1 ≠ 0。

Z6. For all x ∈ ℝ, if x ≠ 0, then − x ≠ 0.

Z6. 对于所有 x ∈ ℝ,如果 x ≠ 0,则 −x ≠ 0。

Proof. Z2. [We try proof by contradiction.] Assume, to get a contradiction, that there exist x , y ∈ ℝ such that x ≠ 0 and y ≠ 0 and x · y = 0. We know x −1 exists because x ≠ 0; so we may multiply both sides of x · y = 0 on the left by x −1 , obtaining x −1 · ( x · y ) = x · 0. Using associativity (M2) and the previous result (Z1), this becomes ( x −1 · x ) · y = 0. By the inverse axiom (M4), we get 1 · y = 0. By the identity axiom (M3), we deduce y = 0. But this contradicts the assumption y ≠ 0. So property Z2 holds.

证明。Z2. [我们尝试用反证法证明。] 假设为了得到矛盾,存在 x, y ∈ ℝ 使得 x ≠ 0 且 y ≠ 0 且 x · y = 0。我们知道 x −1 存在,因为 x ≠ 0;因此,我们可以将等式 x · y = 0 的左侧两边同时乘以 x −1 ,得到 x −1 · (x · y) = x · 0。利用结合律 (M2) 和之前的结果 (Z1),可得 (x −1 · x) · y = 0。根据逆公理 (M4),我们得到 1 · y = 0。根据恒等公理 (M3),我们推导出 y = 0。但这与假设 y ≠ 0 相矛盾。因此性质 Z2 成立。

Z3. The IF-statement here is the contrapositive of the IF-statement in Z2, so Z3 is logically equivalent to Z2.

Z3. 此处的 IF 语句是 Z2 中 IF 语句的逆否命题,因此 Z3 在逻辑上等价于 Z2。

Z4. Fix x , y ∈ ℝ. Using a contrapositive proof, we assume x = 0 or y = 0, and prove x · y = 0. In the case where x = 0, x · y = 0 · y = 0 follows from Z1. Similarly, in the case where y = 0, x · y = x · 0 = 0 follows from Z1.

Z4. 固定 x, y ∈ ℝ。利用逆否命题证明,我们假设 x = 0 或 y = 0,并证明 x · y = 0。当 x = 0 时,由 Z1 可得 x · y = 0 · y = 0。类似地,当 y = 0 时,由 Z1 可得 x · y = x · 0 = 0。

Z5. Assume, to get a contradiction, that there exists x ∈ ℝ with x ≠ 0 (so x −1 exists) and x −1 = 0. Multiply both sides of x −1 = 0 by x to get x · x −1 = x · 0. By the inverse axiom (M4) and Z1, we get 1 = 0. This contradicts the nontriviality axiom N, which states 1 ≠ 0.

Z5. 为了得出矛盾,假设存在 x ∈ ℝ 且 x ≠ 0(因此 x −1 存在),并且 x −1 = 0。将 x −1 = 0 两边同乘以 x,得到 x · x −1 = x · 0。根据逆公理 (M4) 和 Z1,我们得到 1 = 0。这与非平凡性公理 N 相矛盾,因为 N 指出 1 ≠ 0。

Z6. We use a contrapositive proof. Fix x ∈ ℝ; assume − x = 0; prove x = 0. By Theorem 8.11 , x = −( − x ) = −0 = 0.     □

Z6. 我们使用逆否命题证明。固定 x ∈ ℝ;假设 −x = 0;证明 x = 0。根据定理 8.11,x = −( − x) = −0 = 0。□

In abstract algebra, an integral domain is a commutative ring where 1 ≠ 0 and there are no zero divisors. In other words, an integral domain satisfies all the additive and multiplicative axioms omitting M4 and multiplicative inverses, and including Z3 as an additional axiom. Later, we define ℤ (the set of integers) and prove that ℤ is an integral domain. Our proof of Z3 from the field axioms shows that every field is an integral domain , but the converse is false (ℤ provides a counterexample).

在抽象代数中,整环是一个交换环,其中 1 ≠ 0 且不存在零因子。换句话说,整环满足所有加法和乘法公理,但省略了 M4 和乘法逆元,并将 Z3 作为附加公理。接下来,我们定义 ℤ(整数集)并证明 ℤ 是一个整环。我们从域公理出发证明 Z3 表明每个域都是一个整环,但反之不成立(ℤ 提供了一个反例)。

Sign Rules; More Arithmetic

符号规则;更多算术

When first learning about negative numbers, many people are surprised to learn that the product of two negative numbers is positive. This fact follows from the sign rules , which relate the additive inverse operation to multiplication.

初学负数时,许多人会惊讶地发现两个负数的乘积是正数。这是由符号规则推导出来的,符号规则将加法的逆运算与乘法联系起来。

8.18. Theorem on Signs. For all a , b , x , y , z ∈ ℝ:

8.18. 符号定理。对于所有 a, b, x, y, z ∈ ℝ:

(a) First Sign Rule: a · ( − b ) = −( a · b ).

(a)第一符号规则:a · ( − b) = −(a · b)。

(b) Second Sign Rule: ( − a ) · b = −( a · b ).

(b)第二符号规则:( − a) · b = −(a · b)。

(c) Third Sign Rule: ( − a ) · ( − b ) = a · b .

(c)第三符号规则:( − a) ·( − b)= a · b。

(d) Inverses and Multiplication by −1: x = ( − 1) · x .

(d)逆运算和乘以 -1: -x = ( - 1) · x。

(e) Distributive Laws for Subtraction: x · ( y z ) = ( x · y ) − ( x · z ) and ( y z ) · x = ( y · x ) − ( z · x ).

(e)减法分配律:x · (y − z) = (x · y) − (x · z) 和 (y − z) · x = (y · x) − (z · x)。

Proof. Fix a , b ∈ ℝ. For (a), we compute ( a · b ) + ( a · ( − b )) = a · ( b + ( − b )) = a · 0 = 0 using the distributive axiom (D), then additive inverses (A4), then property Z1. Using Theorem 8.11 (b) with x = a · b and y = a · ( − b ), we conclude y = − x , i.e., a · ( − b ) = −( a · b ). The second sign rule is proved similarly. Using both of these sign rules and the double negative rule, we compute

证明。固定 a, b ∈ ℝ。对于 (a),我们利用分配律 (D) 计算 (a · b) + (a · ( − b)) = a · (b + ( − b)) = a · 0 = 0,然后利用加法逆元 (A4),再利用性质 Z1。利用定理 8.11(b),令 x = a · b 且 y = a · ( − b),我们得出 y = −x,即 a · ( − b) = −(a · b)。第二个符号规则的证明类似。利用这两个符号规则和双重否定规则,我们计算

( a 一个 ) ( b b ) = [ ( a 一个 ) b b ] = [ ( a 一个 b b ) ] = a 一个 b b .

To prove (d), fix x ∈ ℝ. By the sign rule and the identity axiom M3, ( − 1) · x = −(1 · x ) = − x . To prove (e), fix x , y , z ∈ ℝ. We compute      □

为了证明 (d),固定 x ∈ ℝ。根据符号规则和恒等公理 M3,( − 1) · x = −(1 · x) = −x。为了证明 (e),固定 x, y, z ∈ ℝ。我们计算 □

x x ( y z z ) = x x ( y + ( z z ) ) = ( x x y ) + ( x x ( z z ) ) = ( x x y ) + ( x x z z ) = ( x x y ) ( x x z z ) ,

by the definition of subtraction, the distributive law for addition, the sign rule, and the definition of subtraction. The other distributive law is proved similarly.

利用减法的定义、加法的分配律、符号规则以及减法的定义,可以证明另一个分配律。

Using the distributive law, the sign rules, and theorems about addition, we can now prove more arithmetical facts about specific real numbers. For example, let us show that 2 × 4 = 8. (Here, x × y is defined to mean x · y .) Assume that we have already proved 2 + 2 = 4, 4 + 2 = 6, and 6 + 2 = 8, using the technique illustrated in Proposition 8.14 . First,

利用分配律、符号规则和加法定理,我们现在可以证明更多关于特定实数的算术事实。例如,让我们证明 2 × 4 = 8。(这里,x × y 定义为 x · y。)假设我们已经使用命题 8.14 中所示的方法证明了 2 + 2 = 4、4 + 2 = 6 和 6 + 2 = 8。首先,

2 × × 2 = 2 × × ( 1 + 1 ) = ( 2 × × 1 ) + ( 2 × × 1 ) = 2 + 2 = 4 ,

using the definition of 2, then the distributive axiom (D), then the identity axiom (M3), then the known result for addition. Second,

首先,利用公式 2 的定义,然后利用分配律公理 (D),接着利用恒等式公理 (M3),最后利用已知的加法结果。其次,

2 × × 3 = 2 × × ( 2 + 1 ) = ( 2 × × 2 ) + ( 2 × × 1 ) = 4 + 2 = 6 ,

using the definition of 3, then the distributive axiom (D), then previously proved results. Third,

首先,利用定义3,然后利用分配律公理(D),最后利用先前已证明的结果。第三,

2 × × 4 = 2 × × ( 3 + 1 ) = ( 2 × × 3 ) + ( 2 × × 1 ) = 6 + 2 = 8 ,

using the definition of 4, then the distributive axiom (D), then previously proved results. Now, the sign rules allow us to conclude

利用 4 的定义,然后是分配律公理 (D),再是先前证明的结果。现在,符号规则允许我们得出结论。

2 × × ( 4 ) = 8 = ( 2 ) × × 4 ; ( 2 ) × × ( 4 ) = 8.

Here is one more illustration of the power of the distributive law.

以下是分配法则威力的又一个例子。

8.19. Theorem: FOIL Rule. For all t , x , y , z ∈ ℝ, ( t + x ) · ( y + z ) = ty + tz + xy + xz .

8.19. 定理:FOIL 法则。对于所有 t, x, y, z ∈ ℝ,(t + x) · (y + z) = ty + tz + xy + xz。

Proof. Fix t , x , y , z ∈ ℝ. Set a = y + z ∈ ℝ. Using the distributive law three times, we find that ( t + x ) · ( y + z ) = ( t + x ) · a = ta + xa = t ( y + z ) + x ( y + z ) = ( ty + tz ) + ( xy + xz ).

证明。固定 t、x、y、z ∈ ℝ。令 a = y + z ∈ ℝ。应用分配律三次,我们发现 (t + x) · (y + z) = (t + x) · a = ta + xa = t(y + z) + x(y + z) = (ty + tz) + (xy + xz)。

Using this rule, we can deduce algebraic identities such as ( x + y ) 2 = x 2 + 2 xy + y 2 and ( x + y )( x y ) = x 2 y 2 ; here z 2 is defined to be z · z , for any real z .

利用这条规则,我们可以推导出代数恒等式,例如 (x + y) 2 = x 2 + 2xy + y 2 和 (x + y)(x − y) = x 2 − y 2 ;这里 z 2 定义为 z · z,其中 z 为任意实数。

Properties of Multiplication and Inverses

乘法和逆运算的性质

We now derive properties of multiplication and multiplicative inverses that are based on the left and right cancellation laws. These properties and their proofs are analogous to the corresponding properties for addition, except we need to ensure that we never take the multiplicative inverse of zero.

现在我们推导基于左右约分法则的乘法及其逆元的性质。这些性质及其证明与加法的相应性质类似,只是我们需要确保永远不取零的乘法逆元。

8.20. Theorem on Multiplication. For all a , b , c , x , y ∈ ℝ:

8.20. 乘法定理。对于所有 a, b, c, x, y ∈ ℝ:

(a) Cancellation Laws for Multiplication: If a ≠ 0 and a · b = a · c , then b = c .

(a)乘法消去法则:如果 a ≠ 0 且 a · b = a · c,则 b = c。

(b) If a ≠ 0 and b · a = c · a , then b = c .

(b)如果 a ≠ 0 且 b · a = c · a,则 b = c。

(c) Uniqueness of Multiplicative Identity: If x ≠ 0 and x · y = x , then y = 1.

(c)乘法单位元的唯一性:如果 x ≠ 0 且 x · y = x,则 y = 1。

(d) Uniqueness of Multiplicative Inverses: If x · y = 1, then x ≠ 0 and y ≠ 0 and y = x −1 .

(d)乘法逆元的唯一性:如果 x · y = 1,则 x ≠ 0 且 y ≠ 0 且 y = x −1

(e) Inverse of Identity: 1 −1 = 1.

(e)单位元的逆元:1 −1 = 1。

(f) Double Inverse Rule: If a ≠ 0, then a −1 ≠ 0 and ( a −1 ) −1 = a .

(f)双重逆规则:如果 a ≠ 0,则 a −1 ≠ 0 且 (a −1 ) −1 = a。

(g) Inverse of a Product: If a ≠ 0 and b ≠ 0, then a · b ≠ 0 and ( a · b ) −1 = ( b −1 ) · ( a −1 ).

(g)乘积的逆:如果 a ≠ 0 且 b ≠ 0,则 a · b ≠ 0 且 (a · b) −1 = (b −1 ) · (a −1 )。

(i) Inverse of a Negative: If a ≠ 0, then − a ≠ 0 and ( − a ) −1 = −( a −1 ).

(i)负数的逆:如果 a ≠ 0,则 −a ≠ 0 且 ( − a) −1 = −(a −1 )。

Proof. We prove (c), (d), and (g) as examples (assuming (a) has already been proved). For (c), fix x , y ∈ ℝ; assume x · y = 1. Since 1 ≠ 0 by axiom N, we deduce x ≠ 0 and y ≠ 0 by Z4. Now we can use the inverse axiom (M4) to write x · y = x · x −1 . Using part (a) to cancel the nonzero real number x , we get y = x −1 . For (d), use axiom N to see that 1 −1 exists; then use M3 and M4 to compute 1 −1 = 1 · 1 −1 = 1. For (g), fix a ∈ ℝ with a ≠ 0. We proved − a ≠ 0 in Z6, so a −1 and ( − a ) −1 both exist. Now ( − a ) · ( − [ a −1 ]) = a · a −1 = 1 using the sign rules and M4. Invoking part (c) with x = − a and y = −[ a −1 ], we conclude y = x −1 , that is, −[ a −1 ] = ( − a ) −1 .

证明。我们以 (c)、(d) 和 (g) 为例进行证明(假设 (a) 已被证明)。对于 (c),固定 x, y ∈ ℝ;假设 x · y = 1。由于根据公理 N,1 ≠ 0,因此根据 Z4,我们推导出 x ≠ 0 且 y ≠ 0。现在我们可以使用逆公理 (M4) 写出 x · y = x · x −1 。利用 (a) 部分消去非零实数 x,我们得到 y = x −1 。对于 (d),利用公理 N 可知 1 −1 存在;然后利用 M3 和 M4 计算 1 −1 = 1 · 1 −1 = 1。对于 (g),固定 a ∈ ℝ 且 a ≠ 0。我们在 Z6 中证明了 −a ≠ 0,因此 a −1 和 ( − a) −1 都存在。现在,利用符号规则和 M4,( − a) · ( − [a −1 ]) = a · a −1 = 1。对 (c) 部分应用 x = −a 和 y = −[a −1 ],我们得出 y = x −1 ,即 −[a −1 ] = ( − a) −1

As in the case of addition, using commutativity (M5) we can write ( x · y ) −1 = x −1 · y −1 when x , y are nonzero real numbers. We can also prove generalized associativity and generalized commutativity of multiplication by repeated use of M2 and M5. This justifies writing expressions such as abcd without any parentheses. We can omit even more parentheses by agreeing that multiplication takes precedence over addition, so that ab + cd means ( ab ) + ( cd ) rather than a ( b + c ) d .

与加法类似,利用交换律(M5),当 x 和 y 为非零实数时,我们可以写成 (x · y) = x · y。我们还可以通过反复运用 M2 和 M5 来证明乘法的广义结合律和广义交换律。这使得像 abcd 这样的表达式可以省略括号。我们还可以省略更多的括号,因为乘法优先于加法,所以 ab + cd 表示 (ab) + (cd) 而不是 a(b + c)d。

8.21 Theorem on Division. For all x , y , z ∈ ℝ, if y ≠ 0 and z ≠ 0, then:

8.21 除法定理。对于任意 x, y, z ∈ ℝ,若 y ≠ 0 且 z ≠ 0,则:

(a) x /1 = x .

(a)x/1 = x。

(b) 1/ y = y −1 .

(b)1/y = y −1

(c) y / y = 1.

(c)y/y = 1。

(d) y / z ≠ 0.

(d)y/z ≠ 0。

(e) ( y / z ) −1 = z / y = y −1 · z .

(e) (y/z) −1 = z/y = y −1 · z。

(f) ( x / y ) · ( y / z ) = x / z .

(f)(x/y)·(y/z)=x/z。

(g) ( x · z )/( y · z ) = x / y .

(g)(x · z)/(y · z) = x/y。

Proof. We prove part (g) as an example. Fix x , y , z ∈ ℝ with y and z nonzero. By Z2, y · z ≠ 0, so the division ( xz )/( yz ) is defined. We compute

证明。我们以 (g) 部分为例进行证明。固定 x, y, z ∈ ℝ,其中 y 和 z 非零。由 Z2 可知,y · z ≠ 0,因此除法 (xz)/(yz) 有定义。我们计算

x x z z y z z = ( x x z z ) ( y z z ) 1 = ( x x z z ) ( z z 1 y 1 ) = x x ( ( z z z z 1 ) y 1 ) = x x ( 1 y 1 ) = x x y 1 = x x y ,

using the definition of division, the rule for the inverse of a product, generalized associativity, the inverse axiom, the identity axiom, and the definition of division.     □

利用除法的定义、积的逆运算规则、广义结合律、逆运算公理、恒等式公理以及除法的定义。□

Fraction Rules

分数法则

We now derive some algebraic rules for executing arithmetic operations on fractions.

现在我们推导出一些对分数进行算术运算的代数规则。

8.22. Theorem on Fractions. For all x , y ∈ ℝ and all nonzero t , u , w ∈ ℝ,

8.22. 分式定理。对于所有 x, y ∈ ℝ 和所有非零 t, u, w ∈ ℝ,

(a) Addition Rules: x u + y u = x + y u , x u + y w = x w + u y u w .

(a)加法规则:xu+yu=x+yu,xu+yw=xw+uyuw。

(b) Negation Rules: x u = x u = x u , x u = x u .

(b)否定规则:−xu=−xu=x−u,−x−u=xu。

(c) Subtraction Rules: x u y u = x y u , x u y w = x w u y u w .

(c) Subtraction Rules: xu−yu=x−yu,xu−yw=xw−uyuw.

(d) Multiplication Rules: x y u = x y u = y x u , x u y w = x y u w .

(d)乘法规则:x⋅yu=x⋅yu=y⋅xu,xu⋅yw=x⋅yu⋅w。

(e) Division Rules: ( t w ) 1 = w t , x / u t / w = x w u t .

(e)除法规则:(tw)−1=wt,x/ut/w=x⋅wu⋅t。

Proof. We prove the addition rules, a multiplication rule, and a division rule as examples. Fix x , y ∈ ℝ and nonzero t , u , w ∈ ℝ. First,

证明。我们以加法规则、乘法规则和除法规则为例进行证明。固定 x, y ∈ ℝ 和非零 t, u, w ∈ ℝ。首先,

( x x / u ) + ( y / u ) = x x u 1 + y u 1 = ( x x + y ) u 1 = ( x x + y ) / u ,

using the definition of division, then the distributive axiom (D), then the definition of division. Second, recall the previous result x u = x w u w and y w = u y u w . Combining this with what we just proved, we see that

首先,利用除法的定义,然后利用分配律公理(D),最后利用除法的定义。其次,回顾之前的结果 xu=xwuw 和 yw=uyuw。结合我们刚刚证明的内容,我们发现

x x u + y w 西 = x x w 西 u w 西 + u y u w 西 = x x w 西 + u y u w 西 ,

as needed. Third, compute

根据需要。第三,计算

x x y u w 西 = ( x x y ) ( u w 西 ) 1 = ( x x y ) ( w 西 1 u 1 ) = ( x x u 1 ) ( y w 西 1 ) = x x u y w 西

using the definition of division, the rule for the inverse of a product, generalized associativity and commutativity, and the definition of division. Finally, using the rule just proved, note that

利用除法的定义、积的逆运算规则、广义结合律和交换律,以及除法的定义。最后,利用刚才证明的规则,注意到:

t t w 西 w 西 t t = t t w 西 w 西 t t = t t w 西 t t w 西 = ( t t w 西 ) ( t t w 西 ) 1 = 1.

By Theorem 8.20 (c), we deduce ( t / w ) −1 = w / t .

根据定理 8.20(c),我们推导出 (t/w) −1 = w/t。

Section Summary

章节概要

1. Properties of Zero. A product of real numbers is zero iff one of the factors is zero. The set of nonzero real numbers is closed under multiplication, inverses, and negatives.

1.零的性质。实数的乘积为零当且仅当其中一个因数为零。非零实数集对乘法、逆运算和负运算封闭。

2. More Algebraic Properties. The sign rules state a ( − b ) = ( − a ) b = −( ab ) and ( − a )( − b ) = ab for all a , b ∈ ℝ. Nonzero factors can be cancelled: if a ≠ 0 and ab = ac , then b = c . The number 1 is the unique multiplicative identity. For a ≠ 0, a −1 is the unique multiplicative inverse of a . When the inverses are defined, we have ( a −1 ) −1 = a , ( ab ) −1 = b −1 a −1 , and ( − a ) −1 = −( a −1 ). The distributive law generalizes to give identities such as ( t + x )( y + z ) = ty + tz + xy + xz .

2.更多代数性质。符号规则表明,对于所有 a, b ∈ ℝ,有 a( − b) = ( − a)b = −(ab) 且 ( − a)( − b) = ab。非零因子可以约分:如果 a ≠ 0 且 ab = ac,则 b = c。数字 1 是唯一的乘法单位元。对于 a ≠ 0,a −1 是 a 的唯一乘法逆元。当逆元定义时,我们有 (a −1 ) −1 = a,(ab) −1 = b −1 a −1 ,以及 ( − a) −1 = −(a −1 )。分配律可以推广为如下恒等式:(t + x)(y + z) = ty + tz + xy + xz。

3. Fraction Rules. We can perform arithmetic on fractions via rules such as ( x / u ) + ( y / w ) = ( xw + uy )/( uw ), ( x / u ) · ( y / w ) = ( xy )/( uw ), and ( t / w ) −1 = w / t (assuming all denominators are nonzero).

3.分数规则。我们可以通过诸如 (x/u) + (y/w) = (xw + uy)/(uw)、(x/u) · (y/w) = (xy)/(uw) 和 (t/w) −1 = w/t 之类的规则对分数进行算术运算(假设所有分母均非零)。

Exercises

练习

1. (a) Prove a R , a 0 = 0 by simplifying ( a · 0) + ( a · a ). (b) Prove the sign rule a , b R , ( a b ) = ( a ) b .

1.(a) 通过简化 (a · 0) + (a · a) 证明 ∀a∈R,a⋅0=0。(b) 证明符号规则 ∀a,b∈R,−(a⋅b)=(−a)⋅b。

2. Finish the proof of Theorem 8.20 .

2.完成定理 8.20 的证明。

3. Use Z3 (no zero divisors) to prove the cancellation property in Theorem 8.20 (a) without using multiplicative inverses. (It follows that this cancellation property holds in all integral domains.)

3. 利用 Z3(无零因子)证明定理 8.20(a) 中的消去性质,不使用乘法逆元。(由此可知,该消去性质在所有整数区域都成立。)

4. Prove: (a) 3 × 3 = 9. (b) 5 × ( − 2) = −10. (c) ( − 2) × ( − 3) = 6.

4.证明:(a)3 × 3 = 9。(b)5 × ( − 2) = −10。(c)( − 2) × ( − 3) = 6。

5. Prove: (a) 1/2 + 2/3 = 7/6. (b) 1/3 − 3/5 = −4/15. (c) (3/8) · (2/3) = 1/4.

5.证明:(a)1/2 + 2/3 = 7/6。(b)1/3 − 3/5 = −4/15。(c)(3/8) · (2/3) = 1/4。

6. Finish the proof of Theorem 8.21 .

6.完成定理 8.21 的证明。

7. Prove Theorem 8.22 parts (b) and (c).

7.证明定理 8.22 的 (b) 和 (c) 部分。

8. Finish the proof of Theorem 8.22 parts (d) and (e).

8.完成定理 8.22 的 (d) 和 (e) 部分的证明。

9. Use the distributive law to prove the following identities for all v , w , x , y , z ∈ ℝ.

9.利用分配律证明下列恒等式对所有 v, w, x, y, z ∈ ℝ 成立。

(a) ( x + y ) 2 = x 2 + 2 x y + y 2

(a) (x+y)2=x2+2xy+y2

(b) ( x y ) 2 = x 2 2 x y + y 2

(b) (x−y)2=x2−2xy+y2

(c) ( x + y ) ( x y ) = x 2 y 2

(c) (x+y)(x−y)=x2−y2

(d) w ( x + y + z ) = w x + w y + w z

(d) w(x+y+z)=wx+wy+wz

(e) ( v + w ) ( x + y + z ) = v x + w x + v y + w y + v z + w z

(e) (v+w)(x+y+z)=vx+wx+vy+wy+vz+wz

(f) ( x + y ) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3 (where u 3 means u ċ u ċ u , for u ∈ ℝ).

(f) (x+y)3=x3+3x2y+3xy2+y3 (其中 u 3 表示 u ċ u ċ u,对于 u ∈ ℝ)。

10. For x , y ∈ ℝ − { − 1}, define x y = x y + x + y . (a) Show: x , y R { 1 } , x y R { 1 } . (b) Prove that axioms M2, M3, and M5 hold with · replaced by and 1 replaced by 0. (c) Define an operation x ′ (analogous to multiplicative inversion) and prove that for all x ∈ ℝ − { − 1}, x ′ ∈ ℝ − { − 1} and x x = 0 = x x .

10.对于 x, y ∈ ℝ − { − 1},定义 x⋆y=x⋅y+x+y。(a)证明:∀x,y∈R−{−1},x⋆y∈R−{−1}。(b)证明将 · 替换为 ⋆,1 替换为 0,公理 M2、M3 和 M5 仍然成立。(c)定义运算 x′(类似于乘法逆元),并证明对于所有 x ∈ ℝ − { − 1},x′ ∈ ℝ − { − 1},且 x⋆x′=0=x′⋆x。

11. Define the set of matrices M 2 (ℝ) to be the set ℝ 4 , where we display ( a , b , c , d ) ∈ ℝ 4 as the 2 × 2 array [ a b c d ] . Define algebraic operations on matrices as follows: for a , b , c , d , w , x , y , z ∈ ℝ, let

11.定义矩阵集合M 2 (ℝ)为集合ℝ 4 ,其中(a, b, c, d) ∈ ℝ 4 表示为2×2数组[abcd]。定义矩阵的代数运算如下:对于a, b, c, d, w, x, y, z ∈ ℝ,令

[ a 一个 b b c c d d ] + [ w 西 x x y z z ] = [ a 一个 + w 西 b b + x x c c + y d d + z z ] , [ a 一个 b b c c d d ] = [ a 一个 b b c c d d ] ,
0 M M 2 ( R ) = [ 0 0 0 0 ] , 0 M M 2 ( R ) = [ 1 0 0 1 ] ,
[ a 一个 b b c c d d ] [ w 西 x x y z z ] = [ a 一个 w 西 + b b y a 一个 x x + b b z z c c w 西 + d d y c c x x + d d z z ] ,
[ a 一个 b b c c d d ] 1 = 1 a 一个 d d a 一个 c c [ d d b b c c a 一个 ] when 什么时候 a 一个 d d b b c c 0.

Use known algebraic properties of ℝ to prove that M 2 (ℝ) satisfies all the algebraic axioms except M5 and the requirement that every nonzero matrix have a multiplicative inverse.

利用 ℝ 的已知代数性质证明 M 2 (ℝ) 满足除 M5 和每个非零矩阵都有乘法逆的要求之外的所有代数公理。

12. Which of the properties Z1 through Z6 hold for matrices in M 2 (ℝ)? Explain. (Assume x −1 exists when considering Z5.)

12.性质Z1至Z6中哪些适用于M 2 (ℝ)中的矩阵?请解释。(考虑Z5时,假设x −1 存在。)

8.3  Natural Numbers, Integers, and Rational Numbers

8.3 自然数、整数和有理数

We are all informally acquainted with the natural numbers ℕ = ℤ >0 = {1, 2, 3, …}, the integers ℤ = {…, − 2, − 1, 0, 1, 2, …}, and the rational numbers ℚ (ratios of integers). It is possible to give a rigorous construction of these number systems using set theory, defining first ℕ, then ℤ, then ℚ, and finally ℝ, and proving that each number system satisfies the appropriate algebraic and ordering properties. The details of this construction are very tedious and intricate, however, and we have chosen instead to begin with an axiom system for the final number system ℝ on our list, which requires no prior knowledge of the more basic number systems.

我们都非正式地熟悉自然数集ℕ = ℤ >0 = {1, 2, 3, …}、整数集ℤ = {…, −2, −1, 0, 1, 2, …} 和有理数集ℚ(整数的比值)。我们可以利用集合论对这些数系进行严格的构造,首先定义ℕ,然后是ℤ,接着是ℚ,最后是ℝ,并证明每个数系都满足相应的代数性质和排序性质。然而,这种构造过程非常繁琐复杂,因此我们选择从最后一个数系ℝ的公理体系入手,这样就不需要任何关于更基础数系的先验知识。

However, we frequently need integers and rational numbers to prove theorems about real numbers and calculus. It turns out we can define ℕ, ℤ, and ℚ as certain subsets of the set ℝ, and then derive the basic properties of these number systems fairly quickly from the axioms for ℝ. In particular, the Induction Axiom for ℕ (discussed in §4.1) is provable from our definition of ℕ given below. Aided by induction, we show that ℕ, ℤ, and ℚ satisfy appropriate closure properties among those listed in axioms A1 and M1. The remaining algebraic and order properties of these number systems then follow automatically from the corresponding properties of ℝ.

然而,我们经常需要整数和有理数来证明关于实数和微积分的定理。我们可以将 ℕ、ℤ 和 ℚ 定义为集合 ℝ 的某些子集,然后从 ℝ 的公理中可以相当迅速地推导出这些数系的基本性质。特别地,ℕ 的归纳公理(在 §4.1 中讨论)可以从我们下面给出的 ℕ 定义中证明。借助归纳法,我们证明了 ℕ、ℤ 和 ℚ 满足公理 A1 和 M1 中列出的相应封闭性质。这些数系的其余代数性质和序性质则可由 ℝ 的相应性质自动得出。

Inductive Sets; Defining Natural Numbers

归纳集;自然数的定义

Intuitively, the set of natural numbers 1 is ℕ = {1, 2, 3, …}. The trouble with this intuitive description is that it does not precisely define what is meant by the … at the end. Given that real numbers and addition are already available to us, we can approach this problem via closure properties in ℝ. Restating the intuitive description, ℕ should be the set of all real numbers that we can reach by starting with 1 and adding 1 repeatedly. This motivates the following definition.

直观地说,自然数集 1 是ℕ = {1, 2, 3, …}。这种直观描述的不足之处在于,它没有精确定义末尾的“…”的含义。鉴于我们已经掌握了实数和加法运算,我们可以利用ℝ的封闭性质来解决这个问题。重新表述一下直观描述,ℕ应该是从1开始并不断加1所能得到的所有实数的集合。由此引出了以下定义。

8.23. Definition: Inductive Sets. Given S ⊆ℝ,

8.23. 定义:归纳集。给定 S⊆ℝ,

S is inductive iff 1 S and x R , x S ( x + 1 ) S .

S 是归纳的,当且仅当 1∈S ∀x∈R,x∈S⇒(x+1)∈S。

Thus, an inductive set must be closed under 1 (meaning 1 is in the set) and closed under adding 1 . For example, ℝ itself is an inductive set due to the closure axioms A1 and M1. Later, when we discuss properties of order, we will see that ℝ >0 and ℝ ≥1 are also inductive sets. Informally, ℤ and ℚ (not yet officially defined) are inductive sets. All of these inductive sets are too big to be ℕ because they contain extra elements not reachable from 1 by adding 1. The following clever device lets us precisely define the idea that ℕ should be the smallest possible inductive subset of ℝ.

因此,归纳集必须在 1 下封闭(即 1 包含在集合中),并且在加 1 的运算下也封闭。例如,ℝ 本身就是一个归纳集,因为满足封闭公理 A1 和 M1。稍后,当我们讨论序的性质时,我们将看到 ℝ >0 和 ℝ ≥1 也是归纳集。非正式地讲,ℤ 和 ℚ(尚未正式定义)也是归纳集。所有这些归纳集都太大,不可能是 ℕ,因为它们包含一些无法通过加 1 从 1 到达的额外元素。以下巧妙的方法可以让我们精确地定义 ℕ 应该是 ℝ 的最小归纳子集这一概念。

8.24. Definition of the Natural Numbers ℕ. Let I be the set of all inductive subsets of ℝ. Define N = S I S .

8.24. 自然数 ℕ 的定义。令 I 为 ℝ 的所有归纳子集的集合。定义 N=⋂S∈IS。

Thus, ℕ is the intersection of all inductive subsets of ℝ. Since ℝ itself is inductive, I is nonempty, and the generalized intersection is defined. Expanding the definition, we see that for all objects x , x iff for every inductive set S , x S .

因此,ℕ 是 ℝ 的所有归纳子集的交集。由于 ℝ 本身是归纳的,所以 I 非空,广义交集定义成立。展开定义,我们看到对于所有对象 x,x∈ℕ 当且仅当对于每个归纳集 S⊆ℝ,x∈S。

8.25. Theorem. ℕ is an inductive set.

8.25. 定理。ℕ 是一个归纳集。

Proof. For every inductive set S , 1 ∈ S . So 1 ∈ ℕ. Next we show ℕ is closed under adding 1. Fix x ∈ ℕ and prove x + 1 ∈ ℕ. For any inductive set S , we know x S , so x + 1 ∈ S since S is inductive. Since this holds for all inductive sets S , we see that x + 1 ∈ ℕ.

证明。对于任意归纳集 S,都有 1 ∈ S。因此 1 ∈ ℕ。接下来,我们证明 ℕ 对加 1 封闭。固定 x ∈ ℕ,并证明 x + 1 ∈ ℕ。对于任意归纳集 S,我们知道 x ∈ S,所以 x + 1 ∈ S,因为 S 是归纳集。由于这对所有归纳集 S 都成立,所以 x + 1 ∈ ℕ。

We now prove two versions of the Induction Axiom for ℕ. The second version (involving open sentences) is the one discussed earlier in §4.1.

现在我们证明 ℕ 的归纳公理的两个版本。第二个版本(涉及开放语句)是前面在 §4.1 中讨论的版本。

8.26. Theorem: Set-Theoretic Induction Principle. Suppose S is a set satisfying these conditions: (a) S ⊆ℕ; (b) 1 ∈ S ; (c) for all x S , x + 1 ∈ S . Then S = ℕ.

8.26. 定理:集合论归纳原理。假设 S 是一个满足以下条件的集合:(a) S⊆ℕ;(b) 1 ∈ S;(c) 对于所有 x ∈ S,x + 1 ∈ S。则 S = ℕ。

Proof. Assusme S satisfies the three conditions. By condition (a), it suffices to prove that ℕ⊆ S . By conditions (b) and (c), S is an inductive set. Thus ℕ, which is the intersection of all inductive sets, must be a subset of S .     □

证明。假设集合 S 满足三个条件。由条件 (a) 可知,只需证明 ℕ⊆ S。由条件 (b) 和 (c) 可知,S 是一个归纳集。因此,作为所有归纳集交集的 ℕ 必定是 S 的子集。□

8.27. Theorem: Induction Principle. Suppose P ( n ) is an open sentence such that P (1) is true, and for all n ∈ ℕ, if P ( n ) is true, then P ( n + 1) is true. Then for all n ∈ ℕ, P ( n ) is true.

8.27. 定理:归纳原理。假设 P(n) 是一个开语句,其中 P(1) 为真,并且对于所有 n ∈ ℕ,如果 P(n) 为真,则 P(n + 1) 也为真。那么对于所有 n ∈ ℕ,P(n) 都为真。

Proof. Given the open sentence P ( n ), form the set S = { n N : P ( n ) is true } . By definition of S and the conditions on P ( n ), S satisfies the three conditions in the Set-Theoretic Induction Principle. For example, to check condition (c), fix x S . We know x ∈ ℕ and P ( x ) is true, so we deduce x + 1 ∈ ℕ and P ( x + 1) is true, hence x + 1 ∈ S . Now the Set-Theoretic Induction Principle tells us that S = ℕ, which means that P ( n ) is true for all n ∈ ℕ.     □

#

证明。给定开语句 P(n),构造集合 S={n∈N:P(n) 为真}。根据 S 的定义和 P(n) 的条件,S 满足集合论归纳原理中的三个条件。例如,为了检验条件 (c),固定 x ∈ S。我们知道 x ∈ ℕ 且 P(x) 为真,因此我们推导出 x + 1 ∈ ℕ 且 P(x + 1) 为真,从而 x + 1 ∈ S。现在,集合论归纳原理告诉我们 S = ℕ,这意味着对于所有 n ∈ ℕ,P(n) 都为真。□

Closure Properties of ℕ

ℕ 的封闭性质

Since ℕ is an inductive set, we know that ℕ contains 1 and is closed under adding 1. Using induction, we can now prove the stronger statements that ℕ is closed under addition and multiplication.

由于 ℕ 是一个归纳集,我们知道 ℕ 包含 1 并且对加 1 封闭。利用归纳法,我们现在可以证明更强的结论:ℕ 对加法和乘法封闭。

8.28. Theorem on Closure of ℕ under Addition. For all m , n ∈ ℕ, m + n ∈ ℕ.

8.28. ℕ 在加法下的封闭性定理。对于所有 m, n ∈ ℕ,m + n ∈ ℕ。

Proof. Let m 0 be a fixed element of ℕ. For n ∈ ℕ, let P ( n ) be the statement “ m 0 + n ∈ ℕ.” We use induction on n to prove P ( n ) is true for every n ∈ ℕ. To see that P (1) is true, note that m 0 + 1 ∈ ℕ because m 0 ∈ ℕ and ℕ is closed under adding 1. Next, fix n 0 ∈ ℕ, assume P ( n 0 ), and prove P ( n 0 + 1). We have assumed m 0 + n 0 ∈ ℕ, and we must prove m 0 + ( n 0 + 1) ∈ ℕ. Using associativity in ℝ (A2), we know m 0 + ( n 0 + 1) = ( m 0 + n 0 ) + 1. Since m 0 + n 0 ∈ ℕ and ℕ is closed under adding 1, ( m 0 + n 0 ) + 1 ∈ ℕ. Thus m 0 + ( n 0 + 1) ∈ ℕ, as needed.      □

证明。设 m 0 为 ℕ 的一个固定元素。对于 n ∈ ℕ,令 P(n) 为命题“m 0 + n ∈ ℕ”。我们对 n 进行归纳,证明对于每个 n ∈ ℕ,P(n) 都成立。为了证明 P(1) 成立,注意到 m 0 + 1 ∈ ℕ,因为 m 0 ∈ ℕ,且 ℕ 对加 1 封闭。接下来,固定 n 0 ∈ ℕ,假设 P(n 0 ) 成立,并证明 P(n 0 + 1) 成立。我们假设 m 0 + n 0 ∈ ℕ,我们需要证明 m 0 + (n 0 + 1) ∈ ℕ。利用 ℝ 中的结合律 (A2),我们知道 m 0 + (n 0 + 1) = (m 0 + n 0 ) + 1。由于 m 0 + n 0 ∈ ℕ 且 ℕ 对加 1 封闭,因此 (m 0 + n 0 ) + 1 ∈ ℕ。因此,m 0 + (n 0 + 1) ∈ ℕ,符合要求。□

8.29. Theorem on Closure of ℕ under Multiplication. For all m , n ∈ ℕ, m · n ∈ ℕ.

8.29. ℕ 在乘法下的封闭性定理。对于所有 m, n ∈ ℕ,m · n ∈ ℕ。

Proof. Let m 0 be a fixed element of ℕ. We prove n N , m 0 n N by induction on n . When n = 1, the identity axiom (M3) gives m 0 · 1 = m 0 ∈ ℕ. Now fix n 0 ∈ ℕ, assume m 0 · n 0 ∈ ℕ, and prove m 0 · ( n 0 + 1) ∈ ℕ. By the distributive law in ℝ and (M3), m 0 · ( n 0 + 1) = m 0 · n 0 + m 0 · 1 = m 0 · n 0 + m 0 . By induction hypothesis, m 0 · n 0 ∈ ℕ. By closure of ℕ under addition, m 0 · n 0 + m 0 ∈ ℕ. So m 0 · ( n 0 + 1) ∈ ℕ, as needed.

证明。设 m 0 为 ℕ 的一个固定元素。我们对 n 进行归纳,证明 ∀n∈N,m0⋅n∈N。当 n = 1 时,恒等公理 (M3) 给出 m 0 · 1 = m 0 ∈ ℕ。现在固定 n 0 ∈ ℕ,假设 m 0 · n 0 ∈ ℕ,并证明 m 0 · (n 0 + 1) ∈ ℕ。根据 ℝ 中的分配律和 (M3),m 0 · (n 0 + 1) = m 0 · n 0 + m 0 · 1 = m 0 · n 0 + m 0 。根据归纳假设,m 0 · n 0 ∈ ℕ。根据 ℕ 对加法的封闭性,m 0 · n 0 + m 0 ∈ ℕ。因此,m 0 · (n 0 + 1) ∈ ℕ,如有必要。

On the other hand, ℕ is not closed under additive inverses, subtraction, multiplicative inverses, or division. To prove these claims, we need order properties to be discussed later.

另一方面,ℕ 对加法逆元、减法逆元、乘法逆元和除法都不封闭。为了证明这些结论,我们需要用到一些序性质,这些性质将在后面讨论。

Defining Integers; Closure Properties of ℤ

整数的定义;ℤ的封闭性质

Informally, we obtain the set of integers from the set of natural numbers (positive integers) by adjoining zero and the negatives of all the positive integers. These negatives are known to exist in ℝ, by the closure axiom A1 for ℝ. Here is the official definition.

通俗地说,我们可以通过将零和所有正整数的负数添加到自然数集(正整数集)中,得到整数集。根据 ℝ 的闭包公理 A1,这些负数在 ℝ 中是存在的。以下是正式定义。

8.30. Definition of Integers. For all x ∈ ℝ, x iff x or x = 0 or - x . Elements of ℤ are called integers .

8.30. 整数的定义。对于所有 x ∈ ℝ,x∈ℤ 当且仅当 x∈ℕ 或 x=0 或 -x∈ℕ。ℤ 中的元素称为整数。

Now we show that ℤ is closed under all the arithmetical operations except multiplicative inversion and division.

现在我们证明,ℤ 在除乘法求逆和除法之外的所有算术运算下都是封闭的。

8.31. Theorem on Closure of ℤ under Negation. For all y ∈ ℤ, − y ∈ ℤ.

8.31. ℤ 在否定下的封闭性定理。对于所有 y ∈ ℤ,−y ∈ ℤ。

Proof. Fix y ∈ ℤ; prove − y ∈ ℤ. We know y ∈ ℕ or y = 0 or − y ∈ ℕ, so consider three cases.

证明。固定 y ∈ ℤ;证明 −y ∈ ℤ。我们知道 y ∈ ℕ 或 y = 0 或 −y ∈ ℕ,因此考虑三种情况。

Case 1. Assume y ∈ ℕ. Then −( − y ) = y ∈ ℕ by the double negative rule, so − y ∈ ℤ because − y satisfies the third alternative in the definition of ℤ (taking x there to be − y ).

情况 1. 假设 y ∈ ℕ。则根据双重否定规则,−( − y) = y ∈ ℕ,因此 −y ∈ ℤ,因为 −y 满足 ℤ 定义中的第三个选项(将 x 取为 −y)。

Case 2. Assume y = 0. Then − y = −0 = 0, so − y ∈ ℤ by the second alternative in the definition of ℤ.

情况 2. 假设 y = 0。则 −y = −0 = 0,因此根据 ℤ 定义的第二种选择,−y ∈ ℤ。

Case 3. Assume − y ∈ ℕ. Then − y ∈ ℤ by the first alternative in the definition of ℤ (taking x there to be − y ).     □

情况 3. 假设 −y ∈ ℕ。则根据 ℤ 定义中的第一种情况(取 x 为 −y),−y ∈ ℤ。□

8.32. Theorem on Closure of ℤ under Multiplication. For all x , y ∈ ℤ, x · y ∈ ℤ.

8.32. ℤ 在乘法下的封闭性定理。对于所有 x, y ∈ ℤ,x · y ∈ ℤ。

Proof. Fix x , y ∈ ℤ; prove x · y ∈ ℤ. We know x ∈ ℕ or x = 0 or − x ∈ ℕ, and similarly for y , giving nine cases. Five of these cases can be handled by the observation that if x = 0 or y = 0, then x · y = 0 ∈ ℤ. Now we examine the remaining four cases.

证明。固定 x, y ∈ ℤ;证明 x · y ∈ ℤ。我们知道 x ∈ ℕ 或 x = 0 或 −x ∈ ℕ,y 的情况类似,共有九种情况。其中五种情况可以通过观察得出:如果 x = 0 或 y = 0,则 x · y = 0 ∈ ℤ。现在我们考察剩余的四种情况。

Case 1. Assume x ∈ ℕ and y ∈ ℕ. We already proved x · y ∈ ℕ, so x · y ∈ ℤ.

情况 1. 假设 x ∈ ℕ 且 y ∈ ℕ。我们已经证明 x · y ∈ ℕ,所以 x · y ∈ ℤ。

Case 2. Assume x ∈ ℕ and − y ∈ ℕ. By the sign rules and closure of ℕ under multiplication, we deduce −( x · y ) = x · ( − y ) ∈ ℕ. So x · y ∈ ℤ by the third alternative in the definition of ℤ.

情况 2. 假设 x ∈ ℕ 且 −y ∈ ℕ。根据符号规则和 ℕ 在乘法下的封闭性,我们推导出 −(x · y) = x · ( − y) ∈ ℕ。因此,根据 ℤ 的定义中的第三种选择,x · y ∈ ℤ。

Case 3. Assume − x ∈ ℕ and y ∈ ℕ. As in Case 2, we get −( x · y ) = ( − x ) · y ∈ ℕ, hence x · y ∈ ℤ.

情况 3. 假设 −x ∈ ℕ 且 y ∈ ℕ。与情况 2 一样,我们得到 −(x · y) = ( − x) · y ∈ ℕ,因此 x · y ∈ ℤ。

Case 4. Assume − x ∈ ℕ and − y ∈ ℕ. The sign rules and closure of ℕ give x · y = ( − x ) · ( − y ) ∈ ℕ, so x · y ∈ ℤ.      □

情况 4. 假设 −x ∈ ℕ 且 −y ∈ ℕ。根据符号规则和 ℕ 的闭包性质,可知 x · y = ( − x) · ( − y) ∈ ℕ,因此 x · y ∈ ℤ。□

Proving closure of ℤ under addition turns out to be trickier; we begin with a lemma.

证明 ℤ 在加法运算下的封闭性其实比较棘手;我们从引理开始。

8.33. Lemma. For all x ∈ ℤ, x − 1 ∈ ℤ.

8.33. 引理。对于所有 x ∈ ℤ,x − 1 ∈ ℤ。

Proof. Fix x ∈ ℤ; prove x − 1 = x + ( − 1) ∈ ℤ. We know x ∈ ℕ or x = 0 or − x ∈ ℕ, so consider three cases.

证明。固定 x ∈ ℤ;证明 x − 1 = x + ( − 1) ∈ ℤ。我们知道 x ∈ ℕ 或 x = 0 或 −x ∈ ℕ,因此考虑三种情况。

Case 1. Assume x ∈ ℕ. We prove x − 1 ∈ ℤ by induction on x . If x = 1, then 1 − 1 = 0 ∈ ℤ. For the induction step, fix x ∈ ℕ, assume x − 1 ∈ ℤ, and prove ( x + 1) − 1 ∈ ℤ. Since ( x + 1) − 1 = x + (1 − 1) = x + 0 = x ∈ ℕ, we can conclude that ( x + 1) − 1 ∈ ℤ without even using the assumption x − 1 ∈ ℤ.

情况 1. 假设 x ∈ ℕ。我们用归纳法证明 x − 1 ∈ ℤ。如果 x = 1,则 1 − 1 = 0 ∈ ℤ。在归纳步骤中,固定 x ∈ ℕ,假设 x − 1 ∈ ℤ,并证明 (x + 1) − 1 ∈ ℤ。由于 (x + 1) − 1 = x + (1 − 1) = x + 0 = x ∈ ℕ,我们甚至无需假设 x − 1 ∈ ℤ 即可得出 (x + 1) − 1 ∈ ℤ。

Case 2. Assume x = 0; prove x − 1 ∈ ℤ. Here, x − 1 = 0 − 1 = −1 ∈ ℤ since −( − 1) = 1 ∈ ℕ.

情况 2. 假设 x = 0;证明 x − 1 ∈ ℤ。这里,x − 1 = 0 − 1 = −1 ∈ ℤ,因为 −( − 1) = 1 ∈ ℕ。

Case 3. Assume − x ∈ ℕ; prove x − 1 ∈ ℤ. Write − x = n ∈ ℕ. Since ℕ is closed under adding 1, we know n + 1 ∈ ℕ. Now x − 1 = −( − x ) + ( − 1) = ( − n ) + ( − 1) = −( n + 1), so −( x − 1) = n + 1 ∈ ℕ, so x − 1 ∈ ℤ by the third alternative in the definition of ℤ.     □

情况 3. 假设 −x ∈ ℕ;证明 x − 1 ∈ ℤ。令 −x = n ∈ ℕ。由于 ℕ 对加 1 封闭,我们知道 n + 1 ∈ ℕ。现在 x − 1 = −( − x) + ( − 1) = ( − n) + ( − 1) = −(n + 1),所以 −(x − 1) = n + 1 ∈ ℕ,因此根据 ℤ 定义的第三种情况,x − 1 ∈ ℤ。□

8.34. Theorem on Closure of ℤ under Addition. For all x , y ∈ ℤ, x + y ∈ ℤ.

8.34. ℤ 在加法下的封闭性定理。对于所有 x, y ∈ ℤ,x + y ∈ ℤ。

Proof. Fix x , y ∈ ℤ. We know x ∈ ℕ or x = 0 or − x ∈ ℕ, and similarly for y . If x = 0, then x + y = 0 + y = y ∈ ℤ. If y = 0, then x + y = x + 0 = x ∈ ℤ. So we can reduce to the following four cases.

证明。固定 x, y ∈ ℤ。我们知道 x ∈ ℕ 或 x = 0 或 −x ∈ ℕ,y 的情况类似。如果 x = 0,则 x + y = 0 + y = y ∈ ℤ。如果 y = 0,则 x + y = x + 0 = x ∈ ℤ。因此,我们可以简化为以下四种情况。

Case 1. Assume x , y ∈ ℕ. We already proved x + y ∈ ℕ, so x + y ∈ ℤ.

情况 1. 假设 x, y ∈ ℕ。我们已经证明 x + y ∈ ℕ,所以 x + y ∈ ℤ。

Case 2. Assume − x , − y ∈ ℕ. Write n = − x and m = − y ; since n , m ∈ ℕ, we know m + n ∈ ℕ. Then −( x + y ) = [ − y ] + [ − x ] = m + n ∈ ℕ, so x + y ∈ ℤ by the third alternative in the definition of ℤ.

情况 2. 假设 −x, −y ∈ ℕ。令 n = −x 且 m = −y;由于 n, m ∈ ℕ,我们知道 m + n ∈ ℕ。那么 −(x + y) = [ − y] + [ − x] = m + n ∈ ℕ,因此根据 ℤ 定义的第三种选择,x + y ∈ ℤ。

Case 3. Assume x ∈ ℕ and − y ∈ ℕ. We prove that x + y ∈ ℤ by induction on n = − y . Since y = − n , our goal is to show that x n ∈ ℤ for each natural number n . If n = 1, then x − 1 ∈ ℤ by the lemma. For the induction step, fix n ∈ ℕ, assume x n ∈ ℤ, and prove x − ( n + 1) ∈ ℤ. We know x − ( n + 1) = x + ( − ( n + 1)) = x + (( − n ) + ( − 1)) = ( x + ( − n )) + ( − 1) = ( x n ) − 1. By hypothesis, x n ∈ ℤ. By the lemma, ( x n ) − 1 ∈ ℤ, completing the induction step for Case 3.

情况 3. 假设 x ∈ ℕ 且 −y ∈ ℕ。我们通过对 n = −y 进行归纳来证明 x + y ∈ ℤ。由于 y = −n,我们的目标是证明对于每个自然数 n,x − n ∈ ℤ。如果 n = 1,则根据引理,x − 1 ∈ ℤ。对于归纳步骤,固定 n ∈ ℕ,假设 x − n ∈ ℤ,并证明 x − (n + 1) ∈ ℤ。我们知道 x − (n + 1) = x + ( − (n + 1)) = x + (( − n) + ( − 1)) = (x + ( − n)) + ( − 1) = (x − n) − 1。根据假设,x − n ∈ ℤ。根据引理,(x − n) − 1 ∈ ℤ,完成了情况 3 的归纳步骤。

Case 4. Assume − x ∈ ℕ and y ∈ ℕ. We can prove ( − x ) + y ∈ ℤ as in Case 3, or we can reduce this case to Case 3 by noting ( − x ) + y = y + ( − x ).

情况 4. 假设 −x ∈ ℕ 且 y ∈ ℕ。我们可以像情况 3 一样证明 ( − x) + y ∈ ℤ,或者我们可以通过注意到 ( − x) + y = y + ( − x) 将此情况简化为情况 3。

It follows that ℤ is closed under subtraction, since x y = x + ( − y ). Since ℤ⊆ℝ by construction, we can also immediately conclude that the set ℤ satisfies all the axioms and theorems discussed so far that do not mention multiplicative inverses, division, or least upper bounds. For instance, since x + y = y + x holds for all real numbers x and y by A5, this equation must also hold for all x and y in the subset ℤ of ℝ. As another example, properties Z2 and Z3 (absence of zero divisors) hold for ℤ even though their proofs used the concept of multiplicative inverses in ℝ. In summary, we have proved that is an ordered integral domain . All the properties of integers proved in Chapter 4 (such as theorems on divisibility, primes, greatest common divisors, and unique prime factorizations) are now officially available for use in our formal development of ℝ.

由此可知,ℤ 在减法运算下封闭,因为 x − y = x + ( − y)。由于 ℤ ⊆ ℝ 由构造可知,我们也可以立即得出结论:集合 ℤ 满足迄今为止讨论的所有公理和定理,除了涉及乘法逆元、除法或最小上界的公理和定理。例如,根据公理 A5,对于所有实数 x 和 y,x + y = y + x 成立,因此该等式对于 ℝ 的子集 ℤ 中的所有 x 和 y 也成立。再举一例,性质 Z2 和 Z3(不存在零因子)对于 ℤ 也成立,尽管它们的证明使用了 ℝ 中的乘法逆元概念。总之,我们已经证明了 ℤ 是一个有序整环。第 4 章中证明的所有整数性质(例如整除性定理、素数定理、最大公约数定理和唯一素因数分解定理)现在都可以正式用于我们对 ℝ 的正式开发。

Defining Rational Numbers

定义有理数

Now that we have formally constructed ℤ within ℝ, the definition of rational numbers given in §2.1 is valid.

既然我们已经在 ℝ 中正式构造了 ℤ,那么 §2.1 中给出的有理数定义就是有效的。

Recall that for all x ∈ ℝ, we defined

回想一下,对于所有 x ∈ ℝ,我们定义了

x x Q m Z Z , n n Z Z , n n 0 x x = m / n n .

Using the fraction rules, we can prove the following closure properties of ℚ.

利用分数规则,我们可以证明 ℚ 的以下封闭性质。

8.35. Theorem on Closure Properties of ℚ

8.35. ℚ 的封闭性质定理

For all x , y ∈ ℚ, x + y ∈ ℚ, − x ∈ ℚ, x y ∈ ℚ, and x · y ∈ ℚ; if y ≠ 0, then y −1 ∈ ℚ and x / y ∈ ℚ. Moreover, ℤ⊆ℚ, so 0 ∈ ℚ and 1 ∈ ℚ.

对于所有 x, y ∈ ℚ,x + y ∈ ℚ,−x ∈ ℚ,x − y ∈ ℚ,以及 x · y ∈ ℚ;如果 y ≠ 0,则 y −1 ∈ ℚ 且 x/y ∈ ℚ。此外,ℤ⊆ℚ,因此 0 ∈ ℚ 且 1 ∈ ℚ。

Proof. We prove closure under subtraction as an example. Fix x , y ∈ ℚ; write x = a / b and y = c / d for some a , b , c , d ∈ ℤ with b , d ≠ 0. By the fraction rules, x y = ( a / b ) − ( c / d ) = ( ad bc )/( bd ). By closure properties of ℤ proved earlier, we know ad , bc , ad bc , and bd are in ℤ with bd ≠ 0 (by Z2). So x y meets the criterion for membership in ℚ, taking m = ad bc and n = bd in the definition above.

证明。我们以证明减法封闭性为例。固定 x, y ∈ ℚ;令 x = a/b 且 y = c/d,其中 a, b, c, d ∈ ℤ,且 b, d ≠ 0。根据分数规则,x − y = (a/b) − (c/d) = (ad − bc)/(bd)。根据前面证明的 ℤ 的封闭性质,我们知道 ad, bc, ad − bc 和 bd 都在 ℤ 中,且 bd ≠ 0(由 Z2 可知)。因此,取 m = ad − bc 且 n = bd,x − y 满足 ℚ 的成员条件。

To prove that ℤ⊆ℚ, note that 1 ∈ ℤ (since 1 ∈ ℕ) and 1 ≠ 0 by axiom N. So for any m ∈ ℤ, writing m = m /1 shows that m ∈ ℚ.     □

为了证明 ℤ⊆ℚ,注意到 1 ∈ ℤ(因为 1 ∈ ℕ),且根据公理 N,1 ≠ 0。因此,对于任意 m ∈ ℤ,令 m = m/1 可知 m ∈ ℚ。□

Having proved all the closure properties, we conclude that is an ordered field ; in other words, the set ℚ with the restricted arithmetical operations and order relation satisfies all the axioms for ℝ (and their consequences) except possibly completeness. In particular, the unique prime factorization theorem for ℚ (see §4.7) is now available. One consequence of that theorem was the fact that for any positive integer n that has an odd exponent in its prime factorization, the equation x 2 = n has no solution x in ℚ. It can be proved from the Completeness Axiom (O7) that this equation does have a solution x in ℝ; see Theorem 8.61 below. These facts combine to show that the number system ℚ does not satisfy the Completeness Axiom. This axiom thus reveals a crucial difference between the rational number system and the real number system. Indeed, many of the central theorems of calculus would not work if we used ℚ instead of ℝ.

在证明了所有封闭性质之后,我们得出结论:ℚ 是一个有序域;换句话说,具有受限算术运算和序关系的集合 ℚ 满足 ℝ 的所有公理(及其推论),除了完备性之外。特别地,ℚ 的唯一素因数分解定理(见 §4.7)现在可以成立。该定理的一个推论是:对于任何素因数分解中指数为奇数的正整数 n,方程 x 2 = n 在 ℚ 中无解 x。根据完备性公理 (O7) 可以证明,该方程在 ℝ 中存在解 x;见下文定理 8.61。这些事实共同表明,数系 ℚ 不满足完备性公理。因此,该公理揭示了有理数系和实数系之间的一个关键区别。事实上,如果我们用 ℚ 代替 ℝ,微积分的许多核心定理将无法成立。

Section Summary

章节概要

  1. Natural Numbers. An inductive set is a subset of ℝ containing 1 and closed under adding 1. The set ℕ is the intersection of all inductive sets. ℕ itself is an inductive set closed under addition and multiplication. To prove n N , P ( n ) , we may prove that P (1) is true and for all n ∈ ℕ, if P ( n ) is true, then P ( n + 1) is true.
    自然数。归纳集是包含 1 且对加 1 封闭的 ℝ 子集。集合 ℕ 是所有归纳集的交集。ℕ 本身也是一个归纳集,对加法和乘法封闭。为了证明对于所有 n∈N,P(n) 成立,我们可以证明 P(1) 为真,并且对于所有 n∈ℕ,如果 P(n) 为真,则 P(n + 1) 也为真。
  2. Integers. The set ℤ consists of all real numbers x such that x ∈ ℕ or x = 0 or − x ∈ ℕ. ℤ is closed under addition, negation, and multiplication. ℤ is an ordered integral domain.
    整数集 ℤ 由所有满足 x ∈ ℕ 或 x = 0 或 −x ∈ ℕ 的实数 x 组成。ℤ 对加法、取反和乘法封闭。ℤ 是一个有序整环。
  3. Rational Numbers. The set ℚ consists of all real numbers x such that there exist m , n ∈ ℤ with n ≠ 0 and x = m / n . ℚ contains ℤ and is closed under addition, negation, multiplication, and inverses of nonzero elements. ℚ is an ordered field that does not satisfy the Completeness Axiom.
    有理数。集合 ℚ 由所有实数 x 组成,满足存在 m, n ∈ ℤ,其中 n ≠ 0 且 x = m/n。ℚ 包含 ℤ,并且对加法、取反、乘法以及非零元素的逆运算封闭。ℚ 是一个有序域,但不满足完备性公理。

Exercises

练习

  1. Finish the proof of Theorem 8.35 .
    完成定理 8.35 的证明。
  2. Prove or disprove: the union of any nonempty collection of inductive subsets of ℝ is an inductive set.
    证明或反证:ℝ 的任意非空归纳子集的并集是一个归纳集。
  3. Decide (with proof) whether each set below is inductive. [You may assume that 1 > 0 has already been proved.] (a) ℤ ≥0 (b) ℤ >1 (c) ℤ ≥1 (d) (e) the set of odd integers (f) { k /3: k ∈ ℤ} (g) {1, 2, 3, 4, 5, 6, 7, 8, 9} (your proof should use 1 > 0 somewhere).
    判断(并证明)以下每个集合是否为归纳集。[你可以假设 1 > 0 已被证明。] (a) ℤ ≥0 (b) ℤ >1 (c) ℤ ≥1 (d) ∅ (e) 奇数集合 (f) {k/3:k ∈ ℤ} (g) {1, 2, 3, 4, 5, 6, 7, 8, 9}(你的证明中应该用到 1 > 0)。
  4. (a) Show that ℝ − ℤ is closed under adding 1. (b) Is ℝ − ℤ an inductive set? (c) Is ℝ − ℤ closed under addition? negation? multiplication? multiplicative inverses? When answering (c), assume we have proved that 2 −1 ∈ ℝ − ℤ.
    (a) 证明 ℝ − ℤ 对加 1 封闭。(b) ℝ − ℤ 是归纳集吗?(c) ℝ − ℤ 对加法、取反、乘法和乘法逆元封闭吗?回答 (c) 时,假设我们已经证明 2 −1 ∈ ℝ − ℤ。
  5. Repeat the previous exercise, replacing ℝ − ℤ by ℝ − ℚ and replacing 2 −1 by 2 .
    重复前面的练习,将 ℝ − ℤ 替换为 ℝ − ℚ,并将 2 −1 替换为 2。
  6. Show that ℤ is the intersection of all subsets of ℝ containing 0 and 1 and closed under addition and negation.
    证明 ℤ 是 ℝ 的所有包含 0 和 1 且对加法和取反运算封闭的子集的交集。
  7. Show that ℤ is the intersection of all subsets of ℝ containing 1 and closed under subtraction.
    证明 ℤ 是 ℝ 的所有包含 1 且对减法封闭的子集的交集。
  8. Show that ℤ is the intersection of all inductive subsets closed under subtracting 1. Use this to formulate an induction principle for proving statements of the form n Z , P ( n ) .
    证明 ℤ 是所有在减 1 下封闭的归纳子集的交集。利用此证明,制定一个归纳原理,以证明形如 ∀n∈Z,P(n) 的陈述。
  9. Show that ℚ is the intersection of all subsets of ℝ containing 0 and 1 and closed under addition, subtraction, multiplication, and division by nonzero elements.
    证明 ℚ 是 ℝ 的所有包含 0 和 1 且对加法、减法、乘法和除以非零元素封闭的子集的交集。
  10. Double Induction. Suppose S is a set such that: (a) S ⊆ℕ × ℕ; (b) (1, 1) ∈ S ; (c) ( x , y ) S , ( x + 1 , y ) S ( x , y + 1 ) S . Prove that S = ℕ × ℕ.
    双重归纳法。假设 S 是一个集合,满足:(a) S⊆ℕ × ℕ;(b) (1, 1) ∈ S;(c) ∀(x,y)∈S,(x+1,y)∈S∧(x,y+1)∈S。证明 S = ℕ × ℕ。
  11. State an induction principle for proving statements of the form

    陈述一个归纳原理,用于证明形如 的命题

    m N , n N , P ( m , n ) . Prove this principle using the previous exercise.

    ∀m∈N,∀n∈N,P(m,n)。利用前面的练习证明这个原理。

  12. We know from the Division Theorem that there is a function MOD : Z × Z > 0 Z given by MOD ( a , b ) = a mod b = the unique remainder when a is divided by b . For n ∈ ℤ >1 , define the set of integers modulo n to be
    Z n = { 0 , 1 , 2 , , n 1 } = { x Z : 0 x < n } .
    Define algebraic operations and constants on ℤ n as follows: for x , y ∈ ℤ n , let x n y = ( x + y ) mod n , n x = ( x ) mod n , x n y = ( x y ) mod n , 0 n = 0, and 1 n = 1. Prove that ℤ n with these operations is a nonzero commutative ring (i.e., satisfies all algebraic axioms excluding M4). It can help to first prove this lemma: for all a , b ∈ ℤ n , a = b iff n divides a b . (An alternative construction of ℤ n , based on the equivalence relation ≡ n , is given in §6.6.)
    根据除法定理,我们知道存在一个函数 MOD:Z×Z>0→Z,其定义为 MOD(a,b)=amodb= a 除以 b 的唯一余数。对于 n ∈ ℤ >1 ,定义模 n 的整数集合为 Zn={0,1,2,…,n−1}={x∈Z:0≤x<n}。在 ℤ n 上定义如下代数运算和常数:对于 x, y ∈ ℤ n ,令 x⊕ny=(x+y)modn,⊖nx=(−x)modn,x⊗ny=(x⋅y)modn,0 n = 0,1 n = 1。证明具有这些运算的 ℤ n 是一个非零交换环(即满足除 M4 之外的所有代数公理)。首先证明以下引理会有所帮助:对于所有 a, b ∈ ℤ n ,a = b 当且仅当 n 整除 a − b。 (第 6.6 节给出了基于等价关系 ≡ n 的 ℤ n 的另一种构造方法。)
  13. Continuing the previous exercise, suppose p is a prime. (a) Use gcds to prove: for all nonzero k ∈ ℤ p , there exists a unique m ∈ ℤ p with k p m = 1 = m p k . Define the multiplicative inverse of k in ℤ p to be this unique m . (b) Show that ℤ p is a field. (c) Show that if n is not prime, then ℤ n does not satisfy Z2 and therefore cannot be a field.
    继续上一题,假设 p 是素数。(a) 利用最大公约数证明:对于所有非零 k ∈ ℤ p ,存在唯一的 m ∈ ℤ p ,使得 k ⊗ p m = 1 = m ⊗ p k。定义 ℤ p 中 k 的乘法逆元为这个唯一的 m。(b) 证明 ℤ p 是一个域。(c) 证明如果 n 不是素数,则 ℤ n 不满足 Z2,因此不可能是域。

8.4  Ordering, Absolute Value, and Distance

8.4 排序、绝对值和距离

Now that we understand the basic algebraic properties of ℝ, we begin to study the consequences of the order axioms O1 through O6. We develop fundamental laws of inequalities as well as various facts about the absolute value function. This leads to a discussion of the concept of distance on the real line, which is defined in terms of absolute value.

现在我们已经理解了ℝ的基本代数性质,接下来我们将研究序公理O1到O6的推论。我们将推导出不等式的基本定律以及关于绝对值函数的各种性质。这最终将引出对实数轴上距离概念的讨论,该概念是用绝对值定义的。

Ordering Properties

排序属性

The basic ordering axioms were stated in § 8.1 . Axioms O1 through O4 postulate that ℝ is totally ordered by the relation ≤. In detail, for all x , y , z ∈ ℝ, x x (reflexivity, axiom O1); if x y and y x , then x = y (antisymmetry, axiom O2); if x y and y z , then x z (transitivity, axiom O3); and x y or y x (total ordering, axiom O4). We also defined x y to mean y x , x < y to mean x y and x y , and x > y to mean y < x .

基本排序公理已在第 8.1 节中阐述。公理 O1 至 O4 假定 ℝ 由关系 ≤ 完全排序。具体而言,对于所有 x, y, z ∈ ℝ,x ≤ x(自反性,公理 O1);如果 x ≤ y 且 y ≤ x,则 x = y(反对称性,公理 O2);如果 x ≤ y 且 y ≤ z,则 x ≤ z(传递性,公理 O3);以及 x ≤ y 或 y ≤ x(完全排序,公理 O4)。我们还定义了 x ≥ y 表示 y ≤ x,x < y 表示 x ≤ y 且 x ≠ y,以及 x > y 表示 y < x。

Our first result derives the properties of the strict order < that correspond to the properties of ≤ listed above. This result applies to any totally ordered set ( X , ≤ ), not just X = ℝ.

我们的第一个结果推导出了严格序 < 的性质,这些性质与上面列出的 ≤ 的性质相对应。该结果适用于任何全序集 (X, ≤ ),而不仅仅是 X = ℝ。

8.36. Theorem on Order Properties of <, >, and ≥. For all x , y , z ∈ ℝ:

8.36. 关于 <、> 和 ≥ 的序性质的定理。对于所有 x, y, z ∈ ℝ:

(a) Irreflexivity: x < x is false.

(a)非自反性:x < x 为假。

(b) Transitivity: If x < y and y < z , then x < z .

(b)传递性:如果 x < y 且 y < z,则 x < z。

(c) x y iff x < y or x = y ; and x y iff x > y or x = y . trichotomy law

(c) x ≤ y 当且仅当 x < y 或 x = y;x ≥ y 当且仅当 x > y 或 x = y。三分律

(d) Trichotomy: Exactly one of these three statements is true: x < y ; x = y ; y < x .

(d)三分法:这三个陈述中只有一个是正确的:x < y;x = y;y < x。

(e) Parts (a), (b), and (d) hold with > in place of <.

(e)第(a)、(b)和(d)部分仍然成立,只是将 < 替换为 >。

(f) Axioms O1 through O4 hold with ≥ in place of ≤.

(f)公理 O1 至 O4 成立,用 ≥ 代替 ≤。

(g) If x < y and y z , then x < z . If x y and y < z , then x < z .

(g)如果 x < y 且 y ≤ z,则 x < z。如果 x ≤ y 且 y < z,则 x < z。

Similar statements hold for > and ≥.

类似的结论也适用于 > 和 ≥。

Proof. Fix x , y , z ∈ ℝ. To prove (a), note that x < x is defined to mean “ x x and x x .” But x = x is true, so this AND-statement is false.

证明。设 x, y, z ∈ ℝ。为了证明 (a),注意到 x < x 的定义是“x ≤ x 且 x ≠ x”。但 x = x 为真,因此这个 AND 语句为假。

To prove (b), assume x < y and y < z ; prove x < z . We have assumed x y and x y and y z and y z ; we must prove x z and x z . On one hand, using x y and y z and transitivity (O3), we get x z . On the other hand, if we had x = z , then x y and y z = x and antisymmetry (O2) would give x = y , which contradicts our assumption. Thus, x z .

为了证明 (b),假设 x < y 且 y < z;证明 x < z。我们已经假设 x ≤ y 且 x ≠ y 且 y ≤ z 且 y ≠ z;我们必须证明 x ≤ z 且 x ≠ z。一方面,利用 x ≤ y 且 y ≤ z 以及传递性 (O3),我们得到 x ≤ z。另一方面,如果 x = z,那么 x ≤ y 且 y ≤ z = x,并且根据反对称性 (O2),我们将得到 x = y,这与我们的假设矛盾。因此,x ≠ z。

To prove (d) assuming (c) has been proved, use total ordering (O4) to see that x y or y x is true; i.e., “ x < y or x = y or y < x or y = x ” is true. Thus at least one of the three statements in (d) must hold. If x < y and x = y were both true, we would deduce x < x , contradicting (a). Similarly x = y and y < x cannot both be true. Finally, if x < y and y < x were both true, (b) would give x < x , contradicting (a). Thus exactly one of the given alternatives holds.

假设 (c) 已证明,为了证明 (d),利用全序关系 (O4) 可知 x ≤ y 或 y ≤ x 为真,即“x < y 或 x = y 或 y < x 或 y = x”为真。因此,(d) 中的三个命题中至少有一个成立。如果 x < y 和 x = y 都为真,则可得出 x < x,这与 (a) 矛盾。类似地,x = y 和 y < x 也不能同时为真。最后,如果 x < y 和 y < x 都为真,则 (b) 可得出 x < x,这与 (a) 矛盾。因此,给定的三个命题中,恰好有一个成立。

We leave the remaining assertions as exercises in manipulating the definitions.

我们将剩余的论断留作对定义的操作练习。

The next result uses trichotomy to deduce rules for forming useful denials of inequalities.

下一个结果利用三分法推导出形成有用的不等式否定的规则。

8.37. Theorem on Denial Rules for Inequalities. For all x , y ∈ ℝ:

8.37. 不等式否定规则定理。对于所有 x, y ∈ ℝ:

(a) ∼( x y ) ⇔ y < x x > y .

(a)∼(x≤y)⇔y<x⇔x>y。

(b) ∼( x < y ) ⇔ y x x y .

(b) ~(x < y) ⇔ y ≤ x ⇔ x ≥ y。

(c) ∼( x y ) ⇔ y > x x < y .

(c) ~(x ≥ y) ⇔ y > x ⇔ x < y。

(d) ∼( x > y ) ⇔ y x x y .

(d) ~(x > y) ⇔ y ≥ x ⇔ x ≤ y。

Proof. We prove (a) as an example. Fix x , y ∈ ℝ. First assume x y is false. By Theorem 8.36 (c), “ x < y or x = y ” is false, so trichotomy forces y < x to be true. This is equivalent to x > y by definition. Conversely, assume y < x is true. By trichotomy, x < y is false and x = y is false, so “ x < y or x = y ” is false. By Theorem 8.36 (c), we see that x y is false.

证明。我们以 (a) 为例进行证明。固定 x, y ∈ ℝ。首先假设 x ≤ y 为假。根据定理 8.36(c),“x < y 或 x = y”为假,因此三分性必然导致 y < x 为真。这等价于 x > y(根据定义)。反之,假设 y < x 为真。根据三分性,x < y 为假且 x = y 为假,因此“x < y 或 x = y”为假。根据定理 8.36(c),我们可知 x ≤ y 为假。

Algebraic Properties of Order

Next we investigate the consequences of axioms O5 and O6, which relate the algebraic operations of addition and multiplication to the ordering relation on ℝ. We recall that the additive axiom (O5) states

x , y , z R , x y x + z y + z ,
and the multiplicative axiom (O6) states
x , y , z R , ( x y 0 z ) x z y z .

8.38. Theorem on Algebraic Properties of Order. For all w , x , y , z ∈ ℝ:

(a) Order Reversal: x y iff − y ≤ − x .

(b) Adding Inequalities: If x y and w z , then x + w y + z .

(c) Multiplying by a Negative Number: If x y and z ≤ 0, then y · z x · z .

(d) Squares are Nonnegative: 0 ≤ w 2 .

(e) Parts (a), (b), (c), axiom O5, and axiom O6 hold with ≤ replaced by < throughout.

In place of (d), we have: w R , w 0 0 < w 2 . Similar properties hold for ≥ and >.

Proof. Fix w , x , y , z ∈ ℝ. To prove (a), first assume x y . Taking z = − x + ( − y ) in (O5), the assumption yields x + ( − x + ( − y )) ≤ y + ( − x + ( − y )). This becomes − y ≤ − x after simplification via known algebra rules. Conversely, assume − y ≤ − x . Adding x + y to both sides (using O5) leads to x y . Alternatively, the converse can be deduced from the direction already proved: from − y ≤ − x we get −( − x ) ≤ −( − y ), so that x y by the double negative rule.     □

To prove (b), assume x y and w z . Adding w to both sides of x y , we get x + w y + w . Adding y to both sides of w z , we get y + w y + z . Now transitivity gives x + w y + z .

To prove (c), assume x y and z ≤ 0. By (a), 0 = −0 ≤ − z . So we may apply (O6) with z there replaced by − z , obtaining x · ( − z ) ≤ y · ( − z ). This becomes −( xz ) ≤ −( yz ) by the sign rules, so yz xz follows from (a).

Next we prove the version of (O6) for <, namely: if x < y and 0 < z , then xz < yz . Assume x < y and 0 < z . Then x y and 0 ≤ z , so the original O6 gives xz yz . To conclude xz < yz , we show that xz = yz cannot occur. Note that 0 < z forces z ≠ 0, by trichotomy. So if xz = yz were true, we could cancel z to get x = y , but this contradicts the assumption x < y . So xz < yz does hold.

Next we prove the version of (d) for <. Assuming w ≠ 0, we must have 0 < w or w < 0 by trichotomy.

Case 1. Assume 0 < w . The version of (O6) just proved (taking x = 0, y = w , and z = w ) gives 0 · w < w · w , so 0 < w 2 .

Case 2. Assume w < 0, so 0 = −0 < − w by (a). Now take x = 0, y = − w and z = − w in the version of (O6) for < to conclude 0 · ( − w ) < ( − w ) · ( − w ), so 0 < w 2 by the sign rules. Thus, 0 < w 2 holds in all cases.

The remaining assertions are left as exercises.     □

Note that 1 2 = 1 · 1 = 1 by the identity axiom (M3). Also 1 ≠ 0 by axiom N. Thus, as a corollary to the strict version of (d) above, we get 0 < 1 . This fact has surprisingly deep consequences, as we will see shortly. As one application, we can repeatedly add 1 to the inequality 0 < 1 and use Theorem 8.38 (a) to obtain

10 < < 4 < 3 < 2 < 1 < 0 < 1 < 2 < 3 < 4 < < 10.
接下来,我们研究公理 O5 和 O6 的推论,它们将加法和乘法的代数运算与 ℝ 上的序关系联系起来。我们回顾一下,加法公理 (O5) 表述为:∀x,y,z∈R,x≤y⇒x+z≤y+z;乘法公理 (O6) 表述为:∀x,y,z∈R,(x≤y∧0≤z)⇒x⋅z≤y⋅z。8.38. 关于序的代数性质的定理。对于所有 w, x, y, z ∈ ℝ:(a) 序反转:x ≤ y 当且仅当 −y ≤ −x。(b) 不等式相加:如果 x ≤ y 且 w ≤ z,则 x + w ≤ y + z。 (c) 乘以负数:如果 x ≤ y 且 z ≤ 0,则 y · z ≤ x · z。(d) 平方数非负:0 ≤ w 2 。(e) 将 (a)、(b)、(c)、公理 O5 和公理 O6 中的 ≤ 替换为 <,则这些结论仍然成立。(d) 的结论可以替换为:∀w∈R,w≠0⇒0<w2。类似的性质也适用于 ≥ 和 >。证明:固定 w, x, y, z ∈ ℝ。为了证明 (a),首先假设 x ≤ y。在 (O5) 中取 z = −x + ( − y),则有 x + ( − x + ( − y)) ≤ y + ( − x + ( − y))。通过已知的代数规则简化后,可得 −y ≤ −x。反之,假设 −y ≤ −x。两边同时加上 x + y(利用 O5)可得 x ≤ y。或者,也可以从已证明的方向推导出逆命题:由 −y ≤ −x 可得 −( − x) ≤ −( − y),因此根据双重否定规则,x ≤ y。□ 为了证明 (b),假设 x ≤ y 且 w ≤ z。两边同时加上 w,可得 x + w ≤ y + w。两边同时加上 y,可得 y + w ≤ y + z。现在,根据传递性可得 x + w ≤ y + z。为了证明 (c),假设 x ≤ y 且 z ≤ 0。由 (a) 可知,0 = −0 ≤ −z。因此,我们可以应用 (O6),并将其中的 z 替换为 −z,得到 x · ( − z) ≤ y · ( − z)。根据符号规则,这变为 −(xz) ≤ −(yz),因此 yz ≤ xz 由 (a) 得出。接下来,我们证明 (O6) 式关于 < 的情况,即:如果 x < y 且 0 < z,则 xz < yz。假设 x < y 且 0 < z。则 x ≤ y 且 0 ≤ z,因此原 O6 式得出 xz ≤ yz。为了得出 xz < yz 的结论,我们证明 xz = yz 不可能成立。注意,根据三分性,0 < z 必然导致 z ≠ 0。因此,如果 xz = yz 成立,我们可以约掉 z 得到 x = y,但这与假设 x < y 矛盾。所以 xz < yz 成立。接下来,我们证明 (d) 式关于 < 的情况。假设 w ≠ 0,根据三分性,我们必须有 0 < w 或 w < 0。情况 1:假设 0 < w。刚才证明的 (O6) 版本(取 x = 0,y = w,z = w)得出 0 · w < w · w,因此 0 < w 2 。情况 2. 假设 w < 0,则由 (a) 可知 0 = −0 < −w。现在,在 (O6) 版本中取 x = 0,y = −w 和 z = −w,对于 < 可知 0 · ( − w) < ( − w) · ( − w),因此由符号规则可知 0 < w 2 。因此,0 < w 2 在所有情况下都成立。其余断言留作练习。□ 注意,由恒等公理 (M3) 可知 1 2 = 1 · 1 = 1。根据公理 N,1 ≠ 0。因此,作为上述 (d) 的严格形式的推论,我们得到 0<1。正如我们稍后将看到的,这一事实具有令人惊讶的深刻意义。作为一项应用,我们可以反复地将 1 加到不等式 0 < 1 中,并利用定理 8.38(a) 得到 −10<⋯<−4<−3<−2<−1<0<1<2<3<4<⋯<10。

Positive and Negative Numbers

正数和负数

Now we can define positive and negative real numbers and study how these concepts relate to inequalities.

现在我们可以定义正实数和负实数,并研究这些概念与不等式的关系。

8.39. Definition. For all x ∈ ℝ, x is (strictly) positive iff x > 0 ; x is (strictly) negative iff x < 0 x is weakly positive iff x 0 ; x is weakly negative iff x 0 . We write ℝ >0 for the set of positive real numbers; ℝ <0 , ℝ ≥0 , ℝ ≤0 , ℤ >0 , etc., are defined similarly.

8.39. 定义。对于所有 x ∈ ℝ,xi 是(严格)正当且仅当 x>0;xi 是(严格)负当且仅当 x<0;xi 是弱正当且仅当 x≥0;xi 是弱负当且仅当 x≤0。我们用 ℝ >0 表示正实数集;ℝ <0 、ℝ ≥0 、ℝ ≤0 、ℤ >0 等的定义类似。

Using the denial rules for inequalities, one sees that x is weakly positive iff x is nonnegative (not negative), and x is weakly negative iff x is nonpositive. For us, the unqualified term “positive” means “strictly positive.” We warn the reader that some texts adopt the opposite convention: for them, “positive” means “weakly positive.”

利用不等式的否定规则,我们可以看出,x 是弱正当且仅当 x 非负,x 是弱负当且仅当 x 非正。对我们而言,不加限定的术语“正”指的是“严格正”。我们提醒读者,有些文献采用相反的约定:在它们看来,“正”指的是“弱正”。

Here are some basic properties of positive numbers. You can formulate and prove analogous properties of negative numbers, nonnegative numbers, and nonpositive numbers.

以下是一些正数的基本性质。你可以推导出并证明负数、非负数和非正数的类似性质。

8.40. Theorem on Positive Numbers.

8.40. 正数定理。

(a) For all x , y ∈ ℝ >0 , x + y ∈ ℝ >0 , xy ∈ ℝ >0 , and x −1 , y −1 ∈ ℝ >0 .

(a)对于所有 x, y ∈ ℝ >0 , x + y ∈ ℝ >0 , xy ∈ ℝ >0 , 和 x −1 , y −1 ∈ ℝ >0

Also, x < y iff y −1 < x −1 .

此外,x < y 当且仅当 y −1 < x −1

(b) For all a ∈ ℝ, a is positive iff − a is negative; and a is negative iff − a is positive.

(b)对于所有 a ∈ ℝ,a 为正当且仅当 −a 为负;并且 a 为负当且仅当 −a 为正。

(c) For all a , b ∈ ℝ, ab ∈ ℝ >0 iff a and b are both positive or both negative.

(c)对于所有 a, b ∈ ℝ,ab ∈ ℝ >0 当且仅当 a 和 b 均为正数或均为负数。

Proof. We prove (c) and part of (a) as examples. For (a), fix x , y ∈ ℝ >0 . By trichotomy applied to x , x ≠ 0, so x −1 exists. By trichotomy applied to x −1 , we know x −1 < 0 or x −1 = 0 or x −1 > 0. In the case x −1 < 0, we can multiply both sides of this inequality by the positive quantity x , obtaining x · x −1 < x · 0, or 1 < 0. This contradicts trichotomy applied to 1, since we proved 0 < 1 above. So x −1 < 0 is impossible. Similarly, in the case x −1 = 0, multiplying both sides by x leads to 1 = 0, which violates axiom N. Thus, the third case x −1 > 0 must hold. Similarly, y −1 > 0. Now assume x < y . Multiply both sides by the positive quantity x −1 to get 1 < x −1 y . Then multiply both sides by the positive quantity y −1 to get y −1 < x −1 . Conversely, assume y −1 < x −1 . The result just proved and the double inverse rule lead to x < y .

证明。我们以 (c) 和 (a) 的一部分为例进行证明。对于 (a),固定 x, y ∈ ℝ >0 。根据三分性定理,x ≠ 0,因此 x −1 存在。根据三分性定理,x −1 存在,因此 x −1 < 0 或 x −1 = 0 或 x −1 > 0。对于 x −1 < 0 的情况,我们可以将不等式两边乘以正数 x,得到 x · x −1 < x · 0,即 1 < 0。这与三分性定理对 1 的定理相矛盾,因为我们上面证明了 0 < 1。所以 x −1 < 0 是不可能的。类似地,当 x −1 = 0 时,两边同乘以 x 得到 1 = 0,这违反了公理 N。因此,第三种情况 x −1 > 0 必然成立。类似地,y −1 > 0 也成立。现在假设 x < y。两边同乘以正数 x −1 得到 1 < x −1 y。然后两边同乘以正数 y −1 得到 y −1 < x −1 。反之,假设 y −1 < x −1 。刚才证明的结果以及双重逆规则得出 x < y。

We prove (c) using proof by cases, assuming (b) has been proved. Fix a , b ∈ ℝ. If a and b are positive, then so is ab by part (a). If a and b are negative, then − a and − b are positive by (b), so ab = ( − a )( − b ) is positive by (a) and the sign rules. If a or b is zero, then ab is zero, so ab is not positive. If a is positive and b is negative, then − b is positive by (b), so − ab = a ( − b ) is positive by the sign rules, so ab is negative by (b), so ab is not positive by trichotomy. The same conclusion holds if a is negative and b is positive. By trichotomy, all possible cases have now been considered, so (c) holds.      □

我们假设 (b) 已证明,并使用分情况证明法来证明 (c)。固定 a, b ∈ ℝ。如果 a 和 b 均为正,则根据 (a) 可知 ab 也为正。如果 a 和 b 均为负,则根据 (b) 可知 −a 和 −b 均为正,因此根据 (a) 和符号规则,ab = ( − a)( − b) 为正。如果 a 或 b 为零,则 ab 为零,因此 ab 不为正。如果 a 为正且 b 为负,则根据 (b) 可知 −b 为正,因此根据符号规则,−ab = a( − b) 为正,因此根据 (b) 可知 ab 为负,因此根据三分法可知 ab 不为正。如果 a 为负且 b 为正,结论相同。根据三分法,所有可能的情况都已考虑在内,因此 (c) 成立。□

We can now deduce that ℕ (the set of natural numbers, as defined in § 8.3 ) coincides with ℤ >0 (the set of positive integers). Another useful technical fact contained in the next lemma is that there can be no integer strictly between two consecutive integers k and k + 1.

现在我们可以推断,ℕ(自然数集,定义见§8.3)与ℤ >0 (正整数集)重合。下一个引理中包含的另一个有用的技术事实是,不存在严格介于两个连续整数k和k+1之间的整数。

8.41. Lemma on ℕ and ℤ >0 .

8.41.关于ℕ和ℤ >0 的引理。

(a) For all n ∈ ℕ, 1 ≤ n .

(a)对于所有 n ∈ ℕ,1 ≤ n。

(b) For all m ∈ ℤ − ℕ, m ≤ 0.

(b)对于所有m∈ℤ−ℕ,m≤0。

(c) N = Z R > 0 = Z > 0 .

(c)N=Z∩R>0=Z>0。

(d) For all n , k ∈ ℤ, k < n < k + 1 is false.

(d)对于所有 n, k ∈ ℤ,k < n < k + 1 为假。

(e) For all m , n ∈ ℤ, m n + 1 iff m n or m = n + 1.

(e)对于所有m,n∈ℤ,m≤n+1当且仅当m≤n或m=n+1。

Proof. We prove (a) by induction on n . For n = 1, 1 ≤ 1 is true by reflexivity (O1). Fix n ∈ ℕ, assume 1 ≤ n , and prove 1 ≤ n + 1. We know 0 ≤ 1; adding this inequality to the assumption 1 ≤ n gives 1 + 0 ≤ n + 1, or 1 ≤ n + 1.

证明。我们对 n 进行归纳证明 (a)。当 n = 1 时,根据自反性 (O1),1 ≤ 1 成立。固定 n ∈ ℕ,假设 1 ≤ n,并证明 1 ≤ n + 1。我们知道 0 ≤ 1;将此不等式加到假设 1 ≤ n 上,得到 1 + 0 ≤ n + 1,即 1 ≤ n + 1。

To prove (b), fix m ∈ ℤ − ℕ. By definition of ℤ, we know m ∈ ℕ or m = 0 or − m ∈ ℕ. The first case cannot occur since we assumed m N . In the second case, m = 0, and 0 ≤ 0 by reflexivity (O1). In the third case, 0 < 1 ≤ − m by part (a), so 0 < − m and hence m < 0 by order reversal. Thus m ≤ 0 in this case as well.

为了证明 (b),固定 m ∈ ℤ − ℕ。根据 ℤ 的定义,我们知道 m ∈ ℕ 或 m = 0 或 −m ∈ ℕ。第一种情况不可能出现,因为我们假设 m∉N。在第二种情况下,m = 0,并且根据自反性 (O1),0 ≤ 0。在第三种情况下,根据 (a) 部分,0 < 1 ≤ −m,所以 0 < −m,进而根据顺序反转,m < 0。因此,在这种情况下,m ≤ 0。

To prove (c), fix n ∈ ℕ. Then n ∈ ℤ by definition, and n ≥ 1 > 0, so n ∈ ℝ >0 . Thus, N Z R > 0 . On the other hand, fix k Z R > 0 . If k were not in ℕ, we would have k ≤ 0 by (b). But we assumed k > 0 also, so this would contradict trichotomy. Thus, k ∈ ℕ, so Z R > 0 N . The equality Z R > 0 = Z > 0 follows from the definition ℤ >0 = { x ∈ ℤ: x > 0}.

为了证明 (c),固定 n ∈ ℕ。根据定义,n ∈ ℤ,且 n ≥ 1 > 0,所以 n ∈ ℝ >0 。因此,N⊆Z∩R>0。另一方面,固定 k∈Z∩R>0。如果 k 不在 ℕ 中,根据 (b),我们将有 k ≤ 0。但我们假设 k > 0,所以这将与三分性矛盾。因此,k ∈ ℕ,所以 Z∩R>0⊆N。等式 Z∩R>0=Z>0 由定义 ℤ >0 = {x ∈ ℤ:x > 0} 得出。

We prove (d) by contradiction. Assume, to get a contradiction, that there exist n , k ∈ ℤ with k < n < k + 1. Adding − k to these inequalities, we get 0 < n k < 1, where m = n k is an integer by closure of ℤ under subtraction. Since m > 0, we must have m ∈ ℕ by (c). But then m ≥ 1 by (a), and now m < 1 ≤ m contradicts trichotomy.

我们用反证法证明 (d)。假设存在 n, k ∈ ℤ 使得 k < n < k + 1,从而得到矛盾。将 −k 加到这些不等式上,我们得到 0 < n − k < 1,其中 m = n − k 是一个整数,这是由 ℤ 对减法的封闭性得出的。由于 m > 0,根据 (c),我们必须有 m ∈ ℕ。但这样一来,根据 (a),m ≥ 1,现在 m < 1 ≤ m 与三分性矛盾。

To prove (e), fix m , n ∈ ℤ. First assume m n + 1, and prove m n or m = n + 1. To do so, further assume m n is false, and prove m = n + 1. We have assumed n < m n + 1. Since n < m < n + 1 is impossible by (d), we conclude that m = n + 1. For the converse, assume m n or m = n + 1, and prove m n + 1. In the case where m n , note that n < n + 1 (as we see by adding n to 0 < 1), so m n + 1 by transitivity. In the case m = n + 1, m n + 1 holds by reflexivity (O1).

为了证明 (e),固定 m, n ∈ ℤ。首先假设 m ≤ n + 1,并证明 m ≤ n 或 m = n + 1。为此,进一步假设 m ≤ n 不成立,并证明 m = n + 1。我们已经假设 n < m ≤ n + 1。由于根据 (d) 可知 n < m < n + 1 不可能成立,因此我们得出 m = n + 1。对于逆命题,假设 m ≤ n 或 m = n + 1,并证明 m ≤ n + 1。在 m ≤ n 的情况下,注意到 n < n + 1(正如我们通过将 n 加到 0 < 1 上所看到的),因此根据传递性,m ≤ n + 1。在 m = n + 1 的情况下,根据自反性 (O1),m ≤ n + 1 成立。

Absolute Value

绝对值

#

We define a function ABS : R R , denoted ABS ( x ) = | x | for x ∈ ℝ, using definition by cases.

我们定义一个函数 ABS:R→R,记为 ABS(x)=|x|,其中 x ∈ ℝ,采用分情况定义。

8.42. Definition: Absolute Value. For x ∈ ℝ, let | x | = x if 0 ≤ x , and let | x | = − x if x ≤ 0.

8.42. 定义:绝对值。对于 x ∈ ℝ,令 |x| = x,如果 0 ≤ x;令 |x| = −x,如果 x ≤ 0。

By total ordering (axiom O4), every real x satisfies at least one of the two cases in the definition. Moreover, the two cases overlap consistently: if 0 ≤ x and x ≤ 0 both hold, then x = 0 by antisymmetry (O2), and x = − x = 0, so |0| = 0. Thus, we do have a well-defined (single-valued) absolute value function. The next theorem gives the basic properties of this function.

根据全序关系(公理 O4),每个实数 x 至少满足定义中的两种情况之一。此外,这两种情况始终重叠:如果 0 ≤ x 和 x ≤ 0 同时成立,则根据反对称性(O2),x = 0,且 x = −x = 0,因此 |0| = 0。由此,我们得到一个定义良好的(单值)绝对值函数。下一个定理给出了该函数的基本性质。

8.43. Theorem on Absolute Value. For all x , y , r ∈ ℝ:

8.43. 绝对值定理。对于所有 x, y, r ∈ ℝ:

(a) 0 ≤ | x |, and | x | = 0 iff x = 0.

(a)0 ≤ |x|,且 |x| = 0 当且仅当 x = 0。

(b) | x | = | − x |.

(b)|x| = | − x|。

(c) | x · y | = | x | · | y |.

(c)|x · y| = |x| · |y|。

(d) | x | ≤ y iff − y x y (similarly for <).

(d)|x| ≤ y 当且仅当 −y ≤ x ≤ y(对于 < 也类似)。

(e) | x y | ≤ r iff y r x y + r (similarly for <).

(e)|x − y| ≤ r 当且仅当 y − r ≤ x ≤ y + r(对于 < 也类似)。

(f) − | x | ≤ x ≤ | x |.

(f)− |x| ≤ x ≤ |x|。

Proof. We prove (b), (d) and (f), leaving the other parts as exercises. For (b), fix x ∈ ℝ. We know 0 ≤ x or x ≤ 0, so consider two cases.

证明。我们证明了 (b)、(d) 和 (f),其余部分留作练习。对于 (b),固定 x ∈ ℝ。我们知道 0 ≤ x 或 x ≤ 0,因此考虑两种情况。

Case 1. Assume 0 ≤ x , so − x ≤ 0. Using the definition of absolute value twice, we see that | x | = x and | − x | = −( − x ) = x , so | x | = | − x |.

情况 1. 假设 0 ≤ x,则 −x ≤ 0。两次使用绝对值的定义,我们发现 |x| = x 且 | − x| = −( − x) = x,因此 |x| = | − x|。

Case 2. Assume x ≤ 0, so 0 ≤ − x . This time, the definition of absolute value gives | x | = − x and | − x | = − x , so | x | = | − x |.

情况 2. 假设 x ≤ 0,则 0 ≤ −x。此时,绝对值的定义给出 |x| = −x 且 | − x| = −x,因此 |x| = | − x|。

For (d), fix x , y ∈ ℝ. First, assume | x | ≤ y , and prove − y x and x y . By (a), 0 ≤ | x |, so 0 ≤ y by transitivity, and − y ≤ 0. In the case where 0 ≤ x , we know | x | = x , so x y by assumption, and moreover − y ≤ 0 ≤ x gives − y x by transitivity. In the case where x ≤ 0, we know | x | = − x , so the assumption becomes − x y , or − y x . Also x ≤ 0 ≤ y gives x y by transitivity. Thus, − y x y in both cases.

对于 (d),固定 x, y ∈ ℝ。首先,假设 |x| ≤ y,并证明 −y ≤ x 和 x ≤ y。由 (a) 可知,0 ≤ |x|,因此根据传递性,0 ≤ y,且 −y ≤ 0。当 0 ≤ x 时,我们知道 |x| = x,因此根据假设,x ≤ y,并且 −y ≤ 0 ≤ x 由传递性可知 −y ≤ x。当 x ≤ 0 时,我们知道 |x| = −x,因此假设变为 −x ≤ y,即 −y ≤ x。此外,x ≤ 0 ≤ y 由传递性可知 x ≤ y。因此,在两种情况下,−y ≤ x ≤ y 都成立。

Conversely, assume − y x y , and prove | x | ≤ y . In the case where 0 ≤ x , we know | x | = x , so we must prove x y . We have just assumed this. In the case where x ≤ 0, we know | x | = − x , so we must prove − x y . We get this from the assumption − y x by order reversal and the double negative rule.

反之,假设 −y ≤ x ≤ y,并证明 |x| ≤ y。当 0 ≤ x 时,我们知道 |x| = x,所以必须证明 x ≤ y。我们刚刚已经假设了这一点。当 x ≤ 0 时,我们知道 |x| = −x,所以必须证明 −x ≤ y。这可以通过将假设 −y ≤ x 的顺序颠倒以及双重否定规则得到。

Finally, (f) follows from (d) by taking y = | x | in (d). Since | x | ≤ | x | is known by reflexivity (O1), the equivalence in (d) lets us deduce that −| x | ≤ x ≤ | x |.      □

最后,(f) 由 (d) 可得,只需在 (d) 中令 y = |x| 即可。由于 |x| ≤ |x| 由自反性 (O1) 可知,(d) 中的等价关系可推导出 −|x| ≤ x ≤ |x|。□

Part (a) of the next theorem contains a key property of absolute value called the Triangle Inequality . The other parts of this theorem give variations of this inequality that are frequently used in analysis.

下一个定理的 (a) 部分包含了绝对值的一个关键性质,称为三角不等式。该定理的其他部分给出了该不等式的变体,这些变体在分析中经常用到。

8.44. Theorem on Triangle Inequality and Variations. For all a , b , x , y , z ∈ ℝ:

8.44. 三角不等式及其变分定理。对于所有 a, b, x, y, z ∈ ℝ:

(a) | a + b | ≤ | a | + | b |.

(a)|a + b| ≤ |a| + |b|。

(b) | x y | ≤ | x | + | y |.

(b)|x − y| ≤ |x| + |y|。

(c) | x + y + z | ≤ | x | + | y | + | z |.

(c)|x + y + z| ≤ |x| + |y| + |z|。

(d) | x y | ≥ | x | − | y |.

(d)|x − y| ≥ |x| − |y|。

(e) | x + y | ≥ | x | − | y |.

(e)|x + y| ≥ |x| − |y|。

(f) | x z | ≤ | x y | + | y z |.

(f)|x − z| ≤ |x − y| + |y − z|。

(g) || x | − | y || ≤ | x y |.

(g)||x| - |y|| ≤ |x − y|。

Proof. We prove (a), (e), and (f), leaving the other parts as exercises. For (a), fix a , b ∈ ℝ. Part (d) of the previous theorem (with x = a + b and y = | a | + | b |) tells us that the Triangle Inequality | a + b | ≤ | a | + | b | is logically equivalent to −(| a | + | b |) ≤ a + b ≤ | a | + | b |. Using part (f) of the previous theorem with x = a and then x = b , we know that −| a | ≤ a ≤ | a | and −| b | ≤ b ≤ | b |. Adding these inequalities and using algebra, we get −(| a | + | b |) ≤ a + b ≤ | a | + | b |, as needed.

证明。我们证明 (a)、(e) 和 (f),其余部分留作练习。对于 (a),固定 a, b ∈ ℝ。前一个定理的 (d) 部分(令 x = a + b 且 y = |a| + |b|)告诉我们,三角不等式 |a + b| ≤ |a| + |b| 在逻辑上等价于 −(|a| + |b|) ≤ a + b ≤ |a| + |b|。利用前一个定理的 (f) 部分(令 x = a 和 x = b),我们知道 −|a| ≤ a ≤ |a| 且 −|b| ≤ b ≤ |b|。将这些不等式相加并运用代数运算,我们得到 −(|a| + |b|) ≤ a + b ≤ |a| + |b|,满足要求。

To prove (e), fix x , y ∈ ℝ. The trick is to notice that x = ( x + y ) + ( − y ), so that part (a) (with a = x + y and b = − y ) gives | x | ≤ | x + y | + | − y | = | x + y | + | y |. Adding −| y | to both sides and rearranging gives | x + y | ≥ | x | − | y |, as needed.

为了证明 (e),固定 x, y ∈ ℝ。关键在于注意到 x = (x + y) + ( − y),因此 (a) 部分(其中 a = x + y,b = −y)可得 |x| ≤ |x + y| + | − y| = |x + y| + |y|。两边同时加上 −|y| 并重新整理,即可得到 |x + y| ≥ |x| − |y|,满足要求。

To prove (f), fix x , y , z ∈ ℝ. Take a = x y and b = y z in part (a), noting that a + b = x z , to see that | x z | ≤ | x y | + | y z |.     □

为了证明 (f),固定 x, y, z ∈ ℝ。在 (a) 部分中,取 a = x − y 和 b = y − z,注意到 a + b = x − z,即可看出 |x − z| ≤ |x − y| + |y − z|。□

Distance

距离

We can use absolute value to define the distance between any two real numbers, as follows.

我们可以使用绝对值来定义任意两个实数之间的距离,如下所示。

8.45. Definition: Distance in ℝ. For all x , y ∈ ℝ, define the distance from x to y to be d ( x , y ) = | x y |.

8.45. 定义:ℝ中的距离。对于所有x,y∈ℝ,定义从x到y的距离为d(x,y)=|x-y|。

The distance function satisfies the following properties.

距离函数满足以下性质。

8.46. Theorem on Distance. For all x , y , z ∈ ℝ,

8.46. 距离定理。对于所有 x, y, z ∈ ℝ,

(a) Positivity: d ( x , y ) ≥ 0, and d ( x , y ) = 0 iff x = y .

(a)正性:d(x, y) ≥ 0,且 d(x, y) = 0 当且仅当 x = y。

(b) Symmetry: d ( x , y ) = d ( y , x ).

(b)对称性:d(x, y) = d(y, x)。

(c) Triangle Inequality: d ( x , z ) ≤ d ( x , y ) + d ( y , z ).

(c)三角不等式:d(x, z) ≤ d(x, y) + d(y, z)。

Proof. These properties all follow readily from corresponding properties of absolute value. For example, given x , y ∈ ℝ, d ( x , y ) = | x y | = | − ( x y )| = | y x | = d ( y , x ).

证明。这些性质均可由绝对值的相应性质直接得出。例如,给定 x, y ∈ ℝ,d(x, y) = |x − y| = | − (x − y)| = |y − x| = d(y, x)。

In analysis, a set X together with a function d : X × X → ℝ satisfying the three conditions in the last theorem (for all x , y , z X ) metric space is called a metric space . The real number system ℝ provides one of the most basic examples of a metric space. More generally, one can show that ℝ n (with the Euclidean distance function d ( x , y ) = i = 1 n ( x i y i ) 2 ) is a metric space, although the proof of the Triangle Inequality is harder.

在分析学中,集合 X 以及满足最后一个定理中三个条件(对于所有 x, y, z ∈ X)的函数 d: X × X → ℝ 被称为度量空间。实数系 ℝ 提供了一个最基本度量空间的例子。更一般地,可以证明 ℝ n (具有欧几里得距离函数 d(x→,y→)=∑i=1n(xi−yi)2)是一个度量空间,尽管三角不等式的证明比较困难。

Section Summary

章节概要

  1. Ordering Properties. The relation < on ℝ satisfies irreflexivity, transitivity, and trichotomy; for x , y ∈ ℝ, x < y iff x y and x y ; x y iff x < y or x = y ; ∼( x y ) iff x > y ; x is positive iff − x is negative; ℝ >0 is closed under addition, multiplication, and multiplicative inversion; 0 < x < y iff 0 < y −1 < x −1 ; and so on.
    排序性质。ℝ 上的关系 < 满足非自反性、传递性和三分性;对于 x, y ∈ ℝ,x < y 当且仅当 x ≤ y 且 x ≠ y;x ≤ y 当且仅当 x < y 或 x = y;∼(x ≤ y) 当且仅当 x > y;x 为正当且仅当 −x 为负;ℝ >0 对加法、乘法和乘法逆元封闭;0 < x < y 当且仅当 0 < y −1 < x −1 ;等等。
  2. Properties Relating Order and Algebra. Inequalities are preserved when we add a constant or multiply by a positive constant; they are reversed when we multiply by a negative constant. Nonzero squares are strictly positive, so 0 < 1.
    序与代数的关系。不等式在加上常数或乘以正常数时保持不变;乘以负常数时,不等式方向改变。非零平方项严格为正,因此 0 < 1。
  3. Ordering of Integers. Every n ∈ ℕ satisfies n ≥ 1, whereas every m ∈ ℤ − ℕ satisfies m ≤ 0. So ℕ = ℤ >0 ; there are no integers k , n satisfying k < n < k + 1; and for all m , n ∈ ℤ, m n + 1 iff m n or m = n + 1.
    整数的排序。每个 n ∈ ℕ 满足 n ≥ 1,而每个 m ∈ ℤ − ℕ 满足 m ≤ 0。因此 ℕ = ℤ >0 ;不存在整数 k, n 满足 k < n < k + 1;并且对于所有 m, n ∈ ℤ,m ≤ n + 1 当且仅当 m ≤ n 或 m = n + 1。
  4. Absolute Value. For all real x , if x ≥ 0, then | x | = x ; if x ≤ 0, then | x | = − x . For x , y , r ∈ ℝ, 0 ≤ | x |, | x | = | − x |, | xy | = | x | · | y |, and | x y | ≤ r iff y r x y + r . The Triangle Inequality says that | a + b | ≤ | a | + | b | for all a , b ∈ ℝ.
    绝对值。对于所有实数 x,如果 x ≥ 0,则 |x| = x;如果 x ≤ 0,则 |x| = −x。对于 x, y, r ∈ ℝ,0 ≤ |x|,|x| = | − x|,|xy| = |x| · |y|,且 |x − y| ≤ r 当且仅当 y − r ≤ x ≤ y + r。三角不等式指出,对于所有 a, b ∈ ℝ,|a + b| ≤ |a| + |b|。
  5. Distance. For x , y ∈ ℝ, the distance from x to y is d ( x , y ) = | x y |. Distance satisfies positivity, symmetry, and the Triangle Inequality.
    距离。对于 x, y ∈ ℝ,x 到 y 的距离为 d(x, y) = |x − y|。距离满足正性、对称性和三角不等式。

Exercises

练习

  1. Finish the proof of Theorem 8.36 .
    完成定理 8.36 的证明。
  2. Finish the proof of Corollary 8.37 .
    完成推论 8.37 的证明。
  3. Finish the proof of Theorem 8.38 .
    完成定理 8.38 的证明。
  4. (a) Finish the proof of Theorem 8.40 . (b) State and prove analogous properties for negative numbers, nonnegative numbers, and nonpositive numbers.
    (a)完成定理 8.40 的证明。(b)陈述并证明负数、非负数和非正数的类似性质。
  5. Finish the proof of Theorem 8.43 .
    完成定理 8.43 的证明。
  6. Finish the proofs of Theorem 8.44 and Theorem 8.46 .
    完成定理 8.44 和定理 8.46 的证明。
  7. Disprove: for all a , b ∈ ℝ, | a b | ≤ | a | − | b |.
    反证:对于所有 a, b ∈ ℝ,|a − b| ≤ |a| − |b|。
  8. For all m , n ∈ ℤ, prove: (a) m < n + 1 iff m n ; (b) m n − 1 iff m n or m = n − 1; (c) m > n − 1 iff m n . (d) Prove these results do not always hold when m , n ∈ ℝ.
    对于所有 m, n ∈ ℤ,证明:(a) m < n + 1 当且仅当 m ≤ n;(b) m ≥ n − 1 当且仅当 m ≥ n 或 m = n − 1;(c) m > n − 1 当且仅当 m ≥ n。(d) 证明当 m, n ∈ ℝ 时,这些结果并非总是成立。
  9. (a) Assume < is a relation on ℝ satisfying transitivity and trichotomy (properties (b) and (d) in Theorem 8.36 ). For x , y ∈ ℝ, define x y to mean x < y or x = y . Prove that ≤ satisfies axioms O1, O2, O3, and O4. (b) Assume < satisfies axioms O5 and O6. Prove that ≤ as defined in part (a) satisfies axioms O5 and O6.
    (a) 假设 < 是 ℝ 上的一个关系,满足传递性和三分性(定理 8.36 中的性质 (b) 和 (d))。对于 x, y ∈ ℝ,定义 x ≤ y 表示 x < y 或 x = y。证明 ≤ 满足公理 O1、O2、O3 和 O4。(b) 假设 < 满足公理 O5 和 O6。证明 (a) 部分定义的 ≤ 满足公理 O5 和 O6。
  10. In some developments of ℝ, the set ℝ >0 of positive real numbers is taken as an undefined concept. Axioms O1 through O6 are replaced by these axioms: (P1) For all x , y ∈ ℝ >0 , x + y ∈ ℝ >0 . (P2) For all x , y ∈ ℝ >0 , x · y ∈ ℝ >0 . (P3) For all x ∈ ℝ, exactly one of these statements is true: x ∈ ℝ >0 ; x = 0; − x ∈ ℝ >0 . For x , y ∈ ℝ, x < y is defined to mean y x ∈ ℝ >0 , and x y is defined to mean x < y or x = y . Show that axioms O1 through O6 follow from these definitions and axioms P1, P2, P3. [Use the previous problem.]
    在某些 ℝ 的演化中,正实数集 ℝ >0 被视为未定义概念。公理 O1 至 O6 被以下公理所取代:(P1) 对于所有 x, y ∈ ℝ >0 ,x + y ∈ ℝ >0 。(P2) 对于所有 x, y ∈ ℝ >0 ,x · y ∈ ℝ >0 。(P3) 对于所有 x ∈ ℝ,以下陈述中恰好有一个为真:x ∈ ℝ >0 ;x = 0;−x ∈ ℝ >0 。对于 x, y ∈ ℝ,x < y 定义为 y − x ∈ ℝ >0 ,x ≤ y 定义为 x < y 或 x = y。证明公理 O1 至 O6 可由这些定义以及公理 P1、P2、P3 推导得出。[利用上一题。]
  11. Define the set of complex numbers ℂ to be the product set ℝ × ℝ. Define algebraic operations on complex numbers as follows: for all x , y , s , t ∈ ℝ, let
    ( x , y ) + ( s , t ) = ( x + s , y + t ) , ( x , y ) = ( x , y ) , 0 C = ( 0 , 0 ) ,
    1 C = ( 1 , 0 ) , ( x , y ) ( s , t ) = ( x s y t , x t + y s ) ,
    ( x , y ) 1 = ( x x 2 + y 2 , y x 2 + y 2 ) for ( x , y ) 0 C .
    Use known algebraic properties of ℝ to show that ℂ satisfies the field axioms (A1 through A5, M1 through M5, D, and N). It follows that all algebraic rules proved for ℝ from these axioms also hold in ℂ.
    定义复数集 ℂ 为 ℝ × ℝ 的乘积集。定义复数上的代数运算如下:对于任意 x, y, s, t ∈ ℝ,令 (x,y)+(s,t)=(x+s,y+t),−(x,y)=(−x,−y),0C=(0,0),1C=(1,0),(x,y)⋅(s,t)=(xs−yt,xt+ys),(x,y)−1=(xx²+y²,−yx²+y²),其中 (x,y)≠0C。利用 ℝ 的已知代数性质证明 ℂ 满足域公理(A1 至 A5、M1 至 M5、D 和 N)。由此可知,所有基于这些公理证明的 ℝ 的代数规则也适用于 ℂ。
  12. (a) Show that the equation z 2 = −1 has exactly two solutions z ∈ ℂ. (b) Hence show that there does not exist an ordering relation ≤ on the field ℂ satisfying axioms O1 through O6.
    (a)证明方程 z 2 = −1 恰好有两个解 z ∈ ℂ。(b)由此证明域 ℂ 上不存在满足公理 O1 到 O6 的排序关系 ≤。
  13. Show that for p prime, there does not exist on ordering relation ≤ on the field ℤ p satisfying axioms O1 through O6. (See the exercises of § 8.3 for the definition of ℤ p and the algebraic operations on this set.)
    证明对于素数 p,域 ℤ p 上不存在满足公理 O1 至 O6 的单序关系 ≤。(参见 §8.3 的练习,了解 ℤ p 的定义以及该集合上的代数运算。)
  14. Prove that ℤ is not a field.
    证明 ℤ 不是一个域。

8.5  Greatest Elements, Least Upper Bounds, and Completeness

8.5 最大元素、最小上界和完备性

Every nonempty finite subset of ℝ has a greatest element and a least element; this can be proved by induction from the total ordering axiom (O4). However, an infinite subset of ℝ may or may not have maximum and minimum elements. This leads to the concept of a least upper bound for a set of real numbers, which generalizes the idea of the maximum element of a set. The Completeness Axiom (O7) guarantees that least upper bounds exist whenever possible. This axiom has many striking consequences about the structure of the real number system, and it is absolutely fundamental for proving the major theorems of calculus.

ℝ 的每个非空有限子集都存在最大元素和最小元素;这可以通过归纳法从全序公理 (O4) 证明。然而,ℝ 的无限子集可能存在也可能不存在最大元素和最小元素。这引出了实数集的最小上界的概念,它推广了集合最大元素的概念。完备性公理 (O7) 保证了最小上界在任何可能的情况下都存在。该公理对实数系统的结构有着许多显著的影响,并且对于证明微积分的主要定理至关重要。

After discussing the basic definitions, we use completeness to prove some familiar but technically subtle facts about how the number systems ℤ and ℚ appear within ℝ. In particular, we show that given any real number there is a larger integer (the Archimedean Property); any half-open interval [ x , x + 1) of length 1 contains exactly one integer; and every open interval ( a , b ) contains a rational number. We also prove some strong versions of completeness that apply to sets of integers, such as the Well-Ordering Principle and the fact that if a nonempty set of integers is bounded above, then that set has a greatest element (not just a least upper bound). Finally, we use completeness to prove the Nested Interval Property of real numbers and the fact that every positive real number has a real square root.

在讨论了基本定义之后,我们利用完备性证明了一些关于数系ℤ和ℚ如何在ℝ中出现的常见但技术上较为微妙的事实。具体来说,我们证明了对于任意实数,都存在一个更大的整数(阿基米德性质);任意长度为1的半开区间[x, x + 1]恰好包含一个整数;并且每个开区间(a, b)都包含一个有理数。我们还证明了一些适用于整数集合的更强的完备性,例如良序原理以及如果一个非空整数集合有上界,则该集合存在最大元素(而不仅仅是最小上界)。最后,我们利用完备性证明了实数的嵌套区间性质以及每个正实数都存在实数平方根。

Maximum and Minimum Elements

最大值和最小值

We recall the definitions of the maximum element and minimum element of a set; these definitions make sense in any partially ordered set.

我们回顾一下集合中最大元素和最小元素的定义;这些定义在任何偏序集中都有意义。

8.47. Definition: Maximum and Minimum Elements. Fix S ⊆ℝ and z 0 ∈ ℝ.

8.47. 定义:最大元素和最小元素。固定 S⊆ℝ 和 z 0 ∈ ℝ。

(a) z 0 = max S iff z 0 S and x S , x z 0 .

(a)z0=maxS 当且仅当 z0∈S 且∀x∈S,x≤z0。

(b) S has a greatest element iff z S , x S , x z .

(b)存在最大元素当且仅当存在 z∈S,∀x∈S,x≤z。

(c) z 0 = min S iff z 0 S and x S , z 0 x .

(c)z0=minS 当且仅当 z0∈S 且∀x∈S,z0≤x。

(d) S has a least element iff z S , x S , z x .

(d)存在最小元素当且仅当存在 z∈S,∀x∈S,z≤x。

We stress that max S and min S need not exist; see the examples below. But if S has a greatest element, that element is unique. For suppose y , z ∈ ℝ are both greatest elements of S . Since y is a greatest element and z S , we deduce z y . Since z is a greatest element and y S , we deduce y z . By antisymmetry (axiom O2), y = z . Similarly, min S is unique when it exists.

我们强调,最大集合 S 和最小集合 S 不一定存在;参见以下示例。但如果 S 存在最大元素,则该元素是唯一的。例如,假设 y, z ∈ ℝ 都是 S 的最大元素。由于 y 是最大元素且 z ∈ S,我们得出 z ≤ y。由于 z 是最大元素且 y ∈ S,我们得出 y ≤ z。根据反对称性(公理 O2),y = z。类似地,当最小集合 S 存在时,它也是唯一的。

8.48. Example: Intervals. Fix a < b in ℝ. The closed interval S = [ a , b ] = { x ∈ ℝ: a x b } has greatest element b = max S and least element a = min S . On the other hand, we claim the open interval T = ( a , b ) = { x ∈ ℝ: a < x < b } has no greatest element and no least element. To prove that T has no maximum, we negate the definition and prove z T , x T , z < x . Fix z T , so a < z < b . Choose x = ( z + b )/2. You can check that a < z < ( z + b )/2 < b , so z < x and a < x < b . Thus, x is a member of T larger than z . A similar proof shows that T has no least element. Observe, in particular, that a cannot be the least element of T since a does not belong to T .

8.48. 例:区间。在 ℝ 中固定 a < b。闭区间 S = [a, b] = {x ∈ ℝ: a ≤ x ≤ b} 的最大元素为 b = max S,最小元素为 a = min S。另一方面,我们断言开区间 T = (a, b) = {x ∈ ℝ: a < x < b} 没有最大元素也没有最小元素。为了证明 T 没有最大值,我们否定定义,并证明 ∀z∈T, ∃x∈T, z<x。固定 z ∈ T,则 a < z < b。取 x = (z + b)/2。可以验证 a < z < (z + b)/2 < b,所以 z < x 且 a < x < b。因此,x 是 T 中大于 z 的元素。类似的证明表明 T 没有最小元素。特别地,a 不可能是 T 的最小元素,因为 a 不属于 T。

8.49. Example. Every nonempty finite set { x 1 , …, x n }⊆ℝ has a maximum and minimum element, as can be proved by induction on n . However, the empty set has no maximum or minimum element, since the empty set has no elements in it at all.

8.49. 例。每个非空有限集 {x 1 , …, x n ⊆ℝ 都有最大元素和最小元素,这可以通过对 n 进行归纳证明。然而,空集 ∅ 没有最大元素或最小元素,因为空集根本不包含任何元素。

8.50. Example. The entire set ℝ has no least element and no greatest element. For, given any x ∈ ℝ, adding x to the inequalities −1 < 0 < 1 shows that x − 1 < x < x + 1, so there exist elements of ℝ greater than x and less than x . The same proof shows that ℤ has no least element and no greatest element. On the other hand, ℝ ≥0 has least element 0, and ℕ = ℤ >0 has least element 1, by Lemma 8.41 (a).

8.50. 示例。整个集合 ℝ 没有最小元素也没有最大元素。对于任意 x ∈ ℝ,将 x 添加到不等式 −1 < 0 < 1 中,可知 x − 1 < x < x + 1,因此 ℝ 中存在大于 x 和小于 x 的元素。同样的证明也表明 ℤ 没有最小元素也没有最大元素。另一方面,根据引理 8.41(a),ℝ ≥0 的最小元素为 0,ℕ = ℤ >0 的最小元素为 1。

We just saw that the set ℕ of natural numbers has a least element. In fact, ℕ satisfies a much stronger condition called the well-ordering property: every nonempty subset of ℕ has a least element. We use the following terminology in the proof: for each n ∈ ℕ, define [ n ] = { m ∈ ℕ: m n }. We proved earlier that every natural number m satisfies 1 ≤ m , so we can also write [ n ] = { m ∈ ℕ: 1 ≤ m n }. It follows from Lemma 8.41 (e) that [ n + 1 ] = [ n ] { n + 1 } for all n ∈ ℕ.

我们刚刚看到自然数集ℕ存在最小元素。事实上,ℕ满足一个更强的条件,称为良序性质:ℕ的每个非空子集都存在最小元素。我们在证明中使用以下术语:对于每个n∈ℕ,定义[n] = {m ∈ ℕ: m ≤ n}。我们之前证明了每个自然数m都满足1 ≤ m,因此我们也可以写成[n] = {m ∈ ℕ: 1 ≤ m ≤ n}。由引理8.41(e)可知,对于所有n∈ℕ,[n+1]=[n]∪{n+1}。

8.51. Theorem on Well-Ordering of ℕ. For all S ⊆ℕ, if S , then S has a least element.

8.51. ℕ 的良序定理。对于所有 S⊆ℕ,如果 S≠∅,则 S 有最小元素。

Proof. First we prove the following statement by induction on n : for all S ⊆ℕ, if S [ n ] is nonempty then S has a least element. For the base case, consider n = 1. Fix S ⊆ℕ and assume S [ 1 ] is nonempty. Now, [1] = { m ∈ ℕ: 1 ≤ m ≤ 1} = {1} by antisymmetry. The assumption S [ 1 ] forces 1 ∈ S . For any k S , we know k ∈ ℕ, so 1 ≤ k . Thus, 1 is the least element of S in this case.

证明。首先,我们对 n 进行归纳证明以下命题:对于所有 S⊆ℕ,如果 S∩[n] 非空,则 S 存在最小元素。对于基本情况,考虑 n = 1。固定 S⊆ℕ 并假设 S∩[1] 非空。现在,根据反对称性,[1] = {m ∈ ℕ: 1 ≤ m ≤ 1} = {1}。假设 S∩[1]≠∅ 迫使 1 ∈ S。对于任意 k ∈ S,我们知道 k ∈ ℕ,因此 1 ≤ k。因此,在这种情况下,1 是 S 的最小元素。

For the induction step, fix n ∈ ℕ and assume: for all S ⊆ℕ, if S [ n ] then S has a least element. We must prove: for all S ⊆ℕ, if S [ n + 1 ] , then S has a least element. Fix S ⊆ℕ satisfying S [ n + 1 ] . Since [ n + 1 ] = [ n ] { n + 1 } , there are two cases.

在归纳步骤中,固定 n ∈ ℕ 并假设:对于所有 S⊆ℕ,如果 S∩[n]≠∅,则 S 存在最小元素。我们需要证明:对于所有 S⊆ℕ,如果 S∩[n+1]≠∅,则 S 存在最小元素。固定满足 S∩[n+1]≠∅ 的 S⊆ℕ。由于 [n+1]=[n]∪{n+1},因此存在两种情况。

Case 1. Assume S [ n ] . By induction hypothesis, S has a least element.

情况 1. 假设 S∩[n]≠∅。根据归纳假设,S 有最小元素。

Case 2. Assume S [ n ] = , which forces n + 1 ∈ S . We claim n + 1 is the least element of S . To see this, fix k S . By trichotomy, we know n + 1 ≤ k or k < n + 1. The second alternative k < n + 1 is impossible, since it would imply k S [ n + 1 ] = { n + 1 } , leading to the contradiction k = n + 1. Thus, n + 1 ≤ k for each k S , as needed.

情况 2. 假设 S∩[n]=∅,这迫使 n + 1 ∈ S。我们断言 n + 1 是 S 的最小元素。为了证明这一点,固定 k ∈ S。根据三分性,我们知道 n + 1 ≤ k 或 k < n + 1。第二种情况 k < n + 1 是不可能的,因为它意味着 k∈S∩[n+1]={n+1},从而导致矛盾 k = n + 1。因此,对于每个 k ∈ S,都有 n + 1 ≤ k,符合要求。

To prove the theorem itself, fix S ⊆ℕ and assume S is nonempty. Then there exists a natural number n S . It follows that S [ n ] is nonempty since n ∈ [ n ]. We can now conclude that S has a least element.

为了证明定理本身,固定 S⊆ℕ 并假设 S 非空。那么存在一个自然数 n ∈ S。由于 n ∈ [n],因此 S∩[n] 非空。由此我们可以得出结论:S 存在最小元素。

Least Upper Bounds and Completeness

The concept of a least upper bound is a kind of substitute for the greatest element when the latter does not exist. For example, the open interval ( a , b ) has least upper bound b . To make this precise, we need some additional definitions.

8.52. Definition: Upper and Lower Bounds. Fix S ⊆ℝ and z 0 ∈ ℝ.

(a) z 0 is an upper bound for S iff x S , x z 0 .

(b) S is bounded above iff z R , x S , x z .

(d) z 0 is a lower bound for S iff x S , z 0 x .

(d) S is bounded below iff z R , x S , z x .

As an example, consider the interval S = [2, 4). Some upper bounds for S are 5, 7, 21/2, and 4. In fact, the set of all upper bounds of S is the interval [4, ∞), and this set of upper bounds has least element 4. We therefore say 4 is the least upper bound (lub) of S and write 4 = lub S . Similarly, the set of all lower bounds of S is the interval ( − ∞, 2]. This set of lower bounds has greatest element 2, so we say 2 is the greatest lower bound (glb) of S .

Here is a related example that motivates the introduction of the Completeness Axiom. Consider the set S = { x ∈ ℝ ≥0 : x 2 < 2}. Intuitively, for positive real x , the condition x 2 < 2 is equivalent to x < 2 , so the set of upper bounds for S ought to be the interval [ 2 , ) . (This is only an intuitive argument for now, since we have not yet defined or proved the existence of 2 .) Thus lub S should exist and be equal to 2 . On the other hand, suppose we restrict ourselves to using rational numbers only. Intuitively, the restricted set S ′ = { x ∈ ℚ ≥0 : x 2 < 2} has set of upper bounds [ 2 , ) Q . This set of upper bounds has no least element since 2 is not rational, but there are rational numbers arbitrarily close to 2 and larger than 2 . Thus lub S ′ does not exist in the ordered set ℚ, even though S ′ is bounded above. This example reveals that the number system ℚ has a “hole” in it where 2 should be. The Completeness Axiom for ℝ says, intuitively, that ℝ “has no holes” in the sense that least upper bounds exist in every case where this is possible. We now officially define lub S and glb S and state the Completeness Axiom.

8.53. Definition: Least Upper Bounds. Fix S ⊆ℝ and z 0 ∈ ℝ. Define z 0 = lub S iff z 0 is an upper bound of S and for every upper bound y of S , z 0 y .

Some texts denote lub S by sup S (the supremum of S ).

8.54. Definition: Greatest Lower Bounds. Fix S ⊆ℝ and z 0 ∈ ℝ. Define z 0 = glb S iff z 0 is a lower bound of S and for every lower bound y of S , y z 0 .

Some texts denote glb S by inf S (the infimum of S ).

Expanding the definition of a least upper bound, we see that

z 0 = lub S iff [ x S , x z 0 and y R , ( x S , x y ) z 0 y ] .
Taking the contrapositive of the final implication and using trichotomy, we arrive at the following phrasing of the definition that is often more convenient in proofs:
z 0 = lub S [ x S , x z 0 ] [ y R , y < z 0 x S , y < x ] .
In words, z 0 is the least upper bound of S iff z 0 is an upper bound of S and each y < z 0 is not an upper bound of S . Similarly,
z 0 = glb S [ x S , z 0 x ] [ y R , z 0 < y x S , x < y ] .

Note that lub S is the minimum element of the set of upper bounds of S , and glb S is the maximum element of the set of lower bounds of S . Thus, lub S and glb S are unique when they exist. The Completeness Axiom postulates a basic sufficient condition for the existence of lub S .

8.55. Completeness Axiom (O7). Every nonempty subset of ℝ that is bounded above has a least upper bound in ℝ.

As a first illustration of this axiom, we use it to derive the analogous sufficient condition for the existence of greatest lower bounds.

8.56. Theorem on Greatest Lower Bounds. For all S ⊆ℝ, if S is nonempty and bounded below, then S has a greatest lower bound in ℝ.

Proof. Fix S ⊆ℝ that is nonempty and bounded below by x 0 . [The main idea of the proof is to use negation to flip the ordering, turning lower bounds into upper bounds and least upper bounds into greatest lower bounds.] Define a new set T = { − y : y S }, which is nonempty since S is nonempty. For all y S , we know x 0 y , so − y ≤ − x 0 . This means that − x 0 is an upper bound for the set T . By the Completeness Axiom (O7), z 0 = lub T exists in ℝ. We show that − z 0 = glb S . On one hand, for fixed y S , we know − y T , so − y z 0 and hence − z 0 y ; this proves that − z 0 is a lower bound for the set S . On the other hand, consider a fixed real w 0 > − z 0 . Then − w 0 < z 0 , so − w 0 is not an upper bound for T ; so there exists t T with − w 0 < t . Writing t = − y for some y S , we have − w 0 < − y , so y < w 0 , so w 0 is not a lower bound for S . This means that − z 0 is the greatest lower bound for S , as needed.     □

最小上界与完备性 最小上界的概念是最大元素不存在时的一种替代。例如,开区间 (a, b) 的最小上界为 b。为了更精确地表述,我们需要一些额外的定义。8.52. 定义:上界和下界。固定 S⊆ℝ 和 z 0 ∈ ℝ。(a) z0 是 S 的上界当且仅当对于所有 x∈S,x≤z0。(b) S 有上界当且仅当对于所有 x∈S,x≤z∈R。(c) z0 是 S 的下界当且仅当对于所有 x∈S,z0≤x。(d) S 有下界当且仅当对于所有 x∈S,z≤x。例如,考虑区间 S = [2, 4)。集合 S 的一些上界为 5、7、21/2 和 4。事实上,S 的所有上界构成的集合是区间 [4, ∞),且该集合的最小元素为 4。因此,我们称 4 为 S 的最小上界 (lub),记为 4 = lubS。类似地,S 的所有下界构成的集合是区间 (−∞, 2)。该集合的最大元素为 2,因此我们称 2 为 S 的最大下界 (glb)。以下是一个相关的例子,它启发了完备性公理的引入。考虑集合 S = {x ∈ ℝ ≥0 :x 2 < 2}。直观地,对于正实数 x,条件 x 2 < 2 等价于 x<2,因此 S 的上界构成的集合应该是区间 [2,∞)。 (目前这只是一个直观的论证,因为我们尚未定义或证明 2 的存在性。)因此,lubS 应该存在且等于 2。另一方面,假设我们仅使用有理数。直观地,受限集 S′ = {x ∈ ℚ ≥0 :x 2 < 2} 的上界集为 [2,∞)∩Q。由于 2 不是有理数,因此该上界集没有最小元素,但存在任意接近 2 且大于 2 的有理数。因此,即使 S′ 有上界,lubS′ 在有序集 ℚ 中也不存在。这个例子表明,数系 ℚ 中存在一个“空洞”,2 本应位于其中。ℝ 的完备性公理直观地指出,ℝ“没有空洞”,因为在所有可能的情况下,最小上界都存在。我们现在正式定义 lubS 和 glbS,并陈述完备性公理。8.53. 定义:最小上界。固定 S⊆ℝ 和 z 0 ∈ ℝ。定义 z0=lubS 当且仅当 z 0 是 S 的一个上界,且对于 S 的每个上界 y,都有 z 0 ≤ y。有些文献将 lubS 表示为 sup S(S 的上确界)。8.54. 定义:最大下界。固定 S⊆ℝ 和 z 0 ∈ ℝ。定义 z0=glbS 当且仅当 z 0 是 S 的一个下界,且对于 S 的每个下界 y,都有 y ≤ z 0 。有些文献用 inf S(S 的下确界)表示 glbS。展开最小上界的定义,我们看到 z0=lubS 当且仅当[∀x∈S,x≤z0且∀y∈R,(∀x∈S,x≤y)⇒z0≤y]。对最后一个蕴含式取逆否命题,并利用三分法,我们得到以下定义表述,这在证明中通常更方便:z0=lubS⇔[∀x∈S,x≤z0]∧[∀y∈R,y<z0⇒∃x∈S,y<x]。换句话说,z 0 是 S 的最小上界当且仅当 z 0 是 S 的上界,且每个 y < z 0 都不是 S 的上界。类似地,z0=glbS⇔[∀x∈S,z0≤x]∧[∀y∈R,z0<y⇒∃x∈S,x<y]。注意,lubS 是 S 的上界集合的最小元素,而 glbS 是 S 的下界集合的最大值。因此,当 lubS 和 glbS 存在时,它们是唯一的。完备性公理假定了 lubS 存在的基本充分条件。8.55. 完备性公理 (O7)。ℝ 的每个非空且有上界的子集在 ℝ 中都有最小上界。作为该公理的第一个例证,我们用它来推导最大下界存在的类似充分条件。8.56. 最大下界定理。对于所有 S⊆ℝ,如果 S 非空且有下界,则 S 在 ℝ 中存在最大下界。证明。固定 S⊆ℝ,使其非空且下界为 x 0 。[证明的主要思想是利用否定来颠倒顺序,将下界转化为上界,将最小上界转化为最大下界。] 定义一个新的集合 T = { − y: y ∈ S},由于 S 非空,因此 T 非空。对于所有 y ∈ S,我们知道 x 0 ≤ y,所以 −y ≤ −x 0 。这意味着 −x 0 是集合 T 的一个上界。根据完备性公理 (O7),z 0 = lubT 存在于 ℝ 中。我们证明 −z 0 = glbS。一方面,对于固定的 y ∈ S,我们知道 −y ∈ T,因此 −y ≤ z 0 ,进而 −z 0 ≤ y;这证明了 −z 0 是集合 S 的一个下界。另一方面,考虑一个固定的实数 w 0 > −z 0 。那么 −w 0 < z 0 ,因此 −w 0 不是 T 的一个上界;因此存在 t ∈ T 使得 −w 0 < t。令 t = −y,其中 y ∈ S,则有 −w 0 < −y,所以 y < w 0 ,因此 w 0 不是 S 的下界。这意味着 −z 0 是 S 的最大下界,符合要求。□

The Archimedean Ordering Property

阿基米德排序性质

We often visualize ℝ as a number line extending infinitely far to the left and right. Each real number x corresponds to exactly one point on this line, and the order relation x < y means that the point for x is located to the left of the point for y . Our intuition tells us that no matter how far to the right we go on this line, there will always be positive integers even farther to the right. The next theorem uses the Completeness Axiom to justify this pictorial intuition.

我们通常将 ℝ 想象成一条向左右无限延伸的数轴。每个实数 x 都对应这条数轴上的唯一一个点,而序关系 x < y 表示 x 对应的点位于 y 对应的点的左侧。我们的直觉告诉我们,无论沿着这条数轴向右延伸多远,总会有更靠右的正整数。下一个定理将利用完备性公理来证明这种图像上的直觉。

8.57. Theorem on Archimedean Property of ℝ.

8.57. ℝ 的阿基米德性质定理。

(a) x R , n Z > 0 , x < n .

(a)∀x∈R,∃n∈Z>0,x<n。

(b) y R > 0 , n Z > 0 , 1 / n < y .

(b)∀y∈R>0,∃n∈Z>0,1/n<y。

Proof. We prove (a) by contradiction. Assume, to get a contradiction, that x R , n Z > 0 , n x . This assumption is precisely the statement that the set ℤ >0 is bounded above. By the completeness axiom for ℝ, the set ℤ >0 has a least upper bound, say x 0 = lubℤ >0 . Now x 0 − 1 < x 0 , so x 0 − 1 is not an upper bound for ℤ >0 . Thus there must exist a positive integer n with x 0 − 1 < n . Adding 1 to both sides, we get x 0 < n + 1, where n + 1 is also a positive integer. Since x 0 is an upper bound for ℤ >0 , we have n + 1 ≤ x 0 . But now x 0 < n + 1 ≤ x 0 contradicts trichotomy. This proves (a).

证明。我们用反证法证明 (a)。假设存在 x∈R,∀n∈Z>0,n≤x,从而得到矛盾。这个假设恰恰等于集合ℤ >0 有上界。根据 ℝ 的完备性公理,集合ℤ >0 存在一个最小上界,记为 x 0 = lubℤ >0 。现在 x 0 − 1 < x 0 ,所以 x 0 − 1 不是 ℤ >0 的上界。因此,必定存在一个正整数 n,使得 x 0 − 1 < n。两边同时加 1,得到 x 0 < n + 1,其中 n + 1 也是一个正整数。由于 x 0 是 ℤ >0 的一个上界,所以 n + 1 ≤ x 0 。但是现在 x 0 < n + 1 ≤ x 0 与三分性矛盾。这就证明了 (a)。

To prove (b), fix y ∈ ℝ >0 . Then 1/ y is a positive real number, and by (a) there exists a positive integer n with 1/ y < n . By Theorem 8.40 (a), we conclude that 1/ n < 1/(1/ y ) = y , as needed.

为了证明 (b),固定 y ∈ ℝ >0 。则 1/y 是一个正实数,并且由 (a) 可知存在一个正整数 n 使得 1/y < n。由定理 8.40(a) 可知,1/n < 1/(1/y) = y,满足要求。

Ordering Properties of ℤ

ℤ 的排序性质

The Completeness Axiom tells us that any subset S ⊆ℝ that is nonempty and bounded above has a least upper bound. If S happens to be a subset of ℤ, we can show that a stronger result holds: S actually has a maximum element. Similarly, if S ⊆ℤ is nonempty and bounded below, then S has a minimum element. To prove these statements, we reduce to the well-ordering property of ℕ established earlier.

完备性公理告诉我们,任何非空且有上界的子集 S⊆ℝ 都存在最小上界。如果 S 恰好是 ℤ 的子集,我们可以证明一个更强的结论成立:S 实际上存在最大元素。类似地,如果 S⊆ℤ 非空且有下界,那么 S 存在最小元素。为了证明这些结论,我们将其简化为之前已建立的 ℕ 的良序性质。

8.58. Theorem on Strong Completeness of ℤ. (a) For all S ⊆ℤ, if S is nonempty and S is bounded below (in ℝ) then S has a least element. (b) For all S ⊆ℤ, if S is nonempty and S is bounded above (in ℝ) then S has a greatest element.

8.58. ℤ 的强完备性定理。(a) 对于所有 S⊆ℤ,如果 S 非空且 S 在 ℝ 中有下界,则 S 有最小元素。(b) 对于所有 S⊆ℤ,如果 S 非空且 S 在 ℝ 中有上界,则 S 有最大元素。

Proof. For (a), fix S ⊆ℤ and assume S is nonempty and bounded below. Let x ∈ ℝ be a lower bound for S . By the Archimedean Property, there is an integer N with − x < N , so − N < x . [The main idea of this proof is to translate S ⊆ℤ to the right N units to get a subset of ℕ, thus reducing the result in (a) to the Well-Ordering Principle.] Consider the set T = { m + N : m S } obtained from S by adding N to each element. Given any element k = m + N T with m S , note that − N < x m , so 0 < m + N = k , so k ∈ ℤ >0 = ℕ. So T is a subset of ℕ, and T is nonempty since S is nonempty. Since ℕ is well-ordered, T has a least element k 0 = m 0 + N , where m 0 S . For any m S , we have m + N T . The fact that m 0 + N = k 0 m + N implies m 0 m . Thus, m 0 is the least element of S .

证明。对于 (a),固定 S⊆ℤ,并假设 S 非空且有下界。设 x ∈ ℝ 为 S 的一个下界。根据阿基米德性质,存在整数 N 使得 −x < N,因此 −N < x。[此证明的主要思想是将 S⊆ℤ 向右平移 N 个单位,得到 ℕ 的一个子集,从而将 (a) 中的结果简化为良序原理。] 考虑集合 T = {m + N: m ∈ S},它是通过将 N 加到 S 的每个元素上得到的。对于任意元素 k = m + N ∈ T,其中 m ∈ S,注意到 −N < x ≤ m,因此 0 < m + N = k,所以 k ∈ ℤ >0 = ℕ。因此 T 是 ℕ 的一个子集,并且由于 S 非空,T 也非空。由于 ℕ 是良序集,T 存在最小元素 k 0 = m 0 + N,其中 m 0 ∈ S。对于任意 m ∈ S,都有 m + N ∈ T。m 0 + N = k 0 ≤ m + N 意味着 m 0 ≤ m。因此,m 0 是 S 的最小元素。

For (b), fix S ⊆ℤ and assume S is nonempty and bounded above. Let x ∈ ℝ be an upper bound for S . [Here the idea is to flip the order by negating everything, as in the proof of Theorem 8.56 .] Define T = { − m : m S }. Since m x for all m S , we have − x ≤ − m for all − m T . Thus, − x is a lower bound for T . Also T is nonempty since S is nonempty. By (a), T has a smallest element, say − m 0 with m 0 S . Given any m S , we know − m 0 ≤ − m , and hence m m 0 . Thus, m 0 is the largest element of S .

对于 (b),固定 S⊆ℤ,并假设 S 非空且有上界。令 x ∈ ℝ 为 S 的一个上界。[这里思路是通过对所有元素取反来颠倒顺序,如同定理 8.56 的证明。] 定义 T = { − m: m ∈ S}。由于对于所有 m ∈ S,都有 m ≤ x,因此对于所有 −m ∈ T,都有 −x ≤ −m。所以,−x 是 T 的一个下界。此外,由于 S 非空,T 也非空。由 (a) 可知,T 存在最小元素,例如 −m 0 ,其中 m 0 ∈ S。对于任意 m ∈ S,我们知道 −m 0 ≤ −m,因此 m ≤ m 0 。所以,m 0 是 S 的最大元素。

When we draw a number line to represent ℝ, we often make tick marks, spaced one unit apart, to represent the integers. Based on such a picture, it is highly believable that for every real x , the half-open interval [ x , x + 1) must contain exactly one integer. The next theorem justifies this pictorial intuition.

当我们用数轴表示 ℝ 时,通常会用间隔一个单位的刻度线来表示整数。基于这样的图示,我们很容易相信,对于每个实数 x,半开区间 [x, x + 1) 必然恰好包含一个整数。下一个定理将验证这种直观的理解。

8.59. Theorem on Distribution of ℤ in ℝ. x R , ! n Z , x n < x + 1 .

8.59. ℤ 在 ℝ 中的分布定理。∀x∈R,∃!n∈Z,x≤n<x+1。

Proof. Fix x ∈ ℝ. To prove existence of n , let S = { m ∈ ℤ: x m }. By the Archimedean Property, S is nonempty. Also x is a lower bound for S , so we know S has a least element n . To get a contradiction, assume that x + 1 ≤ n . Then x n − 1 where n − 1 is an integer (by closure of ℤ under subtraction). But this means n − 1 ∈ S ; since n − 1 < n , this contradicts the choice of n as the least element of S . Hence, n < x + 1 must hold. We also know x n since n S .

证明。固定 x ∈ ℝ。为了证明 n 的存在性,令 S = {m ∈ ℤ: x ≤ m}。根据阿基米德性质,S 非空。又因为 x 是 S 的一个下界,所以我们知道 S 存在最小元素 n。为了得出矛盾,假设 x + 1 ≤ n。那么 x ≤ n − 1,其中 n − 1 是一个整数(由 ℤ 对减法的封闭性可知)。但这意味着 n − 1 ∈ S;由于 n − 1 < n,这与选择 n 作为 S 的最小元素相矛盾。因此,n < x + 1 必然成立。我们也知道 x ≤ n,因为 n ∈ S。

To prove uniqueness of n , suppose (to get a contradiction) that n 1 and n 2 are distinct integers that both belong to the interval [ x , x + 1). We can choose notation so that n 1 < n 2 , and hence 0 < n 2 n 1 . On the other hand, − n 1 ≤ − x and n 2 < x + 1 combine to give n 2 n 1 < ( x + 1) − x = 1. Now n 2 n 1 is an integer in the open interval (0, 1), which contradicts Lemma 8.41 (d). So the n found above is unique.

为了证明 n 的唯一性,假设(为了得出矛盾)n 1 和 n 2 是不同的整数,它们都属于区间 [x, x + 1)。我们可以选择符号,使得 n 1 < n 2 ,因此 0 < n 2 − n 1 。另一方面,−n 1 ≤ −x 和 n 2 < x + 1 结合起来,可得 n 2 − n 1 < (x + 1) − x = 1。现在,n 2 − n 1 是开区间 (0, 1) 内的整数,这与引理 8.41(d) 矛盾。因此,上面找到的 n 是唯一的。

Density of ℚ in ℝ; Real Square Roots

ℚ 在 ℝ 中的密度;实数平方根

The next result says that ℚ is dense in ℝ, meaning that arbitrarily small open intervals contain rational numbers.

下一个结果表明 ℚ 在 ℝ 中稠密,这意味着任意小的开区间都包含有理数。

8.60. Theorem on Density of ℚ in ℝ. a , b R , a < b q Q , a < q < b .

8.60. ℚ 在 ℝ 中的密度定理。∀a,b∈R,a<b⇒∃q∈Q,a<q<b。

Proof. Fix a , b ∈ ℝ with a < b . Since 0 < b a , we can use the Archimedean property to find a positive integer n with 1/ n < b a . Adding a − (1/ n ) to both sides, we find that a < b − (1/ n ); and b − (1/ n ) < b since 0 < 1/ n . [The key idea is to magnify the interval [ b − 1/ n , b ) by a factor of n to get the interval [ nb − 1, nb ) of length 1.] Applying Theorem 8.59 to the real number x = nb − 1, we obtain a (unique) integer m with nb − 1 ≤ m < nb . Multiplying these inequalities by the positive quantity 1/ n , we get b − (1/ n ) ≤ m / n < b . Since a < b − (1/ n ), transitivity gives a < m / n < b . As m and n are integers, q = m / n is a rational number in ( a , b ).

证明。固定 a, b ∈ ℝ,且 a < b。由于 0 < b − a,我们可以利用阿基米德性质找到一个正整数 n,使得 1/n < b − a。两边同时加上 a − (1/n),我们发现 a < b − (1/n);并且由于 0 < 1/n,所以 b − (1/n) < b。[关键思想是将区间 [b − 1/n, b) 放大 n 倍,得到长度为 1 的区间 [nb − 1, nb)。] 将定理 8.59 应用于实数 x = nb − 1,我们得到一个(唯一的)整数 m,使得 nb − 1 ≤ m < nb。将这些不等式乘以正数 1/n,我们得到 b − (1/n) ≤ m/n < b。因为 a < b − (1/n),传递性得出 a < m/n < b。由于 m 和 n 是整数,所以 q = m/n 是 (a, b) 中的有理数。

The set of irrational numbers is also dense in ℝ, but to prove this we must first know that irrational numbers actually exist! We now use the Completeness Axiom to show that all positive real numbers have real square roots. Combining this with the earlier theorem that 2 has no rational square root, we deduce the existence of real numbers that are not rational. We may also conclude that ℚ cannot satisfy the Completeness Axiom (otherwise, the proof given below would apply to the number system ℚ and prove that 2 had a square root in ℚ, which is not true).

无理数集在 ℝ 中也是稠密的,但要证明这一点,我们首先必须知道无理数确实存在!现在我们利用完备性公理证明所有正实数都有实数平方根。结合之前 2 没有有理平方根的定理,我们推断出非有理实数的存在。我们还可以得出结论,ℚ 不能满足完备性公理(否则,下面的证明将适用于数系 ℚ,并证明 2 在 ℚ 中有平方根,这显然是不成立的)。

8.61. Theorem on Existence of Real Square Roots. a R > 0 , x R > 0 , x 2 = a .

8.61. 实平方根存在性定理。∀a∈R>0,∃x∈R>0,x2=a。

Proof. The proof uses the following fact, which is left as an exercise: for all y , z ∈ ℝ ≥0 , y < z iff y 2 < z 2 . We first prove the result for a fixed a ∈ ℝ ≥1 . Since 1 ≤ a and a is positive, we see that a a 2 . [To use the Completeness Axiom to solve x 2 = a , we need a set of real numbers with least upper bound x .] Define the set S = { z ∈ ℝ ≥0 : z 2 < a }. On one hand, S is nonempty because 0 2 = 0 < a , so that 0 ∈ S . On the other hand, we claim that S is bounded above with upper bound a . For given an arbitrary z S , we know z 2 < a a 2 , so z < a by the initial fact.

证明。证明过程中用到以下事实(留作练习):对于所有 y, z ∈ ℝ ≥0 ,y < z 当且仅当 y 2 < z 2 。我们首先证明对于固定的 a ∈ ℝ ≥1 ,结果成立。由于 1 ≤ a 且 a 为正数,我们可知 a ≤ a 2 。[为了利用完备性公理求解 x 2 = a,我们需要一个具有最小上界 x 的实数集。] 定义集合 S = {z ∈ ℝ ≥0 :z 2 < a}。一方面,S 非空,因为 0 2 = 0 < a,所以 0 ∈ S。另一方面,我们断言 S 有上界,上界为 a。对于任意给定的 z ∈ S,我们知道 z 2 < a ≤ a 2 ,所以根据初始事实,z < a。

We have now verified the hypotheses of the Completeness Axiom (O7), so we can conclude that x = lub S exists in ℝ. Since 0 ∈ S , 0 ≤ x . By trichotomy, x 2 < a or a < x 2 or x 2 = a . The third case is the required result; we finish the proof by showing that the first two cases lead to contradictions.

我们已经验证了完备性公理 (O7) 的假设,因此可以得出结论:x = lubS 存在于 ℝ 中。由于 0 ∈ S,所以 0 ≤ x。根据三分法,x 2 < a 或 a < x 2 或 x 2 = a。第三种情况即为所需结果;我们通过证明前两种情况会导致矛盾来完成证明。

First assume that x 2 < a . On one hand, this means x S , so x < a as shown above. On the other hand, for any n ∈ ℕ, we know 1 ≤ n , so n n 2 and 1/ n 2 ≤ 1/ n . Now calculate ( x + 1/ n ) 2 = x 2 + (2/ n ) x + (1/ n 2 ) < x 2 + (2/ n ) a + (1/ n ) ≤ x 2 + (2 a + 1)/ n . By the Archimedean Property applied to the positive number ( a x 2 )/(2 a + 1), we can choose a positive integer n with 1/ n < ( a x 2 )/(2 a + 1). Then x < x + 1/ n and ( x + 1/ n ) 2 x 2 + (2 a + 1)/ n < x 2 + ( a x 2 ) = a . So x + 1/ n belongs to S and is larger than x ; this contradicts the fact that x is an upper bound for the set S .

首先假设 x 2 < a。一方面,这意味着 x ∈ S,因此如上所示,x < a。另一方面,对于任意 n ∈ ℕ,我们知道 1 ≤ n,因此 n ≤ n 2 且 1/n 2 ≤ 1/n。现在计算 (x + 1/n) 2 = x 2 + (2/n)x + (1/n 2 ) < x 2 + (2/n)a + (1/n) ≤ x 2 + (2a + 1)/n。利用阿基米德性质对正数 (a − x)/(2a + 1) 进行定理,我们可以选取一个正整数 n,使得 1/n < (a − x)/(2a + 1)。那么 x < x + 1/n 且 (x + 1/n) ≤ x + (2a + 1)/n < x + (a − x) = a。因此,x + 1/n 属于集合 S 且大于 x;这与 x 是集合 S 的上界这一事实相矛盾。

Now assume that a < x 2 . This time, we calculate ( x − 1/ n ) 2 = x 2 − (2/ n ) x + (1/ n 2 ) ≥ x 2 − (2 x / n ). By the Archimedean Property, there exists a positive integer n with 1/ n < x and 1/ n < ( x 2 a )/(2 x ), so x − (1/ n ) > 0 and ( x − 1/ n ) 2 x 2 − (2 x / n ) > a . Given any y S , note that y 2 < a < ( x − 1/ n ) 2 , so y < x − 1/ n by the initial fact. But this means x − 1/ n is an upper bound for S , contradicting the choice of x as the least upper bound for the set S .

现在假设 a < x 2 。这次,我们计算 (x − 1/n) 2 = x 2 − (2/n)x + (1/n 2 ) ≥ x 2 − (2x/n)。根据阿基米德性质,存在一个正整数 n,使得 1/n < x 且 1/n < (x 2 − a)/(2x),因此 x − (1/n) > 0 且 (x − 1/n) 2 ≥ x 2 − (2x/n) > a。给定任意 y ∈ S,注意到 y 2 < a < (x − 1/n) 2 ,因此根据初始事实,y < x − 1/n。但这意味着 x − 1/n 是 S 的一个上界,这与选择 x 作为集合 S 的最小上界相矛盾。

Finally, we prove the result for fixed a between 0 and 1. Since 1/ a > 1, we have just shown that there exists y ∈ ℝ ≥0 with y 2 = 1/ a . Then x = 1/ y satisfies x 2 = 1/ y 2 = a , as needed.     □

最后,我们证明了当 a 为 0 到 1 之间的固定值时,结果成立。由于 1/a > 1,我们刚刚证明了存在 y ∈ ℝ ≥0 使得 y 2 = 1/a。那么 x = 1/y 满足 x 2 = 1/y 2 = a,符合要求。□

Nested Interval Property

嵌套区间属性

As a final illustration of the Completeness Axiom, we prove a theorem stating that the intersection of any sequence of nested closed intervals is always nonempty. This is another way of formulating the intuitive notion that the real number system ℝ has no holes.

作为完备性公理的最后一个例证,我们证明了一个定理:任意嵌套闭区间序列的交集总是非空的。这可以看作是另一种表述实数系统ℝ没有空洞这一直观概念的方式。

8.62. Nested Interval Theorem. Suppose {[ a n , b n ]: n ∈ ℤ >0 } is a collection of nonempty closed intervals such that [ a n +1 , b n +1 ]⊆ [ a n , b n ] for all n ∈ ℤ >0 . Then n = 1 [ a n , b n ] .

8.62. 嵌套区间定理。假设 {[a n , b n ]:n ∈ ℤ >0 是非空闭区间的集合,使得对于所有 n ∈ ℤ >0 ,[a n +1 , b n +1 ]⊆ [a n , b n ]。则 ⋂n=1∞[an,bn]≠∅。

Proof. For each n ∈ ℤ >0 , we have a n b n since [ a n , b n ] is nonempty, a n a n +1 since a n +1 ∈ [ a n +1 , b n +1 ]⊆ [ a n , b n ], and b n +1 b n since b n +1 ∈ [ a n +1 , b n +1 ]⊆ [ a n , b n ]. By induction, we readily deduce the following consequences of these inequalities: for all positive integers i < j , a i a j and b j b i ; and for all positive integers k , n , a k b n .

证明。对于每个 n ∈ ℤ >0 ,我们有 a n ≤ b n ,因为 [a n , b n ] 非空;a n ≤ a n +1 ,因为 a n +1 ∈ [a n +1 , b n +1 ]⊆ [a n , b n ];以及 b n +1 ≤ b n ,因为 b n +1 ∈ [a n +1 , b n +1 ]⊆ [a n , b n ]. 通过归纳法,我们很容易推导出这些不等式的以下结果:对于所有正整数 i < j,a i ≤ a j 且 b j ≤ b i ;对于所有正整数 k, n,a k ≤ b n

Now consider the set S = { a n : n ∈ ℤ >0 }, which is a nonempty subset of ℝ. This set is bounded above by b 1 , since any a n S satisfies a n b 1 . By the Completeness Axiom (O7), there exists r ∈ ℝ with r = lub S . We claim r n = 1 [ a n , b n ] , which proves that this intersection is nonempty. Fix n ∈ ℤ >0 ; we must show r ∈ [ a n , b n ], which means a n r and r b n . Since r is an upper bound for S , a n r is true. Assume, to get a contradiction, that r b n is false. Then b n < r = lub S , so b n is not an upper bound for S . So there exists a k S with b n < a k , contradicting the previous paragraph. Thus, r b n is true.     □

现在考虑集合 S = {a n :n ∈ ℤ >0 ,它是 ℝ 的一个非空子集。该集合的上界为 b 1 ,因为任何 a n ∈ S 都满足 a n ≤ b 1 。根据完备性公理 (O7),存在 r ∈ ℝ 使得 r = lubS。我们断言 r∈⋂n=1∞[an,bn],这证明了该交集非空。固定 n ∈ ℤ >0 ;我们必须证明 r ∈ [a n , b n ],这意味着 a n ≤ r 且 r ≤ b n 。由于 r 是 S 的一个上界,因此 a n ≤ r 为真。为了得出矛盾,假设 r ≤ b n 为假。那么 b n < r = lubS,所以 b n 不是 S 的一个上界。因此存在 a k ∈ S 使得 b n < a k ,这与上一段矛盾。因此,r ≤ b n 为真。□

One consequence of the Nested Interval Theorem is that for any sequence ( z n : n ∈ ℤ >0 ) and any interval [ a , b ], there is a point in [ a , b ] different from all z n ; in other words, [ a , b ] is an uncountable set (see §7.4). The Nested Interval Theorem can also be used to prove Intermediate Value Theorem the Intermediate Value Theorem for continuous real-valued functions. Given a continuous f :[ a , b ] → ℝ and c ∈ ℝ between f ( a ) and f ( b ), the idea is to find x ∈ ℝ solving f ( x ) = c by repeatedly bisecting the interval [ a , b ] and taking x in the intersection of the resulting sequence of nested closed intervals.

嵌套区间定理的一个推论是,对于任意序列 (z n :n ∈ ℤ >0 ) 和任意区间 [a, b],在 [a, b] 中存在一个点,它不属于所有 z n ;换句话说,[a, b] 是一个不可数集(参见 §7.4)。嵌套区间定理也可用于证明连续实值函数的介值定理。给定一个连续函数 f:[a, b] → ℝ 和介于 f(a) 和 f(b) 之间的 c ∈ ℝ,其思路是通过反复二分区间 [a, b] 并取所得嵌套闭区间序列的交集,找到满足 f(x) = c 的 x ∈ ℝ。

Section Summary

章节概要

  1. Definitions. For S ⊆ℝ and z 0 ∈ ℝ, z 0 is an upper bound of S iff x S , x z 0 ; z 0 = max S iff z 0 S and z 0 is an upper bound of S ; z 0 = lub S (the least upper bound of S ) iff z 0 is an upper bound of S and every y < z 0 is not an upper bound of S . Lower bounds, min S , and the greatest lower bound glb S are defined similarly.
    定义。对于 S⊆ℝ 和 z 0 ∈ ℝ,z 0 是 S 的上界当且仅当对于所有 x∈S,x≤z0;z 0 = max S 当且仅当 z 0 ∈ S 且 z 0 是 S 的上界;z 0 = lubS(S 的最小上界)当且仅当 z 0 是 S 的上界且所有 y < z 0 都不是 S 的上界。下界、min S 和最大下界 glbS 的定义类似。
  2. Completeness and Related Properties. The Completeness Axiom states that every nonempty subset of ℝ that is bounded above has a least upper bound. Consequently, every nonempty subset of ℝ that is bounded below has a greatest lower bound. A nonempty subset of ℤ that is bounded above (resp. below) has a greatest (resp. least) element. Every nonempty finite subset of ℝ has a greatest and a least element. For a general subset S of ℝ, max S , min S , lub S , and glb S need not exist; but each of these numbers is unique when it exists.
    完备性及相关性质。完备性公理指出,ℝ 的每个非空有上界的子集都存在最小上界。因此,ℝ 的每个非空有下界的子集都存在最大下界。ℤ 的每个非空有上界(或有下界)的子集都存在最大(或最小)元素。ℝ 的每个非空有限子集都存在最大元素和最小元素。对于 ℝ 的一般子集 S,maxS、minS、lubS 和 glbS 不一定存在;但如果存在,则每个数都是唯一的。
  3. Archimedean Property. For each x ∈ ℝ, there is a positive integer n with x < n . For each y ∈ ℝ >0 , there is a positive integer n with 1/ n < y .
    阿基米德性质。对于每个 x ∈ ℝ,存在一个正整数 n,使得 x < n。对于每个 y ∈ ℝ >0 ,存在一个正整数 n,使得 1/n < y。
  4. Distribution of ℤ and ℚ in ℝ. For each x ∈ ℝ, there is exactly one integer n such that x n < x + 1. For each a < b in ℝ, there is a rational number q with a < q < b .
    ℤ 和 ℚ 在 ℝ 中的分布。对于每个 x ∈ ℝ,恰好存在一个整数 n,使得 x ≤ n < x + 1。对于 ℝ 中的每个 a < b,存在一个有理数 q,使得 a < q < b。
  5. Existence of Real Square Roots. For each a ∈ ℝ >0 , there exists x ∈ ℝ >0 with x 2 = a . This statement does not hold for ℚ >0 , so ℚ does not satisfy the Completeness Axiom.
    实数平方根的存在性。对于每个 a ∈ ℝ >0 ,存在 x ∈ ℝ >0 使得 x 2 = a。该命题对 ℚ >0 不成立,因此 ℚ 不满足完备性公理。
  6. Nested Interval Property. Any collection of nonempty closed intervals [ a n , b n ] that are nested (i.e., [ a n +1 , b n +1 ]⊆ [ a n , b n ] for all n ∈ ℤ ≥0 ) must have nonempty intersection in ℝ.
    嵌套区间性质。任何非空闭区间 [a n , b n ] 的嵌套集合(即,对于所有 n ∈ ℤ ≥0 ,[a n +1 , b n +1 ]⊆ [a n , b n ])在 ℝ 中必须有非空交集。

Exercises

练习

  1. (a) Prove: for all y , z ∈ ℝ ≥0 , y < z iff y 2 < z 2 . (b) Does either implication in (a) hold for all y , z ∈ ℝ?
    (a) 证明:对于所有 y, z ∈ ℝ ≥0 ,y < z 当且仅当 y 2 < z 2 。(b) (a) 中的两个蕴含式是否对所有 y, z ∈ ℝ 成立?
  2. (a) Prove: for all x , y ∈ ℝ, if x < y , then x < ( x + y )/2 < y . (b) More generally, prove x , y , t R , if x < y and 0 < t < 1, then x < tx + (1 − t ) y < y . (c) Prove: for all a , b ∈ ℝ, if a < b , then the open interval ( a , b ) has no least element.
    (a) 证明:对于任意 x, y ∈ ℝ,若 x < y,则 x < (x + y)/2 < y。(b) 更一般地证明:对于任意 x, y, t ∈ R,若 x < y 且 0 < t < 1,则 x < tx + (1 − t)y < y。(c) 证明:对于任意 a, b ∈ ℝ,若 a < b,则开区间 (a, b) 没有最小元素。
  3. Prove that ℝ <0 has no maximum and no minimum.
    证明 ℝ <0 没有最大值也没有最小值。
  4. Let S = {3 − 1/ n : n ∈ ℤ >0 }. Find (with proof) lub S and glb S . Does max S exist? Does min S exist?
    设 S = {3 − 1/n:n ∈ ℤ >0 。求 (并证明) lubS 和 glbS。max S 是否存在?min S 是否存在?
  5. (a) Prove: For all S ⊆ℝ, if S is bounded above and bounded below, then glb S ≤ lub S . (b) Find all sets S ⊆ℝ such that glb S and lub S exist and are equal.
    (a) 证明:对于所有 S⊆ℝ,如果 S 有上界且有下界,则 glbS ≤ lubS。(b) 求所有集合 S⊆ℝ,使得 glbS 和 lubS 存在且相等。
  6. Recall that [ n ] = {1, 2, …, n } = { m ∈ ℤ >0 : m n }. Prove: for all n ∈ ℤ >0 and all functions f :[ n ] → ℝ, the image of f has a greatest element and a least element. Deduce that every nonempty finite subset of ℝ has a maximum and a minimum element.
    回想一下,[n] = {1, 2, …, n} = {m ∈ ℤ >0 :m ≤ n}。证明:对于所有 n ∈ ℤ >0 和所有函数 f:[n] → ℝ,f 的像有一个最大元素和一个最小元素。由此推断,ℝ 的每个非空有限子集都有一个最大元素和一个最小元素。
  7. (a) If S is not bounded above, what is the set U of all upper bounds of S ? Does U have a least element? Does lub S exist? (b) Repeat part (a) for S = (the empty set).
    (a)如果 S 没有上界,那么 S 的所有上界构成的集合 U 是什么?U 是否有最小元素?lubS 是否存在?(b)对 S=∅(空集)重复(a)部分。
  8. Prove: y R > 0 , ϵ R > 0 , n Z > 0 , n ϵ > y .
    证明:∀y∈R>0,∀ϵεR>0,∃n∈Z>0,nϵ>y。
  9. (a) Where is the flaw in this proposed proof of Theorem 8.57 (a)? “Fix x 0 in ℝ. Choose n = x 0 + 1. Since we know 0 < 1, adding x 0 to both sides gives x 0 < x 0 + 1 = n , as needed.” (b) Where is the flaw in this proposed proof of Theorem 8.60 ? “Fix a , b ∈ ℝ with a < b . Choose q = ( a + b )/2. Since a < b , an earlier exercise gives a < q < b , as needed.”
    (a) 定理 8.57(a) 的证明中哪里有缺陷?“固定 ℝ 中的 x 0 。选择 n = x 0 + 1。由于我们知道 0 < 1,两边同时加上 x 0 可得 x 0 < x 0 + 1 = n,符合要求。” (b) 定理 8.60 的证明中哪里有缺陷?“固定 a, b ∈ ℝ 且 a < b。选择 q = (a + b)/2。由于 a < b,之前的练习可知 a < q < b,符合要求。”
  10. Prove: for all x ∈ ℝ, there exists a unique n ∈ ℤ with x < n x + 1.
    证明:对于所有 x ∈ ℝ,存在唯一的 n ∈ ℤ,使得 x < n ≤ x + 1。
  11. Give a careful proof that {[ n , n + 1): n ∈ ℤ} is a set partition of ℝ.
    请给出仔细的证明,证明 {[n, n + 1):n ∈ ℤ} 是 ℝ 的一个集合划分。
  12. Prove: for all a , b ∈ ℝ, if a < b , then there exists x ∈ ℝ − ℚ with a < x < b . [ Hint: We know 2 R Q , and we know there is a rational number in ( a + 2 , b + 2 ) .]
    证明:对于任意 a, b ∈ ℝ,若 a < b,则存在 x ∈ ℝ − ℚ 使得 a < x < b。[提示:已知 2 ∈ R−Q,且 (a+2, b+2) 中存在有理数。]
  13. Prove that every a ∈ ℝ >0 has a unique positive square root.
    证明每个 a ∈ ℝ >0 都有唯一的正平方根。
  14. Let ( X , ≤ ) be any partially ordered set satisfying the Completeness Axiom (O7). Prove that for all S X , if S is nonempty and bounded below, then S has a greatest lowest bound. (This gives another proof of Theorem 8.56 not relying on the algebraic structure of ℝ.)
    设 (X, ≤) 为满足完备性公理 (O7) 的任意偏序集。证明对于所有 S⊆ X,如果 S 非空且有下界,则 S 存在最大下界。(这给出了定理 8.56 的另一种证明,该证明不依赖于 ℝ 的代数结构。)
  15. For x = ( x 1 , …, x n ) and y = ( y 1 , …, y n ) in ℝ n , define
    d ( x , y ) = max { | x 1 y 1 | , , | x n y n | } .
    Prove that this distance function satisfies the metric space axioms in Theorem 8.46 .
    对于 ℝ n 中的 x = (x 1 , …, x n ) 和 y = (y 1 , …, y n ),定义 d(x,y)=max{|x1−y1|,…,|xn−yn|}。证明该距离函数满足定理 8.46 中的度量空间公理。
  16. For x = ( x 1 , …, x n ) and y = ( y 1 , …, y n ) in ℝ n , define d ( x , y ) = k = 1 n | x k y k | . Prove that this distance function satisfies the metric space axioms in Theorem 8.46 .
    对于 ℝ n 中的 x = (x 1 , …, x n ) 和 y = (y 1 , …, y n ),定义 d(x,y)=∑k=1n|xk−yk|。证明该距离函数满足定理 8.46 中的度量空间公理。

1 In the next few sections, we use the notation ℕ, rather than ℤ >0 , to avoid assuming in advance that natural numbers are the same as positive integers . This fact is proved as a theorem later.

在接下来的几节中,我们将使用符号ℕ,而不是ℤ >0 ,以避免预先假设自然数与正整数相同。这一事实将在后面作为定理进行证明。

#

Suggestions for Further Reading

延伸阅读建议

In this text, we have covered fundamental concepts of logic, proofs, set theory, integers, relations, functions, cardinality, and real numbers that are absolutely essential for further pursuit of advanced mathematics. Each of the subjects we have studied is a vast subdiscipline of mathematics that goes far beyond the introductory material discussed here. We now suggest some books that interested readers may consult to learn more about these fascinating topics.

本文涵盖了逻辑、证明、集合论、整数、关系、函数、基数和实数等基本概念,这些概念对于进一步学习高等数学至关重要。我们所学习的每个主题都是数学的一个庞大分支,远远超出了本文所讨论的入门内容。现在,我们推荐一些书籍,供感兴趣的读者参考,以深入了解这些引人入胜的主题。

Logic and Set Theory

逻辑与集合论

Mathematical logic is a beautiful and amazing subject that applies the precise tools of mathematical reasoning to explore the notions of proof, truth, and computability.

数理逻辑是一门优美而神奇的学科,它运用数学推理的精确工具来探索证明、真理和可计算性的概念。

Axiomatic set theory rigorously develops properties of sets beginning with an explicit list of axioms. We sketched the beginnings of this subject in the optional Section 3.7 .

公理化集合论从明确的公理列表出发,严谨地推导集合的性质。我们在选修课第 3.7 节中概述了这一主题的雏形。

Analysis

分析

Analysis is the branch of mathematics that rigorously develops the calculus of real-valued functions and their generalizations. This subject begins with the ordered field axioms for real numbers, which are covered in Chapter 8 of this text.

分析学是数学的一个分支,它严谨地发展实值函数及其推广的微积分。本学科从实数的有序域公理开始,这些公理将在本书第八章中介绍。

Number Theory and Algebra

数论与代数

Number theory studies properties of the natural numbers. Our discussion of divisibility, primes, unique prime factorizations of integers, and modular arithmetic barely scratch the surface of this ancient subject, which has many unexpected practical applications to modern cryptography.

数论研究自然数的性质。我们对整除性、素数、整数的唯一素因子分解以及模运算的讨论仅仅触及了这门古老学科的皮毛,而它在现代密码学中有着许多意想不到的实际应用。

Abstract algebra starts with familiar properties of integers and real numbers (like associativity) and considers these properties in far more general settings. What results is an elegant abstract theory that applies to any structure satisfying the initial axioms.

抽象代数从整数和实数的常见性质(例如结合律)出发,并在更一般的框架下考察这些性质。最终得到的是一个优雅的抽象理论,它适用于任何满足初始公理的结构。

#

Index

指数

( a , b ), see ordered pair or open interval

(a,b),参见有序对或开区间

( a , b , c ), see ordered triple

(a,b,c),参见有序三元组

2, 3,…, 10 defined, 350

2、3、…、10 已定义,350

A − B, see set difference

A − B,参见集合差

A B , see intersection

A∩B,参见交点

A B , see union

A∪B,参见并集

A B , see subset

A⊆B,参见子集

A B , see superset

A⊇B,参见超集

A × B , see product set

A×B,参见产品套装

A B , see symmetric difference

A△B,参见对称差

I X , see identity relation

I X ,参见身份关系

P Q , see if

P⇒Q,看看是否

P Q , see exclusive-OR

P⊕Q,参见独占或

P Q , see or

P∨Q,参见或

P Q , see and

P∧Q,参见和

R [ C ], see image of set under relation

R[C],参见关系下的集合图像

R −1 , see inverse of a relation

R −1 ,参见关系的逆

Y X (set of functions), 338

Y X (函数集),338

ℂ, see complex numbers

ℂ,参见复数

ℕ, see natural numbers

ℕ,参见自然数

ℚ, see rational numbers

ℚ,参见有理数

ℝ, see real numbers

ℝ,参见实数

2 , 134

2 , 134

ℤ, see integers

ℤ,参见整数

Z / n (integers mod n ), 296 , 308

Z/n(整数模 n),296,308

ℤ × ℤ is countably infinite, 324

ℤ × ℤ 是可数无穷的,324

i I A i , see general intersection

⋂i∈IAi,参见一般交集

a , see formal union

⋃a,参见正式联盟

i I A i , see general union

⋃i∈IAi,参见一般并集

χ A , see characteristic function

χA,参见特征函数

δ, see delta

δ,参见 delta

, see empty set

∅,参见空集

n (congruence mod n ), 284

n (模 n 同余),284

x , see existential quantifier

∃x,参见存在量词

! x , see uniqueness symbol

∃!x,参见唯一性符号

x , see universal quantifier

∀x,参见全称量词

glb S , see greatest lower bound

glb S,参见最大下限

glue ( f i : i I ) , see gluing

glue(fi:i∈I),参见 gluing

Id A , see identity function

ID A ,参见恒等函数

∈-transitive set, 131

∈传递集,131

inf S , see greatest lower bound

inf S,参见最大下限

lex , 307

≤lex,307

lub S , see least upper bound

lub S,参见最小上界

max S , 303

最大 S,303

min S , 303

我的 S,303

P , see not

∼P,参见 not

k = 1 n a k (product notation), 160

∏k=1nak(乘积表示法),160

P ( A ) , see power set

P(A),参见幂集

sgn( x ), see sign function

sgn(x),参见符号函数

P (equivalence relation induced by P ), 296

∼P(由 P 诱导的等价关系),296

f (equivalence relation induced by f ), 283

∼f(由f诱导的等价关系),283

k = 1 n a k (summation notation), 160

∑k=1nak(求和符号),160

sup S , see least upper bound

sup S,参见最小上界

| A | = | B | (cardinal equality), 322

|A|=|B|(基数相等),322

| A | | B | (cardinal ordering), 322

|A|≤|B|(基数排序),322

| X | = n , 315

|X|=n,315

| x | , see absolute value

|x|,参见绝对值

{ x : P ( x ) } , see set-builder notation

{x:P(x)},参见集合构造器符号

a / b (formal construction), 311

a/b(形式构造),311

a | b ( a divides b ), 185

a|b(a 整除 b),185

f [ A ], see direct image

f[A],参见直接图像

f −1 , see inverse function

f −1 ,参见反函数

f 1 [ B ] , see preimage

f−1[B],参见原像

f | S , see restriction of a function

f|S,参见函数的限制

g : A B (function notation), 231 , 233

g:A→B(函数表示法),231,233

g f , see function composition

g∘f,参见函数组合

j A, B , see inclusion function

j A, B ,参见包含函数

n ! (factorial), 159

n!(阶乘),159

n th root, 200

n 次方根,200

absolute adjective, 43

绝对形容词,43

absolute value, 78 , 371

绝对值,78,371

absorption law, 9 , 120

吸收定律,9,120

abstract algebra, 384

抽象代数,384

addition, 240 , 349

此外,240,349

axioms, 350

公理,350

cancellation law, 352

取消法,352

in ℚ, 311

在ℚ,311

inverse, 240 , 349

倒数,240,349

mod n , 308

反对 n,308

set closed under, 338

设置关闭下方,338

table, 355

表格,355

additivity of sums, 164

和的可加性,164

advice for set proofs, 123

关于集合证明的建议,123

algebraic properties of order, 369

序的代数性质,369

algebraic structure, 351

代数结构,351

algorithm for gcds, 186

求最大公约数的算法,186

all

全部

definition of, 27

定义,27

inference rule, 50

推理规则,50

proofs via generic elements, 82

通过通用元素的证明,82

analysis, 349 , 384

分析,349,384

and

definition of, 3

定义,3

proof template for, 64

证明模板,64

antecedent, 6

前件,6

antisymmetry, 301

反对称性,301

of cardinal ordering, 336

基数排序,336

Archimedean Property of ℝ, 378

阿基米德性质ℝ,378

arcsine, 270

反正弦,270

arithmetic mean, 51

算术平均值,51

arithmetical fact, 353

算术事实,353

arrow diagram, 211 , 246 , 260 , 339

箭头图,211、246、260、339

and inverses, 219

及其逆运算,219

of restriction, 248

限制,248

associativity, 9 , 49 , 52 , 242 , 247

结合律,9,49,52,242,247

generalized, 354

概括而言,354

of composition of relations, 226

关系的构成,226

axiom, 48 , 51

公理,48,51

definitional, 49

定义,49

for ℤ, 49

对于ℤ,49

for addition in ℝ, 350

对于以ℝ为单位的加法,350

for groups, 52

对于团体而言,52

for multiplication in ℝ, 350

对于 ℝ 中的乘法,350

for positive real numbers, 374

对于正实数,374

of ZFC set theory, 156

ZFC集合论,156

Axiom of Abstraction, 149

抽象公理,149

Axiom of Choice, 154

选择公理,154

Axiom of Completeness, 351 , 377

完备性公理,351,377

Axiom of Extension, 113 , 150

扩展公理,113,150

Axiom of Foundation, 129 , 155

基础公理,129,155

Axiom of Induction, 160

归纳公理,160

Axiom of Infinity, 153

无穷公理,153

Axiom of Pairs, 152

配对公理,152

Axiom of Power Sets, 152

幂集公理,152

Axiom of Regularity, 156

正则性公理,156

Axiom of Replacement, 154

替换公理,154

Axiom of Specification, 151

规范公理,151

Axiom of the Empty Set, 153

空集公理,153

Axiom of Unions, 152

联合公理,152

axiomatic set theory, 149 , 383

公理化集合论,149,383

backward direction, 65

向后方向,65

backwards induction, 169

逆向归纳法,169

base b expansion, 184

基础b扩展,184

base case, 161

基本情况,161

biconditional, 6

双条件,6

bijection, 260 , 261

双射,260,261

and inverses, 269

及其逆运算,269

between equivalence relations and set partitions, 297

等价关系与集合划分之间,297

composition of, 264

组成,264

bijective function, 261

双射函数,261

binary union, 152

二元并集,152

bisquare, 62

双正方形,62

bit, 13

位,13

block of a set partition, 295

集合分区块,295

bounded above, 351 , 376

上界为 351、376

bounded below, 376

下限为 376

cancellation law, 52 , 352

取消法,52,352

Cantor’s diagonal argument, 333 , 334

康托尔的对角线论证,333,334

Cantor’s Theorem, 334

康托尔定理,334

cardinal equality, 322

基数相等,322

cardinal ordering, 322

基数排序,322

cardinality, 315

基数,315

definition review, 341 , 344

定义审查,341,344

of power set, 334

功率集,334

Cartesian graph, 339

笛卡尔坐标图,339

cases, 76

病例,76例

chain in a poset, 303

链式偏序集,303

chain proof, 121

链条检验,121

characteristic function, 243

特征函数,243

properties, 243

房产,243

characterization

特征描述

of divisors, 197

除数,197

of subsets, 120

子集,120

Choice Axiom, 154

选择公理,154

circle proof, 119

圆的证明,119

closed interval, 132

闭区间,132

is uncountable, 334 , 381

是不可数名词,334,381

closed set, 148

封闭集,148

closure, 49 , 52 , 350

关闭,49,52,350

and irrational numbers, 79

以及无理数,79

of R 0 , 356

R≠0,356

of integers, 364

整数,364

of natural numbers, 363

自然数中,363

of rational numbers, 365

有理数,365

Closure Theorem for Relations, 226

关系封闭定理,226

codomain, 231 , 233

共域,231,233

enlarging, 238

放大,238

commutative group, 351

交换群,351

commutative ring, 351

交换环,351

commutativity, 9 , 49 , 242

交换律,9,49,242

generalized, 354

概括而言,354

complete induction, 173

完全诱导,173

Completeness Axiom, 351 , 377

完备性公理,351,377

completeness of ℤ, 378

ℤ 的完备性,378

Completeness Theorem, 52

完备性定理,52

complex numbers, 28

复数,28

formal construction of, 374

正式构造,374

component of ordered pair, 132

有序对的组成部分,132

composite number, 179

综合数字,179

composition

作品

and inverses, 268

及其逆运算,268

associativity, 226

结合律,226

inverse of, 227

227 的倒数

of functions, 246

函数,246

of injections, etc., 264

注射等,264

of relations, 221

关系,221

properties, 247

房产,247

concatenation poset, 303

连接偏序集,303

conclusion, 6 , 10

结论,6,10

conditional, 6

条件,6

congruence modulo n , 284

模 n 同余,284

conjunction, 6

连词,6

consequent, 6

因此,6

consistent theory, 52

一致的理论,52

constant, 27

常数,27

generic vs. specific, 83

通用与特定,83

constant function, 239

常数函数,239

continuity, 88

连续性,88

contradiction, 21

矛盾,21

contradiction proof, 70

矛盾证明,70

contrapositive, 12

逆否命题,12

inference rule for IF, 50

IF 推理规则,50

contrapositive proof, 63

逆否命题证明,63

converse, 12

匡威,12

converse error, 50

逆误差,50

converse implication, 65

逆命题,65

conversion rules for quantifiers, 32

量词转换规则,32

countability of ℚ, 325

ℚ 的可数性,325

countability of ℤ × ℤ, 324

ℤ × ℤ 的可数性,324

countability of finite products, 325

有限积的可数性,325

countable set, 328

可数集,328

product of, 329

的产物,329

union of, 330

联盟,330

countably infinite set, 323

可数无限集,323

cube sum, 164

立方和,164

de Morgan’s laws, 10

德摩根定律,10

decimal expansion, 184

十进制展开式,184

defined term, 47

定义术语,47

definition of ordered pair, 136

有序对的定义,136

definition text, 48

定义文本,48

definitional axiom, 49

定义公理,49

delta (Greek letter), 85

希腊字母 delta,85

denial, 6 , 35

否认,6,35

practice, 40

练习,40

rules, 97

规则,97

rules for inequalities, 368

不等式规则,368

density of ℚ in ℝ, 379

ℚ 在 ℝ 中的密度,379

diagonal argument, 333 , 334

对角论证,333,334

difference, 349

差值,349

digraph, 339

二合字母,339

digraph of a relation, 275

关系的有向图,275

direct image, 254

直接图像,254

properties, 255

房产,255

direct implication, 65

直接影响,65

direct proof of IF-statement, 58

IF 语句的直接证明,58

disjoint sets, 115 , 316

不相交集合,115,316

disjunction, 6

析取,6

disproof, 79

反驳,79

disproving a false statement, 79

驳斥错误陈述,79

distance

距离

in ℝ n , 373

在 ℝ n , 373

in ℝ, 372

在 ℝ, 372 中

distribution of ℤ in ℝ, 379

ℤ 在 ℝ 中的分布,379

distributive law, 9 , 49 , 242 , 247

分配法,9、49、242、247

for relations, 227

对于关系,227

divides, 47 , 185

分割,47,185

division, 180 , 349

部门,180,349

for negative integers, 181

对于负整数,181

properties, 358

房产,358

divisor, 185

除数,185

and prime factorization, 197

以及质因数分解,197

domain, 231 , 233

域,231,233

double induction, 366

双感应,366

Double Inverse Rule, 226

双倒数法则,226

dyadic rational, 62

二元理性,62

EI, see existential instantiation

EI,参见存在实例化

eight, 350

八,350

either, 19

19

elementary row operation, 188

基本行运算,188

Elements (Euclid’s textbook), 185

《几何原本》(欧几里得的教科书),185

eliminating the uniqueness symbol, 43

去掉唯一性符号,43

ellipsis, 159

省略号,159

empty relation, 213

空关系,213

empty set, 106 , 153

空集,106,153

proving a set is empty, 115

证明一个集合为空集,115

Empty Set Axiom, 153

空集公理,153

enlarging the codomain, 238 , 242

扩大共域,238,242

Epimenides’s Paradox, 1

埃庇米尼得斯悖论,1

equality

平等

of functions, 241 , 339

函数,241,339

of sets, 113

集合,113

proof via chain proof, 121

通过链式证明进行证明,121

equality relation, 282

等式关系,282

equivalence class, 288 , 342

等价类,288,342

criterion for equality, 291

平等准则,291

disjointness property, 291

不相交性质,291

equivalence relation, 282 , 342

等价关系,282,342

and set partitions, 298 , 345

并设置分区,298,345

equivalence classes of, 288

等价类,288

induced by a function, 283

由函数诱导,283

induced by a set partition, 296

由集合划分诱导,296

intersection lemma, 285

交集引理,285

Euclid, 185

欧几里得,185

gcd algorithm, 186

最大公约数算法,186

lemma on primes, 193

关于素数的引理,193

Theorem on Primes, 179

素数定理,179

Euclidean distance function, 373

欧氏距离函数,373

even, 47

甚至,47

function, 245

功能,245

integer, 169

整数,169

every integer is even or odd, 169 , 182

每个整数要么是偶数,要么是奇数,例如 169、182

exclusive-OR, 3

异或,3

definition of, 3

定义,3

English translation, 5

英文翻译,5

proof template, 72

证明模板,72

properties, 21

属性,21

exhaustion, 82

疲惫,82

existence of square roots in ℝ, 379

ℝ 中平方根的存在性,379

existential instantiation, 59 , 92

存在实例化,59,92

existential quantifier, 27

存在量词,27

proof template, 57

证明模板,57

exponent

指数

laws of, 168 , 170

第168、170条法律

negative, 171

阴性,171

notation, 160

符号,160

Extension Axiom, 150

扩展公理,150

F (false), 2

F(假),2

factorial, 159

阶乘,159

factorization,

因式分解

of a product, 197

产品的197

uniqueness of, 193

独特性,193

favorite mathematics text, 384

最喜欢的数学教材,384

fiber of a function, 289

功能纤维,289

Fibonacci number, 191

斐波那契数列,191

Fibonacci sequence, 173 , 177

斐波那契数列,173,177

closed formula, 178

封闭式,178

field, 351

领域,351

integers mod a prime, 367

整数模素数,367

is an integral domain, 357

是一个整环,357

finite set, 315

有限集,315

five, 350

五,350

FOIL Rule, 16 , 357

FOIL规则,16,357

formal construction of ℤ from Z 0 , 314

从 Z≥0 到 ℤ 的形式构造,314

formal language, 51

正式语言,51

formal union, 152

正式联盟,152

formula, 51

公式,51

forward direction, 65

前进方向,65

Foundation Axiom, 155

基础公理,155

four, 350

四,350

fraction, 311

分数,311

in lowest terms, 195 , 313

最简算,195,313

rules for computing, 359

计算规则,359

function

功能

arithmetic operations, 240

算术运算,240

associated equivalence relation, 283

相关等价关系,283

bijective, 261

双射,261

characteristic, 243

特征,243

composition, 246

组成,246

constant, 239

常数,239

definition review, 340

定义审查,340

direct image, 254

直接图像,254

equality, 241 , 339

平等,241,339

even, 245

甚至,245

formal definition, 233

正式定义,233

gluing, 250

粘合,250

graph of, 232 , 233

图 232、233

identity, 239

身份,239

inclusion, 239

包容性,239

informal definition, 231

非正式定义,231

injective, 261

注射剂,261

inverse, 268 , 342

倒数,268,342

level set, 258

水平集,258

linear, 245

线性,245

odd, 245

奇数,245

one-to-one, 261

一对一,261

onto, 261

到,261

piecewise, 251

分段,251

pointwise operations, 240

逐点运算,240

polynomials, 241

多项式,241

preimage, 254

原像,254

proving equality, 241

证明等式,241

restriction, 248

限制,248

set of, 338

一套,338

surjective, 261

满射,261

theorem review, 343

定理回顾,343

well-defined, 235 , 339

定义明确,235,339

Fundamental Theorem of Arithmetic, 193

算术基本定理,193

further reading, 383

延伸阅读,383

Gaussian elimination, 188

高斯消元法,188

gcd algorithm, 186

最大公约数算法,186

general intersection, 141

一般交叉路口,141

formal construction, 154

正式建筑,154

properties, 142

房产,142

general union, 141

一般工会,141

formal construction, 154

正式建筑,154

properties, 142

房产,142

generalized associativity, 354

广义结合律,354

generalized commutativity, 354

广义交换律,354

generic element, 82

通用元素,82

geometric mean, 51

几何平均值,51

gluable family, 250

可粘合家族,250

gluing, 250

粘合,250

Goldbach’s Conjecture, 34

哥德巴赫猜想,34

graph,

图形,

Cartesian, 339

笛卡尔式,339

of a function, 232 , 233

函数,232,233

of a relation, 212

关系,212

of inverse relation, 220

反比关系,220

of restriction, 248

限制,248

greatest common divisor, 185

最大公约数,185

as linear combination, 192

作为线性组合,192

matrix reduction algorithm, 188

矩阵简化算法,188

prime factorization of, 198

198 的质因数分解

greatest element, 303 , 375

最大元素,303,375

greatest lower bound, 377

最大下限,377

existence of, 377

存在,377

group, 351

组,351

axioms, 52

公理,52

group axioms, 52

群公理,52

Haase diagram, 302

哈斯图,302

half-open interval, 132

半开音程,132

hidden quantifier, 33

隐藏量词,33

Horizontal Line Test, 262

水平线测试,262

hypothesis, 6 , 10

假设,6,10

idempotent law, 9

幂等律,9

identity, 49 , 52 , 242 , 247

身份,49,52,242,247

identity function, 239

恒等函数,239

identity relation, 220

恒等关系,220

if

如果

contrapositive proof, 63

逆否命题证明,63

definition, 10

定义,10

direct proof, 58

直接证据,58

inference rule, 50

推理规则,50

logical properties, 12

逻辑属性,12

iff (if and only if), 20

当且仅当,20

English translations, 20

英文译本,20

proof template, 65

证明模板,65

proof via chain proof, 121

通过链式证明进行证明,121

image

图像

of set under relation, 213

在关系下的集合,213

properties of, 215

属性,215

implication, 6

含义,6

inclusion function, 239

包含函数,239

inclusive-OR, 3

包含性或,3

incomparable elements, 302

无可比拟的元素,302

Incompleteness Theorem, 383

不完备性定理,383

inconsistent theory, 52

不一致的理论,52

index, 141

索引,141

index set, 141

索引集,141

indexed intersection, 141

索引交集,141

indexed union, 141

索引联合,141

induction

就职

backwards, 169

倒着,169

double induction, 366

双感应,366

formulation using sets, 363

使用集合的公式,363

proof template, 161

证明模板,161

starting anywhere, 166

从任何地方开始,166

stepping by 2, 171

步数为 2,171

strong induction, 173

强感应,173

Induction Axiom, 160 , 345

归纳公理,160,345

proof from axioms for ℝ, 363

由 ℝ 的公理证明,363

induction hypothesis, 161

归纳假设,161

induction step, 161

诱导步骤,161

inductive set, 362

感应装置,362

inequality denial rules, 368

不平等否认规则,368

inference rule, 50 , 51

推理规则,50,51

Inference Rule for ALL, 50 , 92

全部、50、92 的推理规则

Inference Rule for IF, 50

IF 推理规则,50

infimum of a set, 377

一组中最低的,377

infinite interval, 132

无限区间,132

infinite sequence, 332

无穷序列,332

infinite set, 315

无限集,315

infinitely many primes, 179

无穷多个素数,179

Infinity Axiom, 153

无穷公理,153

infix notation, 275

中缀表示法,275

informal set theory, 103

非形式集合论,103

initial condition, 159

初始条件,159

injection, 260 , 261

注射,260,261

composition of, 264

组成,264

injective function, 261

单射函数,261

input, 211 , 231

输入,211,231

instance, 48 , 59 , 70

例如,48、59、70

integer division, 180

整数除法,180

Integer Division Theorem, 180 , 182

整数除法定理,180,182

integers, 28

整数,28

and natural numbers, 371

以及自然数,371

are even or odd, 182

是偶数还是奇数,182

axioms for, 49

公理,49

closure properties, 364

封闭性质,364

completeness properties, 378

完备性,378

definition as subset of ℝ, 364

定义为 ℝ 的子集,364

distribution within ℝ, 379

ℝ 内的分布,379

form countably infinite set, 323

构成可数无限集,323

form infinite set, 321

形成无限集,321

formal construction from Z 0 , 314

从 Z≥0 开始的形式构造,314

mod n , 290 , 296 , 308 , 367

反对 n、290、296、308、367

review, 203

评论,203

integral domain, 357

积分域,357

Intermediate Value Theorem, 381

介值定理,381

interpretation, 51

解释,51

intersection, 104 , 105

交叉路口,104,105

formal construction, 151

正式建筑,151

of equivalence relations, 285

等价关系,285

of indexed collection of sets, 141

索引集合,141

intersection A , 157

交叉点 ⋂A,157

interval, 132

间隔,132

inverse, 49 , 52 , 242 , 349

倒数,49,52,242,349

in ℤ p , p prime, 367

在 ℤ p 中,p 为素数,367

of a composition, 227

作品,227

of a relation, 219

关系,219

of functions, 342

函数,342

inverse function, 268

反函数,268

and bijections, 269

以及双射,269

and composition, 268

以及组成,268

left inverse, 270

左反转,270

properties, 270

房产,270

right inverse, 271

右倒数,271

uniqueness, 269

独特性,269

inverse image, 254

反转图像,254

inverse of P Q , 16

P⇒Q 的倒数,16

invertible function, 268

可逆函数,268

IRA, see Inference Rule for ALL

IRA,请参阅所有适用的推断规则

irrational numbers, 79

无理数,79

irrationality of 2 , 71

非理性程度为 2,71

irrationality of roots, 200

根的非理性,200

irreflexivity, 307

非自反性,307

knowledge template for subsets, 111

子集知识模板,111

law of exponents, 168 , 170

指数定律,168,170

least common multiple (lcm), 198

最小公倍数(lcm),198

least element, 165 , 303 , 375

最小元素,165,303,375

least upper bound, 351 , 377

最小上限,351,377

Left Cancellation Law, 52 , 274 , 352

左侧取消法,52,274,352

left inverse, 270

左反转,270

Lemma on ℕ and Z > 0 , 371

关于 ℕ 和 Z>0 的引理,371

Lemma on Divisibility by Primes, 193

素数整除性引理,193

Lemma on Subsets of Z > 0 , 328

关于 Z>0 的子集的引理,328

Lemma on the Minimum, 198

最小值引理,198

level set, 258

水平集,258

lexicographic ordering, 307

词典排序,307

linear combination

线性组合

and greatest common divisors, 192

以及最大公约数,192

of integers, 187

整数,187

linear function, 245

线性函数,245

linearity, 252

线性度,252

linearity of sums, 164

和的线性,164

logical axiom, 351

逻辑公理,351

logical connective, 2

逻辑连接词,2

logical equivalence, 9

逻辑等价性,9

logical symbols, 97

逻辑符号,97

logically equivalent forms, 5

逻辑等价形式,5

lower bound, 376

下限,376

lowest terms for a fraction, 195

195 的最简分数

mathematical logic, 383

数理逻辑,383

matrix

矩阵

elementary row operation, 188

基本行运算,188

reduction algorithm for gcds, 188

最大公约数的约简算法,188

maximum, 198

最大值,198

maximum element of a set, 375

集合中的最大元素,375

meaning, 51

意思是,51

member of a set, 103

集合中的成员,103

meta-truth table, 14

元真值表,14

metric space, 373

度量空间,373

minimum, 198

最低,198

minimum element of a set, 375

集合的最小元素,375

mixed quantifiers, 84

混合量词,84

model, 52

型号,52

modus ponens, 51

营造气氛,51

modus tollens, 51

否定后件式,51

monotonicity

单调性

direct images, 255

直接图像,255

image of relations, 215

关系图,215

of product set operations, 134

产品集运算,134

of set operations, 106

集合运算,106

of unions and intersections, 142

联合体和交叉点,142

preimages, 256

原像,256

relation operations, 227

关系运算,227

multiple, 185

多份,185

multiple quantifiers, 83

多个量词,83

multiplication, 240 , 349

乘法,240,349

axioms, 350

公理,350

by zero, 356

零,356

in ℚ, 311

在ℚ,311

inverse, 240 , 349

倒数,240,349

inverse in ℤ p , 367

在 ℤ p 中取反,367

mod n , 308

反对 n,308

properties, 358

房产,358

set closed under, 338

设置关闭下方,338

natural numbers, 362

自然数,362

closure properties, 363

封闭性质,363

compared to positive integers, 371

与正整数相比,371

definition as subset of ℝ, 362

定义为 ℝ 的子集,362

well-ordering of, 376

有序,376

necessary, 18

必要,18

negation, 6 , 35

否定,6,35

of quantifier, 34

量词,34

rules for, 9

规则 9

negative, 349 , 370

阴性,349,370

negative exponent, 171

负指数,171

Nested Interval Theorem, 334 , 380

嵌套区间定理,334,380

nine, 350

九,350

no set of all sets, 151 , 336

没有所有集合的集合,151,336

No-Universe Theorem, 151 , 336

无宇宙定理,151,336

non-gradable adjective, 43

不可分级的形容词,43

not

不是

definition of, 2

定义,2

proof template, 79

证明模板,79

notation for number systems, 28

数字系统的符号,28

number systems, 28

数字系统,28

number theory, 384

数论,384

review, 203

评论,203

numerals, 350

数字,350

odd, 47

奇数,47

every integer is even or odd, 182

每个整数要么是偶数要么是奇数,182

function, 245

功能,245

integer, 169

整数,169

sum of odd integers, 162

奇数之和,162

one, 349

一,349

one-to-one function, 261

一对一功能,261

only if, 18

只有在18年

onto function, 261

到函数,261

open disk, 148

打开磁盘,148

open interval, 132

开区间,132

open sentence, 27

开放式句子,27

open set, 148

公开组,148

or

或者

definition of, 3

定义,3

proof template, 72

证明模板,72

using known OR-statement, 76

使用已知的 OR 语句,76

order properties for ℝ, 368

ℝ 的有序性质,368

ordered field, 351

有序字段,351

and ℚ, 365

和ℚ,365

ordered pair, 132

有序对,132

definition, 136

定义,136

formal construction, 152

正式建筑,152

Ordered Pair Axiom, 132

有序对公理,132

proof, 137

证明,137

Ordered Pair Theorem, 137

有序对定理,137

ordered ring, 351

订购戒指,351

ordered triple, 139

订购了三份,139

ordering

订购

algebraic properties, 369

代数性质,369

on ℚ, 314

ℚ,314

on ℝ, 349

ℝ,349

relation, 301

关系,301

ordinary induction proof template, 161

普通归纳证明模板,161

origin, 134

起源,134

output, 211

输出,211

pair

一对

ordered, 132

已下令,132

unordered, 125

无序,125

Pair Axiom, 152

配对公理,152

pairwise disjoint sets, 115 , 147 , 316

两两不相交的集合,115,147,316

paradox, 1

悖论,1

partial order, 301

偏序,301

strict, 307

严格,307

piecewise function, 251

分段函数,251

pitfall, 342

陷阱,342

pointwise operations on functions, 240 , 242

函数上的逐点运算,240,242

polynomial, 241

多项式,241

poset, 301

poset,301

concatenation, 303

连接,303

lexicographic order, 307

词典编排顺序,307

product construction, 303

产品结构,303

subposet of, 302

子集,302

well-ordered, 304 , 345

井然有序,304,345

positive, 370

阳性,370

positive real numbers

正实数

axioms for, 374

公理,374

properties, 370

房产,370

postulate, 48

公理,48

power of a function, 252

函数的幂,252

power set, 127 , 334

功率集,127,334

formal construction, 152

正式建筑,152

properties of, 128

属性,128

Power Set Axiom, 152

权力集公理,152

Power Set Rule, 316

力量集规则,316

precedence rules, 6

优先规则,6

preimage, 254

原像,254

properties, 256

房产,256

prime, 179

质数,179

divisibility property, 193

可分性,193

infinitely many, 179

无穷多个,179

infinitely many ≡ 3 (mod 4), 184

无穷多个 ≡ 3 (mod 4), 184

unique factorization, 193

唯一分解,193

prime factorization

质因数分解

consequences of, 197

197 年的后果

existence, 179

存在,179

of gcds, 198

最大公约数,198

of rational numbers, 199

有理数,199

product, 349

产品,349

notation, 160

符号,160

of countable sets, 329

可数集,329

pointwise, 240

按点数计算,240

poset, 303

姿势集,303

Product Rule, 316

产品规则,316

product set, 133 , 152

产品集,133,152

proof template, 135

证明模板,135

properties, 134

房产,134

proof, 50

证明,50

by cases, 76

按病例数计算,76例

by construction, 57

按构造,57

by contradiction, 70

矛盾的是,70

by contrapositive, 63

通过逆否命题,63

by example, 57

例如,57

by exhaustion, 82

因精疲力竭,82

chain proof, 121

链条检验,121

circle proof, 119

圆的证明,119

involving empty set, 115

涉及空集,115

of 0 < 1, 370

0 < 1, 370

of necessity, 65

必要性,65

of sufficiency, 65

充足性,65

proof template, 56

证明模板,56

chain proof, 121

链条检验,121

circle proof, 119

圆的证明,119

contrapositive proof, 63

逆否命题证明,63

direct proof of IF-statement, 58

IF 语句的直接证明,58

for AND-statement, 64

对于 AND 语句,64

for backwards induction, 169

对于逆向归纳法,169

for disproofs, 79

反驳见第79条

for exclusive-OR statement, 72

对于异或语句,72

for existence statements, 57

对于存在性陈述,57

for function equality, 241

对于函数相等性,241

for IFF-statements, 65

对于IFF语句,65

for induction, 161

诱导法,161

for induction starting anywhere, 166

适用于任何位置的感应加热,166

for NOT-statement, 79

对于 NOT 语句,79

for OR-statement, 72

对于 OR 语句,72

for set equality, 113

对于集合相等性,113

for strong induction, 173

对于强感应,173

for subsets, 106

对于子集,106

for uniqueness, 88

独特性方面,88

for universal statements, 82

对于普遍性陈述,82

product set as a subset, 135

产品集作为子集,135

proof by cases, 76

案例证明,76

proof by contradiction, 70

反证法,70

review, 99

评论,99

to prove n Z , P ( n ) , 169

证明 ∀n∈Z,P(n), 169

proper subset, 105

真子集,105

proper superset, 105

正统超级组,105

proposition, 1

命题,1

propositional form, 2

命题形式,2

formal definition, 5

正式定义,5

logical equivalence, 9

逻辑等价性,9

propositional logic, 1 , 98

命题逻辑,1,98

propositional operators and quantifiers, 90

命题运算符和量词,90

propositional variable, 2

命题变量,2

provided, 19

前提是,19

proving an AND-statement, 64

证明一个 AND 语句,64

proving an exclusive-OR statement, 72

证明一个异或语句,72

proving an IFF-statement, 65

证明一个IFF语句,65

proving an OR statement, 72

证明 OR 语句,72

proving function equality, 241

证明函数相等性,241

quantified variable, 83

量化变量,83

quantifier, 27 , 98

量词,27,98

conversion rules, 32

转换规则,32

hidden, 33

隐藏,33

interaction rules, 90

交互规则,90

mixed, 84

混合,84

multiple, 83

多,83

negation rules, 34

否定规则,34

reordering rules, 83 , 92

重新排序规则,83,92

restricted, 28

受限,28

translations, 32

译文,32

quotient, 180

商,180

uniqueness of, 182

独特性,182

quotient set, 295

商集,295

rational, 47

理性,47

dyadic, 62

二元的,62

in lowest terms, 195

最简算,195

rational numbers, 28

有理数,28

are dense in ℝ, 379

在ℝ中密度为379

closure properties, 365

封闭性质,365

definition as subset of ℝ, 365

定义为 ℝ 的子集,365

form countably infinite set, 325 , 329

形成可数无限集,325,329

formal construction, 311

正式建筑,311

ordering of, 314

订购,314

prime factorization, 199

质因数分解,199

reading list, 383

阅读清单,383

real numbers, 28 , 349

实数,28,349

Archimedean property, 378

阿基米德性质,378

closed subset of, 148

的闭子集,148

closure properties, 350

封闭性质,350

distance between, 372

距离,372

open subset of, 148

的开放子集,148

order properties, 368

排序属性,368

ordering relations, 349

排序关系,349

uncountability of, 333

不可数名词,333

undefined terms, 349

未定义术语,349

recursive definition, 5 , 159

递归定义,5,159

recursively defined sequence, 168

递归定义的序列,168

reflexivity, 13 , 276 , 279

反身性,13,276,279

Regularity Axiom, 156

正则性公理,156

relation, 211

关系,211

antisymmetric, 301

反对称,301

closure properties, 226

封闭性质,226

composition, 221

组成,221

definition review, 341

定义审查,341

digraph of, 275

二合字母 of,275

graph of, 212

图 212

identity, 220

身份,220

image of set, 213

套装图片,213

inverse of, 219

219 的倒数

irreflexive, 307

非自反性,307

on a set X , 275

在集合 X 上,275

partial order, 301

偏序,301

properties of, 226

的属性,226

reflexive, 276

反身动词,276

symmetric, 277

对称,277

theorem review, 342

定理回顾,342

total order, 302

总订单量,302

transitive, 278

及物动词,278

remainder, 180

余数,180

uniqueness of, 182

独特性,182

reordering of quantifiers, 83

量词重排序,83

Replacement Axiom, 154

替换公理,154

representative of equivalence class, 289

等价类代表,289

restricted quantifier, 28

受限量词,28

conversion rules, 32

转换规则,32

restriction

限制

of a function, 248

函数,248

of an ordering, 302

订购,302

review

审查

of cardinality, 341 , 344

基数,341,344

of functions, 340 , 343

函数,340,343

of logic and proofs, 97

逻辑与证明,97

of relations, 341 , 342

关系,341,342

of sets and integers, 203

集合与整数,203

Right Cancellation Law, 274 , 352

权利撤销法,274,352

right inverse, 271

右倒数,271

ring, 351

戒指,351

root of real number, 200

实数的根,200

row-reduction of a matrix, 188

矩阵的行简化,188

Russell’s Paradox, 149

罗素悖论,149

Schröder–Bernstein Theorem, 336

施罗德-伯恩斯坦定理,336

semantic concept, 51

语义概念,51

sequence

顺序

finite, 331

有限的,331

infinite, 332

无穷大,332

recursively defined, 168

递归定义,168

set

∈-transitive, 131

∈-及物动词,131

axiomatic development of, 149

公理化发展,149

bounded above, 351 , 376

上界为 351、376

bounded below, 376

下限为 376

closed set, 148

封闭集,148

closed under operations, 338

已关闭运营,338

countable, 328

可数,328

countably infinite, 323

可数无穷,323

difference, 151

差值,151

disjoint, 115 , 316

不相交,115,316

equality, 113

平等,113

informal definition, 103

非正式定义,103

intersection, 151

交叉路口,151

open set, 148

公开组,148

pairwise disjoint, 115 , 147

两两不相交,115,147

power set of, 127

幂集,127

product, 133 , 152

产品,133,152

proof advice, 123

证明建议,123

set-builder notation, 144

集合构建器符号,144

uncountable, 332

不可数名词,332

with n elements, 315

包含 n 个元素,315

with at most 3 elements, 125

最多包含 3 个元素,125

ZFC axioms, 156

ZFC公理,156

set difference, 104 , 105

设置差值,104,105

formal construction, 151

正式建筑,151

set equality, 113

设置相等性,113

proof template, 113

证明模板,113

properties, 113

房产,113

set operations, 103

集合运算,103

set partition, 295 , 342

设置分区,295,342

and equivalence relations, 298 , 345

以及等价关系,298,345

induced by equivalence relation, 295

由等价关系诱导,295

set theory

集合论

axiomatic, 383

公理,383

informal introduction, 103

非正式介绍,103

review, 203

评论,203

set-builder notation, 144

集合构建器符号,144

Set-Theoretic Induction Principle, 363

集合论归纳原理,363

seven, 350

七,350

sign function, 78 , 283

符号函数,78,283

sign rules, 357

标志规则,357

singleton set, 125

单件套装,125

six, 350

六,350

small set, 125

小套装,125

Specification Axiom, 151

规范公理,151

square, 358

正方形,358

sum of, 162

总计 162

square roots in ℝ, 379

ℝ 中的平方根,379

strict partial order, 307

严格偏序,307

strictly negative, 370

严格为负,370

strictly positive, 370

严格为正数,370

strong induction, 173

强感应,173

strong induction hypothesis, 174

强归纳假说,174

Strong Induction Proof Template, 173

强感应证明模板,173

Strong Induction Theorem, 173

强归纳定理,173

subposet, 302

子集,302

subset, 104

子集,104

and power sets, 127

以及功率集,127

characterization of, 120

表征,120

definition, 105

定义,105

of ℤ is countable, 328

ℤ 是可数的,328

proof involving product set, 135

涉及产品集的证明,135

proof template, 106

证明模板,106

properties of, 106

属性,106

substitution property for sets, 114

集合的替换性质,114

substitution rules, 14

替换规则,14

sufficient, 18

足够,18

sum, 349

总计,349

additivity, 164

可加性,164

of cubes, 164

立方体,164

of odd integers, 162

奇数个,162

of squares, 162

正方形,162

pointwise, 240

按点数计算,240

telescoping, 170

伸缩式,170

Sum Rule, 316

求和规则,316

summation index, 160

求和指数,160

summation notation, 160

求和符号,160

superset, 105

超集,105

supremum of a set, 377

一套中的顶部,377

surjection, 260 , 261

满射,260,261

composition of, 264

组成,264

surjective function, 261

满射函数,261

symmetric difference, 124

对称差,124

symmetry, 13 , 277 , 279

对称性,13,277,279

syntactic concept, 51

句法概念,51

T (true), 2

T(真),2

tautology, 21 , 48

同义反复,21,48

telescoping sum, 170

伸缩总和,170

ten, 350

十,350

term, 51 , 332

期限,51,332

theorem, 50

定理,50

Theorem on Absolute Value, 371

绝对值定理,371

Theorem on Addition, 352

加法定理,352

Theorem on Addition and Multiplication mod n , 310

关于模 n 加法和乘法的定理,310

Theorem on Algebraic Operations in ℚ, 312

ℚ 中代数运算定理,312

Theorem on Algebraic Properties of Order, 369

关于序的代数性质的定理,369

Theorem on Archimedean Property of ℝ, 378

关于 ℝ 的阿基米德性质的定理,378

Theorem on Cardinal Equality and Order, 322

关于基数相等和序的定理,322

Theorem on Characteristic Functions, 243

特征函数定理,243

Theorem on Closure for Relations, 226

关于关系封闭性的定理,226

Theorem on Closure Properties of Nonzero Real Numbers, 356

非零实数的封闭性质定理,356

Theorem on Congruence mod n , 284

模 n 同余定理,284

Theorem on Countability of Products, 325

关于乘积可数性的定理,325

Theorem on Countable Sets, 329

可数集定理,329

Theorem on Density of ℚ in ℝ, 379

关于 ℚ 在 ℝ 中的密度定理,379

Theorem on Direct Images, 255

直接像定理,255

Theorem on Distance, 373

距离定理,373

Theorem on Distribution of ℤ in ℝ, 379

关于 ℤ 在 ℝ 中的分布的定理,379

Theorem on Division, 358

除法定理,358

Theorem on Elements of Z / n , 308

关于 Z/n 元素的定理,308

Theorem on Equivalence Classes, 291

等价类定理,291

Theorem on Equivalence Relations and Set Partitions, 298

关于等价关系和集合划分的定理,298

Theorem on Existence of Prime Factorizations, 179

关于素数分解存在性的定理,179

Theorem on Existence of Real Square Roots, 379

实数平方根存在性定理,379

Theorem on Factorization of Divisors, 197

除数分解定理,197

Theorem on Finite Sets, 316

有限集定理,316

Theorem on Fractions, 359

分数定理,359

Theorem on Function Composition, 247

函数复合定理,247

Theorem on General Unions and Intersections, 142

关于一般并集和交集的定理,142

Theorem on Greatest Common Divisors, 185

最大公约数定理,185

Theorem on Greatest Lower Bounds, 377

最大下界定理,377

Theorem on IF, 12

关于IF的定理,12

Theorem on IFF, 20

关于IFF的定理,20

Theorem on Images, 215

关于像的定理,215

Theorem on Infinitude of Primes, 179

素数无穷定理,179

Theorem on Injections, Surjections, and Bijections, 264

关于单射、满射和双射的定理,264

Theorem on Inverses and Bijections, 269

关于逆和双射的定理,269

Theorem on Inverses and Composition, 268

关于逆元和复合的定理,268

Theorem on Invertible Functions, 270

可逆函数定理,270

Theorem on Left Inverses, 271

左逆定理,271

Theorem on Logical Equivalence, 9

逻辑等价定理,9

Theorem on Multiplication, 358

乘法定理,358

Theorem on Multiplication by Zero, 356

关于乘以零的定理,356

Theorem on Nested Intervals, 380

嵌套区间定理,380

Theorem on Order Properties in ℝ, 368

ℝ 中序性质的定理,368

Theorem on Pointwise Operations, 242

逐点运算定理,242

Theorem on Positive Numbers, 370

正数定理,370

Theorem on Power Sets, 128

幂集定理,128

Theorem on Preimages, 256

关于原像的定理,256

Theorem on Prime Factorization of GCDs, 198

最大公约数素因数分解定理,198

Theorem on Prime Factorization of Integers, 194

整数素因数分解定理,194

Theorem on Product Sets, 134

关于乘积集的定理,134

Theorem on Products of Countable Sets, 329

可数集乘积定理,329

Theorem on Products of Integers, 197

整数乘积定理,197

Theorem on Rationality of Roots, 200

根的合理性定理,200

Theorem on Reflexivity, Symmetry, and Transitivity, 279

关于自反性、对称性和传递性的定理,279

Theorem on Relations, 226

关于关系的定理,226

Theorem on Right Inverses, 272

右逆定理,272

Theorem on Set Equality, 113

集合等式定理,113

Theorem on Signs, 357

符号定理,357

Theorem on Strong Completeness of ℤ, 378

关于 ℤ 的强完备性定理,378

Theorem on Subsets, 106

子集定理,106

Theorem on Subtraction, 353

减法定理,353

Theorem on Sum of Odd Integers, 162

奇数和定理,162

Theorem on Sum of Squares, 162

平方和定理,162

Theorem on Triangle Inequality, 372

关于三角形不等式的定理,372

Theorem on Uncountability of Closed Intervals, 334

闭区间不可数性定理,334

Theorem on Uncountable Sets, 332

不可数集定理,332

Theorem on Uncountable Sets of Sequences, 332

关于不可数序列集的定理,332

Theorem on Uncountable Subsets of ℝ, 333

关于 ℝ 的不可数子集的定理,333

Theorem on Unions of Countable Sets, 330

可数集并集定理,330

Theorem on Unique Prime Factorization of Rational Numbers, 199

有理数素因数分解唯一性定理,199

Theorem on Uniqueness of Prime Factorizations, 193

素因子分解唯一性定理,193

Theorem on Well-Ordering of ℕ, 376

ℕ 的良序定理,376

Theorem on Writing GCDs as Linear Combinations, 192

关于将最大公约数表示为线性组合的定理,192

Theorem on XOR, 21

异或定理,21

theorems requiring Axiom of Choice, 155

需要选择公理的定理,155

three, 350

三,350

total ordering, 302

总订单量,302

transitivity, 13 , 122 , 278 , 279

传递性,13,122,278,279

∈-transitive set, 131

∈传递集,131

of IFF, 24

IFF,24

translation of quantifiers, 32

量词的翻译,32

translations of IFF, 20

IFF 的译文,20

Triangle Inequality, 372

三角不等式,372

trichotomy law, 368

三分律,368

triple

三倍

ordered, 139

已下令,139

unordered, 125

无序,125

truth table, 2 , 4

真值表,2,4

two, 350

二,350

two-part proof of IFF-statement, 65

IFF 语句的两部分证明,65

uncountability of ℝ, 333

ℝ 的不可数性,333

uncountability of intervals, 333 , 334

区间的不可数性,333,334

uncountable set, 332

不可数集合,332

undefined concept in set theory, 149

集合论中未定义的概念,149

undefined term, 47 , 349

未定义术语,47,349

uniform continuity, 88

均匀连续性,88

union, 103 , 105

工会,103,105

formal construction, 152

正式建筑,152

of countable sets, 330

可数集,330

of indexed collection of sets, 141

索引集合,141

Union Axiom, 152

联合公理,152

uniqueness, 41 , 88

独特性,41,88

of inverse functions, 269

反函数,269

of prime factorizations, 193

质因数分解,193

of quotient and remainder, 182

商和余数,182

uniqueness symbol, 41 , 47

唯一性符号,41,47

how to eliminate, 43

如何消除,43

universal instantiation, 51

普遍实例化,51

universal quantifier, 27

全称量词,27

universal set, 28 , 150

通用套装,28,150

universal statement about all integers, 169

关于所有整数的普遍陈述,169

universe, 28

宇宙,28

does not exist, 151 , 336

不存在,151,336

unless, 19

除非,19

unordered pair, 125 , 314

无序对,125,314

formal construction, 152

正式建筑,152

unordered triple, 125

无序三元组,125

unrestricted quantifier, 28

不受限制的量词,28

upper bound, 351 , 376

上限,351,376

useful denial, 35

有用的否认,35

using a known subset statement, 111

使用已知子集语句,111

value of function, 231

函数值为 231

variable, 27

变量,27

Venn diagram, 103

维恩图,103

vertical line test, 233

垂直线测试,233

weakly negative, 370

弱阴性,370

weakly positive, 370

弱阳性,370

well-defined function, 235 , 250 , 309 , 339

定义明确的函数,235、250、309、339

in ℚ, 312

在ℚ,312

well-ordered poset, 304 , 345

井然有序的诗句,304,345

well-ordering of Z > 0 , 165 , 305 , 376

Z>0、165、305、376 的良好排序

XOR, see exclusive-OR

XOR,参见异或

Zermelo–Fraenkel–Choice set theory, 149

Zermelo-Fraenkel-选择集理论,149

zero, 349

零,349

divisor, 49 , 356

除数,49,356

multiplication property, 356

乘法性质,356

ZFC set theory, 149

ZFC集合论,149

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